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This handy Math in Focus Grade 8 Workbook Answer Key Chapter 10 Statistics detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 10 Answer Key Statistics

Math in Focus Grade 8 Chapter 10 Quick Check Answer Key

Solve.

Question 1.
The table shows the numbers of students involved in four sports at a high school. Find the relative frequency for each sport.

Answer:
Relative frequency is the probability of an event occurring based on all possible events.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Review Test Answer Key

Concepts and Skills

Find the range, the three quartiles, and the interquartile range.

Question 1.
2, 4, 1, 7, 3, 3, 9, 10, 1, 0, 6, 8, 5, 5, 9
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Ascending order:
0, 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10
Maximum value=10
Minimum value=0
range=10-0
range=10
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Data set: 0, 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10
Q2=5
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
n=15 (odd)
0, 1, 1, 2, 3, 3, 4
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=2
Q3 data set: 5, 6, 7, 8, 9, 9, 10
Q3=8
Interquartile range=Q3-Q1
Interquartile range=8-2
Interquartile range=6

Question 2.
34, 66, 90, 25, 46, 81, 40, 67, 95, 104, 36, 49
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Ascending order:
25, 34, 36, 40, 46, 49, 66, 67, 81, 90, 95, 104
Maximum value=104
Minimum value=25
Range=104-25
Range=79
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Data set=25, 34, 36, 40, 46, 49, 66, 67, 81, 90, 95, 104
Q2=49+66/2
Q2=115/2
Q2=57.5
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 25, 34, 36, 40, 46, 49
Q1=36+40/2
Q1=76/2
Q1=38
Q1 is the median of the lower half of the data.
Q3 data set: 66, 67, 81, 90, 95, 104
Q3=81+90/2
Q3=171/2
Q3=85.5
Interquartile range=Q3-Q1
Interquartile range=85.5-38
Interquartile range=47.5

Question 3.
1.23, 1.45, 1.09, 1.78, 1.55, 1.67, 1.37, 1.05, 1.23, 1.11
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Ascending order:
1.05, 1.09, 1.11, 1.23, 1.23, 1.37, 1.45, 1.55, 1.67, 1.78
Maximum value=1.78
Minimum value=1.05
Range=1.78-1.05
Range=0.73
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Data set: 1.05, 1.09, 1.11, 1.23, 1.23, 1.37, 1.45, 1.55, 1.67, 1.78
Q2=1.23+1.37/2
Q2=2.6/2
Q2=1.3
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 1.05, 1.09, 1.11, 1.23, 1.23
Q1=1.11
Q3 data set: 1.37, 1.45, 1.55, 1.67, 1.78
Q3=1.55
Interquartile range=Q3-Q1
Interquartile range=1.55-1.11
Interquartile range=0.44

Question 4.
162.5, 248.6, 130.7, 344.9, 322.0, 234.2, 150.8, 304.7, 326.4
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Ascending order:
130.7, 150.8, 162.5, 234.2, 248.6, 304.7, 322.0, 326.4, 344.9
Maximum value=344.9
Minimum value=130.7
Range=344.9-130.7
Range=214.2
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Data set: 130.7, 150.8, 162.5, 234.2, 248.6, 304.7, 322.0, 326.4, 344.9
Q2= 248.6
n=9 (odd)
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 130.7, 150.8, 162.5, 234.2
Q1=150.8+162.5/2
Q1=313.3/2
Q1=156.65
Q3 data set: 304.7, 322.0, 326.4, 344.9
Q3=322.0+326.4/2
Q3=648.4/2
Q3=324.2
Interquartile range=Q3-Q1
Interquartile range=324.2-156.65
Interquartile range=167.55

Use the information below to answer the following.

Tara tossed two number dice 24 times. She found the sum of the values for each throw and displayed the sums in a dot plot.

Question 5.
Find the range of the data.
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
According to the dot plot the data set will be:
2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12.
maximum value=12
minimum value=2
Range=12-2
Range=10

Question 6.
Find the 3 quartiles of the data.
Answer:
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
The given data set: 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12.
Q2=7+8/2
Q2=15/2
Q2=7.5
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 7, 7
Q1=5+5/2
Q1=10/2
Q1=5
Q1 is the median of the lower half of the data.
Q3 data set: 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12
Q3=10+10/2
Q3=20/2
Q3=10
Q3 is the median of the upper half of the data.

Question 7.
Find the interquartile range.
Answer:
The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third Quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
where Q₁ is the first quartile and Q₃ is the third quartile of the series.
Q1=5; Q3=10
Interquartile range=Q3-Q1
Interquartile range=10-5
Interquartile range=5

Solve. Show your work.

Question 8.
The map shows the maximum temperature, in degrees Fahrenheit, recorded in 20 cities across the United States in a certain year. Display the data in a stem-and-leaf plot.

Answer:

The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties. Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making:
1. First, determine the smallest and largest number in the data.
2. Identify the stems.
3. Draw a with two columns and name them as “Stem” and “Leaf”.
4. Fill in the leaf data.
5. Remember, a Stem and Leaf plot can have multiple sets of leaves.
The given data:95, 94, 117, 110, 112, 86, 83, 105, 83, 114, 111, 103, 98, 98, 88, 96, 109, 100, 97, 94
Ascending order:83, 83, 86, 88, 94, 94, 95, 96, 97, 98, 98, 100, 103, 105, 109, 110, 111, 112, 114, 117

Question 9.
The table shows the weights of Labrador dogs, in pounds.

a) Draw a stem-and-leaf plot for the data.
Answer:
The given data: 72, 73, 79, 68, 101, 88, 78, 71, 85, 94, 93, 77, 98, 95, 75, 81, 56, 51, 62, 70
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties. Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making:
1. First, determine the smallest and largest number in the data.
2. Identify the stems.
3. Draw a with two columns and name them as “Stem” and “Leaf”.
4. Fill in the leaf data.
5. Remember, a Stem and Leaf plot can have multiple sets of leaves.
Ascending order: 51, 56, 62, 68, 70, 71, 72, 73, 75, 77, 78, 79, 81, 85, 88, 93, 94, 95, 98, 101

5|1 represents 51 pounds.

b) How many Labrador dogs are there?
Answer:
The given data: 72, 73, 79, 68, 101, 88, 78, 71, 85, 94, 93, 77, 98, 95, 75, 81, 56, 51, 62, 70
the number of given data is equal to number of Labrador dogs.
Now count the numbers present in the sata set.
There are 20 Labrador dogs.

c) What is the range?
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data: 72, 73, 79, 68, 101, 88, 78, 71, 85, 94, 93, 77, 98, 95, 75, 81, 56, 51, 62, 70
Ascending order: 51, 56, 62, 68, 70, 71, 72, 73, 75, 77, 78, 79, 81, 85, 88, 93, 94, 95, 98, 101
Maximum value=101
Minimum value=51
range=101-51
range=50

d) What is the mode of the data?
Answer:
The mode is the observation’s value, which occurs most frequently, i.e., an observation with the maximum frequency is called the mode. A data set can have more than one mode, which means more than one observation has the same maximum frequency.
The given data: 72, 73, 79, 68, 101, 88, 78, 71, 85, 94, 93, 77, 98, 95, 75, 81, 56, 51, 62, 70
Ascending order: 51, 56, 62, 68, 70, 71, 72, 73, 75, 77, 78, 79, 81, 85, 88, 93, 94, 95, 98, 101
Therefore, there is no modal weight. No number doesn’t occur more times.

e) What is the median weight?
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set. To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
The formula to calculate the median of the data set is given as follow.
An odd number of observations:
If the total number of observation given is odd, then the formula to calculate the median is:
Median = {(n+1)/2}^thterm; where n is the number of observations
Even number of observations:
If the total number of observations is even, then the median formula is:
Median  = [(n/2)^th term + {(n/2)+1}^th]/2; where n is the number of observations
The given data: 72, 73, 79, 68, 101, 88, 78, 71, 85, 94, 93, 77, 98, 95, 75, 81, 56, 51, 62, 70
Ascending order: 51, 56, 62, 68, 70, 71, 72, 73, 75, 77, 78, 79, 81, 85, 88, 93, 94, 95, 98, 101
Median=77+78/2
Median=155/2
Median=77.5 lb

Find the mean absolute deviation.

Question 10.
57, 60, 31, 30, 26, 46, 52, 40, 35, 60
Answer:
To understand Mean Absolute Deviation, let us split both the words and try to figure out their meaning. ‘Mean’ refers to the average of the observations and deviation implies departure or variation from a preset standard. When put together, we can define mean deviation as the mean distance of each observation from the mean of the data.
formula:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
Calculate MAD:
Steps to find the mean deviation from mean:
(i)Find the mean of the given observations.
(ii)Calculate the difference between each observation and the calculated mean
(iii)Evaluate the mean of the differences obtained in the second step.
This gives you the mean deviation from the mean.
The given data set: 57, 60, 31, 30, 26, 46, 52, 40, 35, 60
First, we have to calculate the mean:
Mean= sum of observations/total number of observations
Mean=57+60+31+30+26+46+52+40+35+60/10
Mean=437/10
mean=43.7
Now calculate the difference between each observation and the calculated mean
|57-43.7|=13.3
|60-43.7|=16.3
|31-43.7|=12.7
|30-43.7|=13.7
|26-43.7|=17.7
|46-43.7|=2.3
|52-43.7|=8.3
|40-43.7|=3.7
|35-43.7|=8.7
|60-43.7|=16.3
Now calculate the mean for obtained differences
MAD=13.3+16.3+12.7+13.7+17.7+2.3+8.3+3.7+8.7+16.3/10
MAD=113/13
MAD=11.3

Question 11.
1.46, 2.03, 3.12, 2.55, 4.25, 1.80, 4.08, 2.87
Answer:
To understand Mean Absolute Deviation, let us split both the words and try to figure out their meaning. ‘Mean’ refers to the average of the observations and deviation implies departure or variation from a preset standard. When put together, we can define mean deviation as the mean distance of each observation from the mean of the data.
formula:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
Calculate MAD:
Steps to find the mean deviation from mean:
(i)Find the mean of the given observations.
(ii)Calculate the difference between each observation and the calculated mean
(iii)Evaluate the mean of the differences obtained in the second step.
This gives you the mean deviation from the mean.
The given data set: 1.46, 2.03, 3.12, 2.55, 4.25, 1.80, 4.08, 2.87
First, we have to calculate the mean:
Mean= sum of observations/total number of observations
Mean=1.46+2.03+3.12+2.55+4.25+1.80+4.08+2.87/8
Mean=22.16/8
mean=2.77
Now calculate the difference between each observation and the calculated mean
|1.46-2.77|=1.31
|2.03-2.77|=0.74
|3.12-2.77|=0.35
|2.55-2.77|=0.22
|4.25-2.77|=1.48
|1.80-2.77|=0.97
|4.08-2.77|=1.31
|2.87-2.77|=0.1
Now calculate the mean for obtained differences
MAD=1.31+0.74+0.35+0.22+1.48+0.97+1.31+0.1/8
MAD=6.48/8
MAD=0.81

Refer to the box plot to answer the following.

The box plot summarizes the heights of bean sprouts, in centimeters.

Question 12.
Find the lower quartile, the median, and the upper quartile.
Answer:
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median.
The box plot summarizes five points. They are:
Lowest value;  Q1 (lower quartile); Q2 (median); Q3 (upper quartile); highest value.
Now observe the box plot and record the corresponding values.
Lowest value=3.9
Q1 (lower quartile)=4.3;
Q2 (median)=4.4;
Q3 (upper quartile)=4.7
Highest value=5.1

Question 13.
Calculate the range and the interquartile range.
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
the data set: 3.9, 4.0, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5.0, 5.1
Maximum value=5.1
Minimum value=3.9
range=5.1-3.9
range=1.2 cms
Interquartile range:
The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third Quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
where Q₁ is the first quartile and Q₃ is the third quartile of the series.
We know Q3=4.7; Q1=4.3 from this we can calculate IQR
IQR=Q3-Q1
IQR=4.7-4.3
IQR=0.4 cms

Question 14.
Math Journal Interpret what the range means in this context.
Answer:
The range in statistics for a given data set is the difference between the highest and lowest values.
Range=maximum observation-minimum observation
how to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
the data set: 3.9, 4.0, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 5.0, 5.1
Maximum value=5.1
Minimum value=3.9
range=5.1-3.9
range=1.2 cms

Question 15.
Math Journal Interpret what the interquartile range means.
Answer:
The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third Quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
where Q₁ is the first quartile and Q₃ is the third quartile of the series.
We know Q3=4.7; Q1=4.3 from this we can calculate IQR
IQR=Q3-Q1
IQR=4.7-4.3
IQR=0.4 cms
Therefore, the spread of the middle 50% heights of the bean sprouts is 0.4 cms.

Use the box plot to answer the following.

The box plot below summarizes the scores obtained by the contestants in a game.

Question 16.
What are the greatest and the least scores?
Answer:
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median.
The box plot summarizes five points. They are:
Lowest value;  Q1 (lower quartile); Q2 (median); Q3 (upper quartile); highest value.
Now observe the box plot and record the corresponding values.
Here asked only greatest and least scores:
Greatest score=10
least score=0

Question 17.
Find the first, second, and third quartiles.
Answer:
Quartiles:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median.
The box plot summarizes five points. They are:
Lowest value;  Q1 (lower quartile); Q2 (median); Q3 (upper quartile); highest value.
Now observe the box plot and record the corresponding values.
Q1 (lower quartile)=4;
Q2 (median)=5.4;
Q3 (upper quartile)=7.8

Question 18.
If there are 160 contestants, how many scored 4 or more points?
Answer:
The number of contenstants=160
The number of members scored 4 or more=X
The maximum score is 10
X=160/10
X=16
Therefore, 16 members scored 4 or more points.

Problem Solving

Use the statistics given ¡n the table to answer questions 19 to 21.

In a population of 200 students taking a science test, the following statistics for the test scores were compiled.

Question 19.
Math Journal By comparing the mean, the median, and the mode, what can you infer about the distribution of the test scores?
Answer:
The middle 50% of the science scores are moderately spread out from 38 to 76 and the mean is quite close to the median, making the distribution fairly symmetrical about the mean.

Question 20.
Math Journal By analyzing the statistics in the table, what can you infer about the variation of the scores?
Answer:
Inbetween the lowest and highest score the scores varied from one to one. The mean is quite close to the median.
Coming to quartiles there is a lot of variation from lower quartile to upper quartile.

Question 21.
Estimate the number of students who scored 76 or less.
Answer: 150 students.
The total number of students=200
76 or less means it can be 76 or 75, 74… up to 26 because it is the least number.
Suppose take 75% of 200
75*200/100
=150
Therefore, 150 students scored 76 or less.

Use the data in the table to answer questions 22 to 27.

The table summarizes the monthly sales figures, in thousands of dollars, for the women’s and men’s clothing departments at a store. For instance, sales in the women’s department in January were $10,000.

Question 22.
Calculate the 5-point summary for each of the two departments.
Answer:
Quartiles mark each 25% of a set of data:
1. The first quartile Q₁ is the 25th percentile
2. The second quartile Q₂ is the 50th percentile
3. The third quartile Q₃ is the 75th percentile
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median.
The box plot summarizes five points. They are:
Lowest value;  Q1 (lower quartile); Q2 (median); Q3 (upper quartile); highest value.
Now observe the box plot and record the corresponding values.
The data set for women: 10, 15, 8, 12, 28, 34, 36, 18, 14, 16, 27, 40
The data set for men: 8, 10, 11, 15, 20, 30, 24, 14, 10, 9, 17, 28
Ascending order for women: 8, 10, 12, 14, 15, 16, 18, 27, 28, 34, 36, 40
Least point=8
Highest point=40
Q2=16+18/2
Q2=34/2
Q2=17
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 8, 10, 12, 14, 15, 16
Q1=12+14/2
Q1=26/2
Q1=13
Q1 is the median of the lower half of the data.
Q3 data set: 18, 27, 28, 34, 36, 40
Q3=28+34/2
Q3=62/2
Q3=31
Q3 is the median of the upper half of the data.
Men’s department:
The data set for men: 8, 10, 11, 15, 20, 30, 24, 14, 10, 9, 17, 28
Ascending order: 8, 9, 10, 10, 11, 14, 15, 17, 20, 24, 28, 30
Lowest value=8
Highest value=30
Q2=14+15/2
Q2=29/2
Q2=14.5
Q1 data set: 8, 9, 10, 10, 11, 14
Q1=10+10/2
Q1=20/2
Q1=10
Q1 is the median of the lower half of the data.
Q3 data set: 15, 17, 20, 24, 28, 30
Q3=20+24/2
Q3=44/2
Q3=22
Q3 is the median of the upper half of the data.

Question 23.
Using the same scale, draw 2 box plots, one for each department.
Answer:
When we display the data distribution in a standardized way using 5 summary – minimum, Q1 (First Quartile), median, Q3(third Quartile), and maximum, it is called a Box plot. It is also termed a box and whisker plot.
A box plot is a chart that shows data from a five-number summary including one of the measures of central tendency. It does not show the distribution in particular as much as a stem and leaf plot or histogram does. But it is primarily used to indicate a distribution is skewed or not and if there are potential unusual observations (also called outliers) present in the data set. Boxplots are also very beneficial when large numbers of data sets are involved or compared.

Question 24.
Math Journal By comparing the two box plots describe the sales performance of the two departments.
Answer:
By comparing two box plots, the sales are much quite closer to each other. There is not much difference between the women’s sales department and men’s sales department.

Question 25.
Calculate the mean sales figure for each of the two departments. Give your answers to the nearest dollar when you can.
Answer:
The data set for women: 10, 15, 8, 12, 28, 34, 36, 18, 14, 16, 27, 40
Mean=sum of observations/total number of observations.
Mean of women sale=10+15+8+12+28+34+36+18+14+16+27+40/12
Mean of women sale=258/12
mean of women sale=21.5
The data set for men: 8, 10, 11, 15, 20, 30, 24, 14, 10, 9, 17, 28
Mean of men’s sale=8+10+11+15+20+30+24+14+10+9+17+28/12
Mean of men’s sale=196/12
mean of men’s sale=16.3

Question 26.
Calculate the mean absolute deviation for each of the two departments. Give your answers to the nearest dollar.
Answer:
To understand Mean Absolute Deviation, let us split both the words and try to figure out their meaning. ‘Mean’ refers to the average of the observations and deviation implies departure or variation from a preset standard. When put together, we can define mean deviation as the mean distance of each observation from the mean of the data.
formula:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
Calculate MAD:
Steps to find the mean deviation from mean:
(i)Find the mean of the given observations.
(ii)Calculate the difference between each observation and the calculated mean
(iii)Evaluate the mean of the differences obtained in the second step.
This gives you the mean deviation from the mean.
The data set for women: 10, 15, 8, 12, 28, 34, 36, 18, 14, 16, 27, 40
Mean=sum of observations/total number of observations.
Mean of women sale=10+15+8+12+28+34+36+18+14+16+27+40/12
Mean of women sale=258/12
mean of women sale=21.5
|10-21.5|=11.5
|15-21.5|=6.5
|8-21.5|=13.5
|12-21.5|=9.5
|28-21.5|=6.5
|34-21.5|=12.5
|36-21.5|=14.5
|18-21.5|=3.5
|14-21.5|=7.5
|16-21.5|=5.5
|27-21.5|=5.5
|40-21.5|=18.5
now calculate mean for obtained differences:
MAD=11.5+6.5+13.5+9.5+6.5+12.5+14.5+3.5+7.5+5.5+5.5+18.5/12
MAD=115/12
MAD of women sale dapartment=9.58
The data set for men: 8, 10, 11, 15, 20, 30, 24, 14, 10, 9, 17, 28
Mean of men’s sale=8+10+11+15+20+30+24+14+10+9+17+28/12
Mean of men’s sale=196/12
mean of men’s sale=16.3
|8-16.3|=8.3
|10-16.3|=6.3
|11-16.3|=5.3
|15-16.3|=1.3
|20-16.3|=3.7
|30-16.3|=13.7
|24-16.3|=7.7
|14-16.3|=2.3
|10-16.3|=6.3
|9-16.3|=7.3
|17-16.3|=0.7
|28-16.3|=11.7
Now calculate the mean for obtained differences:
MAD=8.3+6.3+5.3+1.3+3.7+13.7+7.7+2.3+6.3+7.3+0.7+11.7/12
MAD=74.6/12
MAD of men’s sales department=6.21

Question 27.
Math Journal By comparing the means and the mean absolute deviations of the two clothing departments, what can you infer about their variability in sales?
Answer:
mean of women sale=21.5
mean of men’s sale=16.3
MAD of women sale dapartment=9.58
MAD of men’s sales department=6.21
MAD to mean ratio for women department:
9.58/21.5*100%
=44.6%
MAD to mean ratio for men’s department:
6.21/16.3*100%
=38.1%
By comparing all these, we can say the MAD to mean ratio for the women department is 44.6% and that of the men’s department is 38.1%. These two ratios show that the sales in the women department are slightly more varied than the sales in the men’s department.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Cumulative Review Chapters 10 to 13 to score better marks in the exam.

Math in Focus Grade 1 Cumulative Review Chapters 10 to 13 Answer Key

Concepts and Skills

Fill in the blanks.

Write heavier than, lighter than, or as heavy as.

Question 1.
The book is ___ the ball.
Answer: The book is heavier than the ball.

Question 2.
The doll is ___ the book.
Answer: The doll is lighter than the book.

Question 3.
The ball is ___ the doll.
Answer: The ball is heavier than the doll.

Fill in the blanks.

Question 4.

The ____ is lighter than the carrot.
The ____ is heavier than the carrot.
So, the tomato is lighter than the pineapple.
Answer:
The tomato is lighter than the carrot.
The pineapple is heavier than the carrot.

The picture graph shows the number of cars and trucks in a parking lot.

Fill in the blanks.

Question 5.
There are ____ more cars than trucks.
Answer: There are 2 more cars than trucks.

Question 6.
There are ____ cars and trucks in all.
Answer: There are cars 10 and trucks in all.

Question 7.
There are 2 fewer buses than cars. Draw to show the number of buses.
Answer:

Question 8.
Some cars leave the parking lot. The number of cars and trucks are now the same.
____ cars leave the parking lot.
Answer: 2 cars leave the parking lot.

Complete the tally chart using the data from the picture graph.

Question 9.

Answer:

Make a bar graph using the data from the tally chart.

Question 10.

Answer:

Fill in the blanks.

1 stands for 1 unit.

Question 11.
The weight of the box is __ units.
Answer: The weight of the box is 3 units.

Question 12.

Answer:

Fill in the blanks.

Question 13.
The weight of the toy shovel is about ___ toy bricks.
Answer: The weight of the toy shovel is about 10 toy bricks.

Question 14.
Order the objects from heaviest to lightest.

Answer:

First make tens.
Then count on.
Fill in the missing numbers.

Question 15.

Answer:

Question 16.

Answer:

Write the number.

Question 17.
twenty-five ____
Answer: 25

Question 18.
thirty ____
Answer: 30

Write the number in words.

Question 19.
37 _____
Answer: Thirty seven

Question 20.
40 ____
Answer: Fourty

Fill in the missing numbers.

Question 21.
1 + 20 = ____
Answer: 21

Question 22.
___ + 10 = 36
Answer: 26

Find the missing numbers.

Question 23.

38 = ___ tens ___ ones
Answer: 38 = 3 tens 8 ones

Look at the palce-value chart. Write the number it shows.

Question 24.

Answer:

Question 25.

Answer: 33

Circle the greater number.

Question 26.
25 or 17
Answer: 25 is greater than 17

Question 27.
26 or 29
Answer: 29 is greater than 26

Circle the number that is less.

Question 28.
10 or 29
Answer: 10

Question 29.
38 or 33
Answer: 33

Compare the numbers. Then fill in the blanks.

Question 30.
___ is the least.
Answer: 9 is the least.

Question 31.
____ is the greatest.
Answer: 35 is the greatest.

Question 32.
____ is less than 35 but greater than 16
Answer: 21 is less than 35 but greater than 16

Question 33.
Order the numbers from the least to greatest.

Answer: 9, 16, 21, 35

Complete each number pattern.

Question 34.
17, 20, 23, ___, ____, 32
Answer: 17, 20, 23, 26 , 29, 32
Add 3 to each number to get the next number.

Question 35.
28, ___, 36, 40
Answer: 28, 32, 36, 40
Add 4 to each number to get the next number

Fill in the blanks.

Question 36.
2 less than 25 is ____
Answer: 23

Question 37.
_______ is 3 more than 18.
Answer: 21

Question 38.
24 = _________ tens 4 ones
Answer: 2 tens 4 ones

Question 39.
18 = 1 ten _______ ones
Answer: 1 ten 8 ones

Add or subtract.

Question 40.
21 + 7 = ____
Answer: 28

Question 41.
24 + 10 = ___
Answer: 34

Question 42.
27 – 3 = ____
Answer: 24

Question 43.
38 – 15 = ____
Answer: 23

Question 44.
6 + 3 + 7 = ____
Answer: 16

Question 45.
9 + 8 + 5 = ___
Answer: 22

Question 46.

Answer:

Question 47.

Answer:

Question 48.

Answer:

Question 49.

Answer:

Question 50.

Answer:

Question 51.

Answer:

Question 52.

Answer:

Question 53.

Answer:

Problem Solving

Solve.

Question 54.
Jamal has 20 stamps. Michelle has 4- fewer stamps than Jamal. How many stamps does Michelle have?
Michelle has ____ stamps.
Answer:
Given,
Jamal has 20 stamps,
Michelle has 4 fewer stamps than Jamal,
By subtracting 4 from 20 we get 16,
Therefore, Michelle has 16 stamps.

Question 55.
Nate blows up 30 balloons for his birthday party. Nate blows up 6 fewer balloons than Miguel. How many balloons does Miguel blow up?
Miguel blows up ___ balloons.
Answer:
Given,
Nate blows up 30 balloons for his birthday party,
Nate blows up 6 fewer balloons than Miguel,
By adding 6 with 30 we get 36,
Therefore, Miguel blows up 36 balloons.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 14 Practice 1 Mental Addition to score better marks in the exam.

Math in Focus Grade 1 Chapter 14 Practice 1 Answer Key Mental Addition

Add mentally.
First add the ones.
Then add the ones to the tens.

Example

Question 1.
15 + 2 = ____
Answer: 17

Question 2.
12 + 4 = ____
Answer: 16

Question 3.
35 + 1 = ___
Answer: 36

Question 4.
23 + 5 = ____
Answer: 28

Question 5.
22 + 7 = ___
Answer: 29

Question 6.
31 + 8 = ___
Answer:

Question 7.
6 + 32 = ____
Answer:

Question 8.
5 + 34 = ____
Answer:

Add mentally.
First add the tens.
Then add the tens to the ones.

Example

Question 9.
18 + 10 = ____
Answer:

Question 10.
11 + 20 = ____
Answer: 31

Question 11.
12 + 10 = ____
Answer: 22

Question 12.
14 + 20 = ____
Answer: 34

Question 13.
16 + 10 = ____
Answer: 26

Question 14.
20 + 19 = ____
Answer: 39

Question 15.
10 + 17 = ___
Answer: 27

Question 16.
20 + 13 = ____
Answer: 33

Question 17.
10 + 14 = ____
Answer: 24

Question 18.
20 + 16 = ____
Answer: 36

Add mentally. Use doubles facts.

Example

Question 19.
4 + 5 = ____
Answer: 9
By doubling and adding 4 with 4 we get 8,
Then adding 8 with 1 we get 9.

Question 20.
7 + 8 = _____
Answer: 15
By doubling and adding 7 with 7 we get 14,
Then adding 14 with 1 we get 15.

Question 21.
5 + 6 = ___
Answer: 11
By doubling and adding 5 with 5 we get 10,
Then adding 10 with 1 we get 11.

Question 22.
8 + 9 = ____
Answer: 17
By doubling and adding 8 with 8 we get 16,
Then adding 16 with 1 we get 17.

Solve mentally. Fill in the blanks.

Question 23.

How many stickers will Emily have? ____
Answer:
Given,
Emily have 24 stickers and she needs 5 more,
By adding 24 with 5 we get 29,
Therefore, Emily will have 29 stickers in total.

Solve mentally. Fill in the blanks.

Question 24.

How many marbles are there in the box now? ____
Answer:
Given,
There are 18 marbles in the box,
And added 20 more marbles to it,
By adding 18 with 20 we get 38,
Therefore, there are 38 marbles in the box.

Question 25.

How many muffins does Baker Ross bake in all? _____
Answer:
Given,
There are 11 pecan muffins,
And 6 oat muffins,
By adding 11 with 6 we get 17,
Therefore, there are 17 muffins in total.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Lesson 9.3 Understanding Box Plots and Mean Absolute Deviation detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.3 Answer Key Understanding Box Plots and Mean Absolute Deviation

Math in Focus Grade 7 Chapter 9 Lesson 9.3 Guided Practice Answer Key

Solve.

Question 1.
The box plot summarizes the age distribution of people in a play. State the lower quartile, the median, the upper quartile, the range, and the interquartile range.

Answer:
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
In the median, we can measure the distribution with the help of lesser and higher quartile. Apart from mean and median, there are other measures in statistics, which can divide the data into specific equal parts. A median divides a series into two equal parts. We can partition values of a data set mainly into three different ways.
Quartiles formula:
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median. Consider, we have n number of items in a data set. Then the quartiles are given by;
Q₁ = [(n+1)/4]th item
Q₂ = [(n+1)/2]th item
Q₃ = [3(n+1)/4]th item
In the above-given diagram:
quartiles are given that we need to find out.
The first dot represents the minimum value: 20
The second dot represents the Q1 (lower quartile):26
The third dot represents the Q2 (median):32
The fourth dot represents Q3 (upper quartile):34
The fifth dot represents the maximum value: 40

Construct a box plot.

Question 2.
The table shows the average monthly precipitation of Portland, Oregon, in inches. Draw a box plot of the average monthly rainfall and label it with the 5-point summary.

STEP 1: Arrange the data in ascending order.

STEP 2: Calculate the 5-point summary.
Q₁ = Q₂ = Q₃ =
Lower extreme value = Upper extreme value =

STEP 3: Draw a number line with a scale that covers both extreme values. Place a dot for each value from the 5-point summary.

STEP 4: Draw a box above the number line.

STEP 5: Draw the whiskers and label the box plot with the 5-point summary.
Answer:
The above-given data: 5.1, 4.2, 3.7, 2.6, 2.4, 1.6, 0.7, 0.9, 1.7, 2.9, 5.6, 5.7
Now arrange in ascending order:
0.7, 0.9, 1.6, 1.7, 2.4, 2.6, 2.9, 3.7, 4.2, 5.1, 5.6, 5.7
The minimum point is 0.7
the second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=2.6+2.9/2
Q2=5.5/2
Q2=2.75
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1 data set: 0.7, 0.9, 1.6, 1.7, 2.4, 2.6
Q1=1.6+1.7/2
Q1=3.3/2
Q1=1.65
Q1 is the median of the lower half of the data.
Q3 data set: 2.9, 3.7, 4.2, 5.1, 5.6, 5.7
Q3=4.2+5.1/2
Q3=9.3/2
Q3=4.65
Q3 is the median of the upper half of the data.
The maximum point is 5.7
Step 3:
Draw a number line with a scale that covers both extreme values. Place a dot for each value from the 5-point summary.

Step 4: Draw a box above the number line.

Step 5:
Draw the whiskers and label the box plot with the 5-point summary.

Average monthly rainfall.

Solve.

Question 3.
The scores on a math quiz are 20, 16, 13, 11, 19, 24, 22, 15, 17, and 13.
a) Find the mean absolute deviation.
STEP 1: Find the mean score.
Mean score =
STEP 2: Find the distance of each value from the mean.

STEP 3: Find the sum of the distances.
Sum =

STEP 4: Divide the sum by the number of data values to find the MAD.
MAD =
Answer:
Step 1: First we need to calculate mean
the given data: 20, 16, 13, 11, 19, 24, 22, 15, 17, and 13.
Mean=sum of observations/no.of observations
Mean=20+16+13+11+19+24+22+15+17+13/10
Mean=170/10
Mean=10
Step 2: Finding distance between each value
In statistics, the Mean Absolute Deviation (MAD) of the given data set value is defined as the average deviation between the mean and the data value. It describes the variation in the given data values. The procedure to find the mean absolute deviation is given below:
– Calculate the mean of the given data
– Determine the difference between each data value and the mean
– Now, take the absolute value of each obtained difference
– Finally, find the mean of these differences
|20-17|=3
|16-17|=1
|13-17|=4 Likewise, we need to subtract.

step 3: finding the sum of the above distances:
Sum=3+1+4+6+2+7+5+2+0+4
Sum=34
Step 4:
The formula for MAD:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
MAD=34/10
MAD=3.4

b) What does the mean absolute deviation tell you about this set of data?
Answer:  On average, the data values are 3.4 units from the mean.
This gives us an idea about the deviation of the observations from the measure of central tendency.
Thus we can conclude that,
M.A.D=$\frac{\sum Absolute values of Deviation from cental measure/Total Number of observations.}{}$

Although to calculate mean absolute deviation, any measure of central tendency can be used out generally mean and median are the most common ones.
Technology Activity

Materials

• spreadsheet software
• two sets of 10 data values

USE SPREADSHEET SOFTWARE TO FIND MEAN ABSOLUTE DEVIATION

Work in pairs.

STEP 1: Enter 10 data values in one row of cells.

STEP 2: Choose a cell in a later row for the mean.

STEP 3: Use the spreadsheet software’s function for finding the mean to find the mean of the 10 data values.
See the screen shot below.

STEP 4: Choose a cell in the next row for the mean absolute deviation.

STEP 5: Use the spreadsheet software’s function for finding the mean absolute deviation to find the MAD of the data values.

STEP 6: Explain what the MAD tells you about the data.

STEP 7: Enter a second set of data values and repeat STEP 1 to STEP 6.

Math Journal Compare the two sets of data. Are the data values in each set clustered around the mean, or more spread out? Then compare the mean absolute deviations for the two sets of data. What do you observe?
Answer:

Math in Focus Course 2B Practice 9.3 Answer Key

State Q₁, Q₂, Q₃ the lower extreme, and the upper extreme values shown in the box plots. Then calculate the interquartile range.

Question 1.

Answer:
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
In the median, we can measure the distribution with the help of lesser and higher quartile. Apart from mean and median, there are other measures in statistics, which can divide the data into specific equal parts. A median divides a series into two equal parts. We can partition values of a data set mainly into three different ways.
Quartiles formula:
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median. Consider, we have n number of items in a data set. Then the quartiles are given by;
Q₁ = [(n+1)/4]th item
Q₂ = [(n+1)/2]th item
Q₃ = [3(n+1)/4]th item
In the above-given diagram:
quartiles are given that we need to find out.
The first dot represents the minimum value: 57.5
The second dot represents the Q1 (lower quartile):59.5
The third dot represents the Q2 (median):60.5
The fourth dot represents Q3 (upper quartile):62.5
The fifth dot represents the maximum value: 63.5

Question 2.

Answer:
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
In the median, we can measure the distribution with the help of lesser and higher quartile. Apart from mean and median, there are other measures in statistics, which can divide the data into specific equal parts. A median divides a series into two equal parts. We can partition values of a data set mainly into three different ways.
Quartiles formula:
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median. Consider, we have n number of items in a data set. Then the quartiles are given by;
Q₁ = [(n+1)/4]th item
Q₂ = [(n+1)/2]th item
Q₃ = [3(n+1)/4]th item
In the above-given diagram:
quartiles are given that we need to find out.
The first dot represents the minimum value: 110
The second dot represents the Q1 (lower quartile):115
The third dot represents the Q2 (median):125
The fourth dot represents Q3 (upper quartile):130
The fifth dot represents the maximum value: 140

Question 3.

Answer:
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
In the median, we can measure the distribution with the help of lesser and higher quartile. Apart from mean and median, there are other measures in statistics, which can divide the data into specific equal parts. A median divides a series into two equal parts. We can partition values of a data set mainly into three different ways.
Quartiles formula:
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median. Consider, we have n number of items in a data set. Then the quartiles are given by;
Q₁ = [(n+1)/4]th item
Q₂ = [(n+1)/2]th item
Q₃ = [3(n+1)/4]th item
In the above-given diagram:
quartiles are given that we need to find out.
The first dot represents the minimum value: 20
The second dot represents the Q1 (lower quartile):23
The third dot represents the Q2 (median):24
The fourth dot represents Q3 (upper quartile):31
The fifth dot represents the maximum value: 34

Question 4.

Answer:
Quartiles divide the entire set into four equal parts. So, there are three quartiles, first, second and third represented by Q₁, Q₂ and Q₃, respectively. Q₂ is nothing but the median, since it indicates the position of the item in the list and thus, is a positional average. To find quartiles of a group of data, we have to arrange the data in ascending order.
In the median, we can measure the distribution with the help of lesser and higher quartile. Apart from mean and median, there are other measures in statistics, which can divide the data into specific equal parts. A median divides a series into two equal parts. We can partition values of a data set mainly into three different ways.
Quartiles formula:
Suppose, Q₃ is the upper quartile is the median of the upper half of the data set. Whereas, Q₁ is the lower quartile and median of the lower half of the data set. Q₂ is the median. Consider, we have n number of items in a data set. Then the quartiles are given by;
Q₁ = [(n+1)/4]th item
Q₂ = [(n+1)/2]th item
Q₃ = [3(n+1)/4]th item
In the above-given diagram:
quartiles are given that we need to find out.
The first dot represents the minimum value: 3,050
The second dot represents the Q1 (lower quartile):3,100
The third dot represents the Q2 (median):3,250
The fourth dot represents Q3 (upper quartile):3,350
The fifth dot represents the maximum value: 3,500

Draw a box plot for the given 5-point summary. Include a numeric scale for the number line. Label the box plot with the 5-point summary.

Question 5.
Q₁ = 4, Q₂ = 7, Q₃ = 10, lower extreme value = 1, upper extreme value = 15
Answer:
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 6.
Q₁ = 14, Q₂ = 18, Q₃ = 23, lower extreme value = 10, upper extreme value = 28
Answer:
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 7.
Q₁ = 6.1, Q₂ = 6.6, Q₃ = 7.0, lower extreme value = 5.7, upper extreme value = 7.4
Answer:
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 8.
Q₁ = 320, Q₂ = 360, Q₃ = 380, lower extreme value = 300, upper extreme value = 400
Answer:
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Calculate the mean absolute deviation. Round your answers to 3 significant digits when you can.

Question 9.
2, 2, 5, 3, 9, 6,10, 4, 7, 5
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data: 2, 2, 5, 3, 9, 6,10, 4, 7, 5
First find out the Mean:
Mean=sum of observations/number of observations
Mean=2+2+5+3+9+6+10+4+7+5/10
Mean=53/10
Mean=5.3
– Determine the difference between each data value and the mean
|2-5.3|=3.3
|2-5.3|=3.3
|5-5.3|=0.3
|3-5.3|=2.3
|9-5.3|=3.7
|6-5.3|=0.7
|10-5.3|=4.7
|4-5.3|=1.3
|7-5.3|=1.7
|5-5.3|=0.3
– Now, take the absolute value of each obtained difference
The obtained values are 3.3, 3.3, 0.3, 2.3, 3.7, 0.7, 4.7, 1.3, 1.7, 0.3
– Finally, find the mean of these differences
The formula for MAD:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
MAD=3.3+ 3.3+0.3+ 2.3+ 3.7+0.7+ 4.7+1.3+1.7+0.3/10
MAD=21.6/10
MAD=2.16

Question 10.
58, 62, 45, 39, 40, 60, 67
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data: 58, 62, 45, 39, 40, 60, 67
First find out the Mean:
Mean=58+62+45+39+40+60+67/7
Mean=371/7
Mean=53
– Determine the difference between each data value and the mean
|58-53|=5
|62-53|=9
|45-53|=8
|39-53|=14
|40-53|=13
|60-53|=7
|67-53|=14
– Now, take the absolute value of each obtained difference
The obtained difference values: 5, 9, 8, 14, 13, 7, 14
– Finally, find the mean of these differences
MAD=5+9+8+14+13+7+14/7
MAD=70/7
MAD=7

Question 11.
43.4, 30.5, 20.0, 23.6, 34.5, 36.9, 11.7, 40.2
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data: 43.4, 30.5, 20.0, 23.6, 34.5, 36.9, 11.7, 40.2
First find out the Mean:
Mean=43.4+30.5+20.0+23.6+34.5+36.9+11.7+40.2/8
Mean=240.8/8
Mean=30.1
– Determine the difference between each data value and the mean
|43.4-30.1|=13.3
|30.5-30.1|=0.4
|20.0-30.1|=10.1
|23.6-30.1|=6.5
|34.5-30.1|=4.4
|36.9-30.1|=6.8
|11.7-30.1|=18.4
|40.2-30.1|=10.1
– Now, take the absolute value of each obtained difference
The obtained difference values: 13.3, 0.4, 10.1, 6.5, 4.4, 6.8, 18.4, 10.1
– Finally, find the mean of these differences
MAD=13.3+0.4+10.1+6.5+4.4+6.8+18.4+10.1/8
MAD=70/8
MAD=8.75

Question 12.
1.15, 1.16, 1.25, 1.22, 1.36, 1.38, 1.49, 1.22, 1.11
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data: 1.15, 1.16, 1.25, 1.22, 1.36, 1.38, 1.49, 1.22, 1.11
First find out the Mean:
Mean=1.15+1.16+1.25+1.22+1.36+1.38+1.49+1.22+1.11/9
Mean=11.34/9
Mean=1.26
– Determine the difference between each data value and the mean
|1.15-1.26|=0.11
|1.16-1.26|=0.1
|1.25-1.26|=0.01
|1.22-1.26|=0.04
|1.36-1.26|=0.1
|1.38-1.26|=0.12
|1.49-1.26|=0.23
|1.22-1.26|=0.04
|1.11-1.26|=0.15
– Now, take the absolute value of each obtained difference
The obtained difference values: 0.11, 0.1, 0.01, 0.04, 0.1, 0.12, 0.23, 0.04, 0.15
– Finally, find the mean of these differences
MAD=0.11+0.1+0.01+0.04+0.1+0.12+0.23+0.04+0.15/9
MAD=0.9/9
MAD=0.1

Use the data in the table for the following questions.

The table shows the masses, in grams, of a dozen eggs.

Question 13.
Calculate Q₁, Q₂, and Q₃.
Answer:
The given data: 57, 53, 62, 45, 56, 56, 63, 61, 50, 44, 43, 58
How to calculate quartiles:
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data set in ascending order:
43, 44, 45, 50, 53, 56, 56, 57, 58, 61, 62, 63
Now find the median:
Q2=56+56/2
Q2=112/2
Q2=56
Now find Q1: The lower quartile Q₁ is the median of the lower half of the data.
Q1 data set: 43, 44, 45, 50, 53, 56
Q1=45+50/2
Q1=95/2
Q1=47.5
Q₁ is the median of the lower half of the data.
Now find Q3: The upper quartile Q₃ is the median of the upper half of the data.
Q3 data set: 56, 57, 58, 61, 62, 63
Q3=58+61/2
Q3=119/2
Q3=59.5
Q₃ is the median of the upper half of the data.

Question 14.
Draw a box plot of the masses of the eggs.
Answer:
the above quartiles we got are
Q1=47.5, Q2=56, Q3=59.5, lower value=43, higher value=63
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 15.
Calculate the mean absolute deviation of the masses of the eggs. Round to the nearest hundredth.
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data: 43, 44, 45, 50, 53, 56, 56, 57, 58, 61, 62, 63
First find out the Mean:
Mean=43+44+45+50+53+56+56+57+58+61+61+63/12
Mean=647/12
Mean=53.9
Mean=54 (rounded to nearest number)
– Determine the difference between each data value and the mean
|43-54|=11
|44-54|=10
|45-54|=9
|50-54|=4
|53-54|=1
|56-54|=2
|56-54|=2
|57-54|=3
|58-54|=4
|61-54|=7
|61-54|=7
|63-54|=9
– Now, take the absolute value of each obtained difference
The obtained difference values: 11, 10, 9, 4, 1, 2, 2, 3, 4, 7, 7, 9
– Finally, find the mean of these differences
MAD=11+10+9+4+1+2+2+3+4+7+7+9/12
MAD=69/12
MAD=5.75

Use the information to answer the following questions.

The map shows the amounts of snow, in inches, that fell in different parts of Minnesota during one week in winter.

Question 16.
Calculate Q₁, Q₂, and Q₃.
Answer:
The given data: 13, 6, 8, 15, 9, 18, 18, 20, 24, 16, 11, 10, 13, 15, 10, 12
How to calculate quartiles:
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q₂.
3. At Q₂ split the ordered data set into two halves.
4. The lower quartile Q₁ is the median of the lower half of the data.
5. The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data set in ascending order:
6, 8, 9, 10, 10, 11, 12, 13, 13, 15, 15, 16, 18, 18, 20, 24
Now find the median:
Q2=13+13/2
Q2=26/2
Q2=13
Now find Q1: The lower quartile Q₁ is the median of the lower half of the data.
Q1 data set: 6, 8, 9, 10, 10, 11, 12, 13
Q1=10+10/2
Q1=20/2
Q1=10
Q₁ is the median of the lower half of the data.
Now find Q3: The upper quartile Q₃ is the median of the upper half of the data.
Q3 data set: 13, 15, 15, 16, 18, 18, 20, 24
Q3=16+18/2
Q3=34/2
Q3=17
Q₃ is the median of the upper half of the data.

Question 17.
Draw a box plot of the snow amounts.
Answer:
the above quartiles we got are:
Q1=10, Q2=13, Q3=17, lowest value=6, highest value=24
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 18.
Calculate the mean absolute deviation of the snow amounts. Round your answer to the nearest inch.
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data:
First find out the Mean: 6, 8, 9, 10, 10, 11, 12, 13, 13, 15, 15, 16, 18, 18, 20, 24
Mean=sum of observations/number of observations
Mean=6+8+9+10+10+11+12+13+13+15+15+16+18+18+20+24/16
Mean=218/16
Mean=13.265
Mean=14(nearest inch)
– Determine the difference between each data value and the mean
|6-14|=8
|8-14|=6
|9-14|=5
|10-14|=4
|10-14|=4
|11-14|=3
|12-14|=2
|13-14|=1
|13-14|=1
|15-14|=1
|15-14|=1
|16-14|=2
|18-14|=4
|18-14|=4
|20-14|=6
|24-14|=10
– Now, take the absolute value of each obtained difference
The obtained values are 8, 6, 6, 4, 4, 3, 2, 1, 1, 1, 1, 2, 4, 4, 6, 10
– Finally, find the mean of these differences
The formula for MAD:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
MAD=8+6+6+4+4+3+2+1+1+1+1+2+4+4+6+10/16
MAD=53/16
MAD=3.3

Use the data in the table for questions 19 and 20.

The table shows the heights, in centimeters, of 5 peanut plants.

Question 19.
Draw a box plot of the heights of the peanut plants.
Answer:
The given data: 17.8, 25.4, 20.6, 18.0, 19.2
A box and whisker plot is a graph that exhibits data from a five-number summary, including one of the measures of central tendency. It does not display the distribution as accurately as a stem and leaf plot or histogram does. But, it is principally used to show whether a distribution is skewed or not and if there are potential unusual observations present in the data set, which are also called outliers. Boxplots are also very useful when huge numbers of data collections are involved or compared. Since the centre, spread and overall range are instantly apparent, using these boxplots the arrangements can be matched easily. A box and whisker plot is a way of compiling a set of data outlined on an interval scale. It is also used for descriptive data interpretation.
Elements of a Box and Whisker Plot:
The elements required to construct a box and whisker plot outliers are given below.
The minimum value (Q₀ or 0th percentile)
The first quartile (Q₁ or 25th percentile)
Median (Q₂ or 50th percentile)
Third quartile (Q₃ or 75th percentile)
The maximum value (Q₄ or 100th percentile)
– When we plot a graph for the box plot, we outline a box from the first quartile to the third quartile. A vertical line that goes through the box is the median. The whiskers (small lines) go from each quartile towards the minimum or maximum value, as shown in the figure below.

Question 20.
Calculate the mean absolute deviation of the heights of the peanut plants.
Answer:
We begin with the definition of the mean absolute deviation, which is also referred to as the average absolute deviation. It may make more sense to consider this formula as a process, or series of steps, that we can use to obtain our statistic.
1. We start with an average, of a data set, which we will denote by m.
2. Next, we find how much each of the data values deviates from m. This means that we take the difference between each of the data values and m.
3. After this, we take the absolute value of each of the differences from the previous step. In other words, we drop any negative signs for any of the differences. The reason for doing this is that there are positive and negative deviations from m. If we do not figure out a way to eliminate the negative signs, all of the deviations will cancel one another out if we add them together.
4. Now we add together all of these absolute values.
5. Finally, we divide this sum by n, which is the total number of data values. The result is the mean absolute deviation.
The above-given data:
First find out the Mean: The given data: 17.8, 25.4, 20.6, 18.0, 19.2
Mean=sum of observations/number of observations
Mean=17.8+25.4+20.6+18.0+19.2/5
Mean=101/5
Mean=20.2
– Determine the difference between each data value and the mean
|17.8-20.2|=2.4
|25.4-20.2|=5.2
|20.6-20.2|=0.4
|18.0-20.2|=2.2
|19.2-20.2|=1
– Now, take the absolute value of each obtained difference
The obtained values are 2.4+5.2+0.4+2.2+1
– Finally, find the mean of these differences
The formula for MAD:
The ratio of the sum of all absolute values of deviation from central measure to the total number of observations.
M.A. D = (Σ Absolute Values of Deviation from Central Measure) / (Total Number of Observations)
MAD=2.4+5.2+0.4+2.2+1/5
MAD=11.2/5
MAD=2.24

Solve.

Question 21.
Math journal Compare the box plot and the mean absolute deviation. Which one is a better measure of variation? State your reason.
Answer:
Mean absolute deviation is a better measure of variation. A box plot is not meaningful with only 5 data values.

Question 22.
Math Journal Describe a situation for which you think a box plot would be useful for measuring variation.
Answer:
You need to have information on the variability or dispersion of the data. A boxplot is a graph that gives you a good indication of how the values in the data are spread out. Although boxplots may seem primitive in comparison to a histogram or density plot, they have the advantage of taking up less space, which is useful when comparing distributions between many groups or datasets.

Question 23.
Math journal Describe a situation for which you think the mean absolute deviation would be useful for measuring variation.
Answer:
the mean absolute deviation is most useful when you are interested in how data are spread from the mean.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Lesson 9.2 Stem-and-Leaf Plots detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.2 Answer Key Stem-and-Leaf Plots

Math in Focus Grade 7 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Display data in a stem-and-leaf plot.

Question 1.
The data show the math quiz scores for 15 students. Display the data in a stem-and-leaf plot.

STEP 1: Identify the least and greatest values, and identify the stems.
STEP 2: Place a leaf on the plot for each item of data.
STEP 3: Arrange the leaves in ascending order for each number on the stem.
STEP 4: Write a title and a key for the stem-and-leaf plot.
Answer:
Step 1: Least value:8; greatest value:44; stems:0, 1, 2, 3, 4
Step 2:  Constructing leaf for item of data

Step 3: Now arrange in ascending order

Step 4: Write the title, Remember, a Stem and Leaf plot can have multiple sets of leaves.
0|8 represents 8
1|0 represents 10
1|1 represents 11
1|1 represents 11
1|2 represents 12
1|3 represents 13
1|5 represents 15
1|8 represents 18 and so on… likewise, we represent upto
4|4 represents 44.

Solve.

Question 2.
The data below show the attendance, in the number of days, for some students during the first month of school.

a) Display the data in a stem-and-leaf plot.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

1|5 represents 15 days
1|8 represents 18 days
1|9 represents 19 days
1|9 represents 19 days
2|0 represents 20 days
2|0 represents 20 days
Likewise, we have to write all the representations.

b) How many students are represented by the data?
Answer: 25
Count the number of leaves present in the Stem-and-leaf plot
5, 8, 9, 9, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3.
1|5 represents 15 days
1|8 represents 18 days
1|9 represents 19 days
1|9 represents 19 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|0 represents 20 days
2|1 represents 21 days
2|1 represents 21 days
2|1 represents 21 days
2|2 represents 22 days
2|2 represents 22 days and so on… up to
2|3 represents 23 days.
Total 25.

c) What is the most common attendance, or mode?
Answer: 23
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
23 has the highest frequency. It occurs more time.

d) What is the mean attendance?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data:
15, 18, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23.
Mean=15+18+19+19+20+20+20+20+20+21+21+21+22+22+22+22+22+23+23+23+23+23+23+23+23/25
Mean=528/25
Mean=21.12 days

e) What is the median attendance?
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data:
15, 18, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23.
The middlemost number is 22, so the median is 22.

Solve. Round your answer to the nearest tenth if necessary.

Question 3.
The stem-and-leaf plot shows the heights, in inches, of the players on two football teams.

a) How many players are on each team?
Answer:
Observe the above-given stem-and-leaf plot.
There are two teams: Team A and Team B
In Team A, count the number of leaves:
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
There are 11 players in Team A.
Now coming to Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
There are 11 players in Team B.

b) What is the median height for each team?
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
Median for Team A:
The middlemost number is 70
Median for Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
The middlemost number is 71

c) What is the mean height for each team?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
Finding Mean for Team A:
The given data in the team A is 67, 68, 68, 69, 70, 70, 70, 71, 71, 72, 73.
Mean for Team A= 67+68+68+69+70+70+70+71+71+72+73/11
Mean for Team A= 769/11
Mean for Team A=69.9
Finding mean for Team B:
The given data is 68, 68, 68, 70, 70, 71, 71, 71, 72, 72, 72
Mean for Team B=68+68+68+70+70+71+71+71+72+72+72/11
Mean for Team B=773/11
Mean for Team B=70.3

d) On average, which team has taller players?
Answer:
Median of Team A:70
Mean of Team A:69.9
Median of Team B:71
Mean of Team B:70.3
According to the above observations, we can write which team has taller players.
On average Team B has taller players because both it’s mean and median are greater. But the means and the medians for both teams are very close, so the players on Team B are not much taller on average.

Solve.

Question 4.
The weights, in pounds, of some newborn babies at a hospital are shown in the stem-and-leaf plot.

a) Find the median weight.
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Median=6.5+6.5/2
Median=13/2
Median=6.5lb

b) Find the mean weight.
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Mean=3.1+6.0+6.1+6.2+6.5+6.5+6.5+6.6+6.8+7/10
Mean=61.3/10
Mean=6.13lb

c) Compare the median and mean. Explain the differences.
Answer:
The median is 6.5lb while the mean is 6.13lb. The value of the median is much greater due to the presence of an outlier with the value of 3.1lb

d) What is the range of the data?
Answer:
In statistics, the range is the spread of your data from the lowest to the highest value in the distribution. It is a commonly used measure of variability.
– The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
The given data:
3.1, 6.0, 6.1, 6.2, 6.5, 6.5, 6.5, 6.6, 6.8, 7.
Range=7-3.1
Range=3.9
Range=4 (rounded to the nearest 10)

e) Explain how the outlier 3.1 pounds affects the range of the data. How does the outlier affect the mean of the data?
Answer:
The range is the difference between the least and the greatest values in the data set. So, the range is greater because of the outlier. The mean is also smaller because of the outlier.

Math in Focus Course 2B Practice 9.2 Answer Key

Display the data using a stem-and-leaf plot. Indicate the range and median of each data set.

Question 1.
The masses, in grams, of russet potatoes, are 89, 110, 81, 92, 100, 96, 101, 109, 105, and 112.
Answer:
The given data: 89, 110, 81, 92, 100, 96, 101, 109, 105, and 112.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

8|1 represents 81 grams.
Range: The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
The given data: 81, 89, 92, 96, 100, 101, 105, 109, 110, 112.
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Highest value:112
Lowest value:81
Range=112-81
Range=31 grams.
Median:
The given data: 81, 89, 92, 96, 100, 101, 105, 109, 110, 112.
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median=100+101/2
Median=201/2
Median=100.5grams.

Question 2.
The lengths of salmon, in inches, are 28, 31, 32, 29, 26, 33, 29, 28, 30, and 31.

Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

2|6 represents 26 inches.
Range: The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
The given data: 26, 28, 28, 29, 29, 30, 31, 31, 32, 33
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Highest value:33
Lowest value:26
Range=33-26
Range=7 inches
Median:
The given data: 26, 28, 28, 29, 29, 30, 31, 31, 32, 33
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median=29+30/2
Median=59/2
Median=29.5 inches

Question 3.
The average hourly rate, in dollars, of workers in 8 professions are 26, 13, 10, 17, 12, 15, 14, and 15.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

1|0 represents 10$
the given data: 10, 12, 13, 14, 15, 15, 17, 26
range=maximum value-minimum value
maximum value=26
minimum value=10
range=26-10
range=16$
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
the given data: 10, 12, 13, 14, 15, 15, 17, 26
Median=14+15/2
Median=29/2
Median=14.5$
Mean = (Sum of all the observations/Total number of observations)
The given data: 26, 13, 10, 17, 12, 15, 14, and 15
Mean=26+13+10+17+12+15+14+15/8
Mean=122/8
Mean=15.25

Question 4.
The scores of a basketball competition are 70, 56, 93, 92, 72, 74, 76, 72, 78, and 59.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

5|6 represents 56 points
The given data: 56, 59, 70, 72, 72, 74, 76, 78, 92, 93
range=maximum value-minimum value
range=93-56
range=37 points
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 56, 59, 70, 72, 72, 74, 76, 78, 92, 93
Median=72+74/2
Median=146/2
Median=73 points.

Question 5.
The ages of passengers on board a bus are 33, 37, 5, 11, 63, 55, 56, 13, 15, and 19.
Answer:
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

0|5 represents 5 years
the given data: 5, 11, 13, 15, 19, 33, 37, 55, 56, 63
Range=maximim value-minimum value
Range=63-5
range=58 years.
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
the given data: 5, 11, 13, 15, 19, 33, 37, 55, 56, 63
Median=19+33/2
Median=52/2
Median=26 years.

Use the data in the table to answer the following.

The table shows the weights, in pounds, of 30 pumpkins.

Question 6.
Draw the stem-and-leaf plot of the data.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.

0|8 represents 8 lb

Question 7.
What is the range of the data?
Answer:
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
In statistics, the range is the spread of your data from the lowest to the highest value in the distribution. It is a commonly used measure of variability.
– The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
The range is the easiest measure of variability to calculate. To find the range, follow these steps:
1. Order all values in your data set from low to high.
2. Subtract the lowest value from the highest value.
Range=22-8
Range=14 lb

Question 8.
What is the mode?
Answer:
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
16 has the highest frequency. It occurs more time.

Question 9.
What is the median?
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Median=15+16/2
Median=31/2
Median=15.5 lb
The median is nothing but Q2. Now we have to find Q1(lower quartile), Q3 (upper quartile)
Q1 data set: 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15
Q1=11
Q1 is the median of lower quartile
Q3 data set: 16, 16, 16, 16, 16, 17, 17, 17, 18, 19, 20, 20, 21, 21, 22
Q3=18
Q2 is the median of upper quartile.

Question 10.
What is the interquartile range?
Answer:
The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third Quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus lower quartile.
Formula:
The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q_­1
where Q₁ is the first quartile and Q₃ is the third quartile of the series.
IQR=Q3-Q1
IQR=18-11
IQR=7 lb

Use the data in the table to answer the following.

The table shows the attendance, in working days, of workers in a manufacturing plant over a period of one year.

Question 11.
Draw a stem-and-leaf plot.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves

2 2|0 represents 220 days

Question 12.
What is the mean number of working days in for this particular year?
Answer:
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
Mean=7103/30
Mean=236.7
Mean=237 (rounded to near number)

Question 13.
What is the mode?
Answer:
Mode definition: A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times.
according to the definition:
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
239 days and 240 days have the highest frequency. It occurs more time.

Question 14.
What is the median?
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 220, 230, 231, 232, 232, 233, 234, 234, 235, 236, 237, 238, 239, 239, 239, 239, 239, 239, 239, 239, 239, 240, 240, 240, 240, 240, 240, 240, 240, 240.
Median=239+239/2
Median=478/2
Median=239 days.

Solve. Show your work.

Question 15.
The weights of 10 pumpkins, in pounds, harvested at a farm is shown in the stem-and-leaf plot.

Compare the median and mean. Explain the difference in the values.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
The given data: 2.1, 2.2, 2.6, 2.7, 3.1, 3.3, 3.6, 3.9, 4.1, 8.1
Median=3.1+3.3/2
Median=6.4/2
Median=3.2 lb
Mean: Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.
Mean = (Sum of all the observations/Total number of observations)
Mean=2.1+2.2+2.6+2.7+3.1+3.3+3.6+3.9+4.1+8.1/10
Mean=35.7/10
Mean=3.57 lb
The median is 3.2 pounds while the mean is 3.57 pounds; the value of the mean is greater due to the presence of the outlier with the value of 8.1 pounds.

Question 16.
The speeds, in miles per hour, tracked by a police patrol car along a certain highway, are shown in the stem-and-leaf plot.

Explain how the outliers 40, 43, and 45 miles per hour affect the range of the data. How do these outliers affect the mean of the data?
Answer:
The range is calculated by subtracting the lowest value from the highest value. While a large range means high variability, a small range means low variability in a distribution.
The formula to calculate the range is:
R=H-L
– R = range
– H = highest value
– L = lowest value
Range=maximum value- minimum value
maximum value=81
minimum value=40
Range=41
The range is the difference between the least and the greatest values in the data set. So, the range is greater because of the outlier.

Question 17.
Math Journal The waiting time, in minutes, at a certain bus interchange for two bus routes on a particular day are recorded in the plot.

With the data shown, discuss some advantages and disadvantages of the stem-and-leaf plot.
Answer:
Advantages:
– It is easy to identify the minimum and maximum values from the data set.
– The range, median, and mode can be identified easily from a stem-and-leaf plot.
– The individual data values are preserved, unlike in a histogram.
Disadvantages:
– When the range is too narrow, there would be too many leaves.
– When the range is wide, there would be some stems that have no leaf, especially when outliers cause the range to widen.
– Bus route B has a relatively small range, with data concentrated within a few stems. Bus route A has a range that is much wider. So, when both bus routes are plotted in a double stem-and-leaf plot, bus route A has many stems without any leaf and the stem is much longer.

Question 18.
Math Journal Describe a data set that could be represented by a stem-and-leaf plot. Explain how using a stem-and-leaf plot could help you analyze the data set.
Answer:
The Stem and Leaf plot is a way of organizing data into a form that makes it easy to see the frequency of different values. In other words, we can say that a Stem and Leaf Plot is a table in which each data value is split into a “stem” and a “leaf.” The “stem” is the left-hand column that has the tens of digits. The “leaves” are listed in the right-hand column, showing all the ones digit for each of the tens, the twenties, thirties, and forties.
Remember that Stem and Leaf plots are a pictorial representation of grouped data, but they can also be called a modal representation. Because, by quick visual inspection at the Stem and Leaf plot, we can determine the mode.
Steps for making Stem-and-Leaf plot:
– First, determine the smallest and largest number in the data.
– Identify the stems.
– Draw a with two columns and name them as “Stem” and “Leaf”.
– Fill in the leaf data.
– Remember, a Stem and Leaf plot can have multiple sets of leaves.
Stem-and-leaf plot graphs are usually used when there are large amounts of numbers to analyze.
To compare two sets of data, you can use a back-to-back stem-and-leaf plot. For instance, if you want to compare the scores of two sports teams, you can use the stem-and-leaf plot.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Lesson 9.1 Interpreting Quartiles and Interquartile Range detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Lesson 9.1 Answer Key Interpreting Quartiles and Interquartile Range

Math in Focus Grade 7 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Calculate.

Question 1.
The table shows the average monthly temperatures in degrees Fahrenheit in New York. Find the range of the these temperatures.

Answer:
Definition of range: In statistics, the range of a data set is a measure of the spread or the dispersion of the observations. It is the difference between the largest and the smallest observed values in a data set.
Formula:
Range=max-min
Where max=maximum value in a data set.
Min=Minimum value in a data set.
How to calculate range steps:
1. List the elements of your data set.
2. Identify the highest and lowest numbers in the set.
3. Subtract the smallest number in your data set from the largest number.
4. Label the range clearly.
The given temperature data set: 31.5, 33.6, 42.4, 52.5, 62.7, 71.6, 76.8, 75.5, 68.2, 57.5, 47.6, 36.6
Now arrange the data set from lowest to highest:
31.5, 33.6, 36.6, 42.4, 47.6, 52.5, 57.5, 62.7, 68.2, 71.6, 75.5, 76.8
Range= 76.8-31.5
Range=45.3

Complete.

Question 2.
The heights, in centimeters, of the tomato seedling plants in a greenhouse are listed below. Find the first, second, and third quartiles of the heights of the seedling plants.

First arrange the heights in ascending order:
Q₂ is the median of the data. So, Q₂ =
Q₁ is the median of the lower half of the data:
Q₁ =
Q₃ is the median of the upper half of the data:
Q₃ =
Answer:
– In statistics, we generally deal with a large amount of numerical data. We have various concepts and formulas in statistics to evaluate the large data. One of such best applications is quartiles.
– Quartiles are the values that divide the given data set into four parts, making three points. Thus, quartiles are the values that divide the given data into three quarters. The middle part of the quartiles tells the central point of distribution. The difference between the higher and lower quartiles so formed gives the interquartile range.
– The quartiles formula is used to divide a given set of numbers into quarters. There are three quartiles formed, dividing the given data into four quarters. We are arranging the data in ascending order. The first quartile lies between the first number and the median, the second quartile is the median, and the third quartile lies between the median and the last number.
When the set of numbers is arranged in ascending order, then the formula used to calculate the quartiles are given below:
1. First quartile(Q1)=(n+1/4)th term
2. Second quartile(Q2)=(n+1/2)th term.
3. Third quartile=(Q3)=[3(n+1)/4)]th term.
The given data set: 9.3, 4.3, 5.2, 3.9, 10, 7, 6, 6.4, 9.5, 7.7, 10.6, 4.8, 8, 3.2
Now arrange the set from lowest to highest:
3.2, 3.9, 4.3, 4.8, 5.2, 6, 6.4, 7, 7.7, 8, 9.3, 9.5, 10, 10.6
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=6.4+7/2
Q2=6.5
Q2 is the median of the data.
The first quartile Q₁ is the median of the lower half not including the value of Q₂. The third quartile Q₃ is the median of the upper half not including the value of Q₂.
Q1=3.2, 3.9, 4.3, 4.8, 5.2, 6, 6.4
Q1=4.8
Q₁ is the median of the lower half of the data
Q3=7, 7.7, 8, 9.3, 9.5, 10, 10.6
Q3=9.3
Q₃ is the median of the upper half of the data

Solve.

Question 3.
Out of 50 multiple choice questions, 19 students scored the following numbers of correct answers.

Find the interquartile range and interpret what it means.
Answer:
the given data set: 36, 49, 45, 30, 27, 50, 45, 44, 20, 18, 14, 19, 31, 50, 26, 33, 10, 42, 30
Now arrange the data into lowest to highest:
10, 14, 18, 19, 20, 26, 27, 30, 30, 31, 33, 36, 42, 44,  45, 45, 49, 50, 50
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
How to calculate Interquartile range:
The procedure to calculate the interquartile range is given as follows:
– Arrange the given set of numbers into increasing or decreasing order.
– Then count the given values. If it is odd, then the centre value is median otherwise obtain the mean value for two centre values. This is known as the Q₂ value. If there are an even number of values, the median will be the average of the middle two values.
– Median equally cuts the given values into two equal parts. They are described as Q₁ and Q₃ parts.
– The median of data values below the median represents Q₁.
– The median of data values above the median value represents Q₃.
– Finally, we can subtract the median values of Q₁ and Q₃.
– The resulting value is the interquartile range.
Q2=31
Q2 is the median of the data.
n=19 (odd)
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=10, 14, 18, 19, 20, 26, 27, 30, 30
Q1=20
Q1 is the lower half of the data
Q3=33, 36, 42, 44,  45, 45, 49, 50, 50
Q3=45
Now we know Q1 and Q3
Interquartile range=Q3-Q1
Interquartile range=45-20
Interquartile range=25

Calculate.

Question 4.
A survey of 50 people asking the number of books they read in a year produced the following results.

a) Find the lower quartile, the upper quartile, and the interquartile range.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.

n=50
n/2=50/2=25
Median=1/2[(n/2)th term + (n/2+1)th term]
Median=1/2[ 25 + 26]
Median=1/2[51]
Median=25.5
Now check for the number, if the number was not there then check for the 25.5 above number and write the certain books
25.5 is not there, the next nearest number is 33 and the number of books they read is 5.
Median Q2=5
Q1 lower quartile=(n/4)th term
Q1=50/4
Q1=12.5
check for the next nearest number for 12.5 which is 13 and the corresponding number is 2
Therefore, Q1 is 2
Q3 upper quartile=(3n/4)th term
Q3=(3*50/4)
Q3=150/4
Q3=37.5
37.5 is not there, the next nearest number is 40 and the corresponding number is 6.
Q3=6
Interquartile range=Q3-Q1
Interquartile range=6-2
Interquartile range=4

b) Interpret what the interquartile range means.
Answer:
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
How to calculate Interquartile range:
The procedure to calculate the interquartile range is given as follows:
– Arrange the given set of numbers into increasing or decreasing order.
– Then count the given values. If it is odd, then the centre value is median otherwise obtain the mean value for two centre values. This is known as the Q₂ value. If there are an even number of values, the median will be the average of the middle two values.
– Median equally cuts the given values into two equal parts. They are described as Q₁ and Q₃ parts.
– The median of data values below the median represents Q₁.
– The median of data values above the median value represents Q₃.
– Finally, we can subtract the median values of Q₁ and Q₃.
– The resulting value is the interquartile range.

Technology Activity

Materials
spreadsheet software

USE SPREADSHEET SOFTWARE TO FIND QUARTILES, INTERQUARTILE RANGE, AND RANGE

Work in pairs.

Step 1: Randomly pick 10 students. Count the number of letters in their first names and enter the values in a row of cells.

Step 2: Choose another 5 cells for Q₁, Q₂, Q₃, interquartile range, and range.

Step 3: Use the spreadsheet software’s function for finding quartiles, maximums, and minimums of data. In the 5 cells you choose in Step 2 generate statistics for the 10 values you entered in Step 1.

Step 4: What is the median of the data? What does this tell you about the first names in the data set? What does the interquartile range tell you about the first names in the data set?

Step 5: Randomly pick another 10 students and repeat the process. Compare the statistics for the two data sets. What differences do you notice in the interquartile range for the two data sets?

Math Journal Suppose you add a data value for a student whose first name has many more letters than anyone else’s. How would adding this data value affect the range and the interquartile range of the data set? Explain your thinking.
Answer:

Math in Focus Course 2B Practice 9.1 Answer Key

Find the range of each set of data values.

Question 1.
Weights of rabbits in pounds: 15, 9, 10, 22, 14.5, 12.5, 16, 8, and 11.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 15, 9, 10, 22, 14.5, 12.5, 16, 8, and 11.
Highest value=22
Lowest value=8
Range=highest value-lowest value
Range=22-8
Range=14

Question 2.
Number of pairs of shoes sold each day for a week at a retail shop: 67, 63, 66, 60, 58, 70, and 75.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 67, 63, 66, 60, 58, 70, and 75.
Highest value=75
Lowest value=58
Range=75-58
Range=17

Question 3.
Time taken in seconds to solve a puzzle by a group of contestants: 20, 25.6, 15.4, 20, 22.8, 18.7, 19, 24, 13.4, and 16.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 20, 25.6, 15.4, 20, 22.8, 18.7, 19, 24, 13.4, and 16
Highest value=25.6
Lowest value=13.4
Range=25.6-13.4
Range=12.2

Question 4.
Average monthly rainfall in centimeters in a city:
18.3, 17.8, 17.3, 8.9, 6.1, 2.0, 1.4, 2.2, 2.8, 5, 12.7, and 14.
Answer:
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
The given data set: 18.3, 17.8, 17.3, 8.9, 6.1, 2.0, 1.4, 2.2, 2.8, 5, 12.7, and 14.
Highest value=18.3
Lowest value=1.4
Range=18.3-1.4
Range=16.9

Find the first, second, and third quartiles of each set of data values.

Question 5.
Science test scores of 13 students:
78, 63, 56, 85, 62, 59, 78, 90, 83, 63, 84, 66, and 63.
Answer:
The given data set: 78, 63, 56, 85, 62, 59, 78, 90, 83, 63, 84, 66, and 63.
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
56,59,62,63,63,63,66,78,78,83,84,85,90
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=66
n=13
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=56,59,62,63,63,63
Q1=62+63/2
Q1=125/2
Q1=62.5
Thus, the first quartile Q1 is 62.5
Q3=78,78,83,84,85,90
Q3=83+84/2
Q3=167/2
Q3=83.5

Question 6.
Scores of a basketball team in a series of games:
82, 66, 70, 68, 54, 77, 80, 70, 82, and 65.
Answer:
The given data set: 82, 66, 70, 68, 54, 77, 80, 70, 82, and 65.
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
54, 65, 66, 68, 70, 70, 77, 80, 82, 82
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2= 70+70/2
Q2=140/2
Q2=70
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
n=10
Q1=54, 65, 66, 68, 70
Q1=66
Q₁ is the median of the lower half of the data
Q3=70, 77, 80, 82, 82
Q3=80
Q3 is the median of the upper half of the data

Question 7.
Ages of 20 children at a party:
6, 9, 5, 12, 7, 8, 9, 10, 7, 7, 7, 6, 9, 9, 12, 10, 8, 10, 11, and 9.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 12, 12
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=9+9/2
Q2=18/2
Q2=9
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
n=20
Q1=5, 6, 6, 7, 7, 7, 7, 8, 8, 9
Q1=7+7/2
Q1=14/2
Q1=7
Q₁ is the median of the lower half of the data
Q3=9, 9, 9, 9, 10, 10, 10, 11, 12, 12
Q3=10+10/2
Q3=20/2
Q3=10
Q3 is the median of the upper half of the data

Question 8.
Heights of 15 tomato plants in inches:
36, 27, 18, 40, 29, 43, 24, 20, 15, 31, 39, 22, 27, 30, and 20.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
15, 18, 20, 20, 22, 24, 27, 27, 29, 30, 31, 36, 39, 40, 43
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=27
n=15
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
Q1=15, 18, 20, 20, 22, 24, 27
Q1=20
Q3=29, 30, 31, 36, 39, 40, 43
Q3=36

Use the information and table below to answer questions 9 to 12.

To create greater public awareness of the environment, 22 volunteers collected cans for recycling. The table shows the number of cans each volunteer collected.

Question 9.
Find the range of the number of cans collected.
Answer:
The given data: 25, 105, 100, 47, 61, 82, 75, 44, 61, 70, 53, 38, 50, 73, 54, 80, 36, 25, 28, 24, 110, 78
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
range=110-24
range=86

Question 10.
Calculate Q₁, Q₂, and Q₃.
Answer:
How to calculate quartiles:
– Order your data set from lowest to highest values
– Find the median. This is the second quartile Q₂.
– At Q₂ split the ordered data set into two halves.
– The lower quartile Q₁ is the median of the lower half of the data.
– The upper quartile Q₃ is the median of the upper half of the data.
Now arrange the data from lowest to highest:
24, 25, 25, 28, 36, 38, 44, 47, 50, 53, 54, 61, 61, 70, 73, 75, 78, 80, 82, 100, 105, 110
The second quartile Q₂ is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
Q2=54+61/2
Q2=115/2
Q2=57.5
n=22 (even number)
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q₂.
Q1=24, 25, 25, 28, 36, 38, 44, 47, 50, 53, 54
Q1=38
Q3=61, 61, 70, 73, 75, 78, 80, 82, 100, 105, 110
Q3=78

Question 11.
Find the interquartile range.
Answer:
Definition of interquartile range: The interquartile range defines the difference between the third and the first quartile. Quartiles are the partitioned values that divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by Q₁ known as the lower quartile, the second Quartile is denoted by Q₂ and the third
A quartile is denoted by Q₃ known as the upper quartile. Therefore, the interquartile range is equal to the upper quartile minus the lower quartile.
Interquartile range formula: The difference between the upper and lower quartile is known as the interquartile range. The formula for the interquartile range is given below:
Interquartile range = Upper Quartile – Lower Quartile = Q_­3 – Q1
Where Q1 is the first quartile, Q3 is the third quartile.
Q1=38; Q3=78
Interquartile range=Q3-Q1
Interquartile range=78-38
Interquartile range=40

Question 12.
Interpret Q₃ and the interquartile range.
Answer:
The third quartile (Q 3), also known as the upper quartile, splits the lowest 75% (or highest 25%) of data. It is the middle value of the upper half. The first quartile is also known as the 25th percentile, the second quartile as 50th percentile, and the third quartile as 75th percentile. The Interquartile range is from Q 1 to Q 3.

Use the information and table below to answer questions 13 to 16.

The table shows the results of a survey asking 50 people how many hours they spent using the Internet in a day.

Question 13.
Find the median, the lower quartile, and the upper quartile of the number of hours spent using the Internet.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.
Cumulative frequency for 1 hour=3
Cumulative frequency for 2 hour=3+0=3
Cumulative frequency for 3 hour=3+9=12
Cumulative frequency for 4 hour=5+12=17
Cumulative frequency for 5 hour=7+17=24
Cumulative frequency for 6 hour=11+24=35
Cumulative frequency for 7 hours=5+35=40
Cumulative frequency for 8 hours=1+40=41
Cumulative frequency for 9 hours=6+41=47
Cumulative frequency for 10 hours=47+3=50
n=50
therefore, n/2=50/2=25
24 is nearer to the 25
median=l+(n/2-cf)/f*c
l=lower limit of median class
n=total number of observations
cf=cumulative frequency of the class
preceding the median class
f=frequency of the median class
c=class length
Median=5+(25-17)/7*1
After calculating the above equation
Median=5+(18)/7*1
Meadian=5+(1.14)*1
Median=6.14
If we round the nearest number, then it will be 6 hours.
Lower quartile=(n/4)th term
Lower quartile=50/4
Lower quartile=12.5
We got 12.5 checks the number which is above the 12.5
We have 17: it is 4 hours
Therefore, Q1=4 hours
Q3=(3n/4)th term
Q3=(3*50/4)
Q3=37.5
Now check the number which is having above 37.5.
So the number is 40: 7 hours

Question 14.
Calculate the interquartile range.
Answer:
Interquartile range=upper quartile-lower quartile
Interquartile range=7-4
Interquartile range=3 hours.

Question 15.
Interpret the data values that are greater than the upper quartile.
Answer:
Upper quartile Q3=(3n/4)th term
Upper quartile Q3=(3*50/4)
Upper quartile Q3=37.5
Now check the number which is having above 37.5.
So the number is 40: 7 hours
25% of the people spent at least 7 hours on the internet.

Question 16.
Math journal A majority of the people surveyed spent 6 hours or less using the Internet.” Do you agree with this statement? Justify your argument by using the quartiles.
Answer:
The above quartiles we got are:
Q1= 4 hours
Q2=6 hours
Q3=7 hours.
yes, according to the quartiles Q1 and Q2 majority of the people surveyed spent 6 hours or less using the Internet.

Use the information and table below to answer questions 17 to 21.

The table shows the lifespans of 29 batteries that were tested.

Question 17.
Find the range of the lifespans.
Answer:
The given data set for lifespan: 2, 3.5, 4.5, 5, 6, 7.5, 8, 9
Range: The range in statistics for a given data set is the difference between the highest and lowest values. For example, if the given data set is {2,5,8,10,3}, then the range will be 10 – 2 = 8.
– Thus, the range could also be defined as the difference between the highest observation and lowest observation. The obtained result is called the range of observation. The range in statistics represents the spread of observations.
Range formula:
The formula of the range in statistics can simply be given by the difference between the highest and lowest value.
Range = Highest Value – Lowest Value (Or)
Range = Highest observation – Lowest observation (Or)
Range = Maximum value – Minimum Value
How to find range:
To find the range in statistics, we need to arrange the given values or set of data or set of observations in ascending order. That means, firstly write the observations from the lowest to the highest value. Now, we need to use the formula to find the range of observations.
Maximum value=9
Minimum value=2
Range=maximum value-minimum value
Range=9-2
Range=7
Therefore, the range of batteries lifespan is 7 hours.

Question 18.
Math Journal Explain why the range is not a good statistic to use for evaluating the lifespans of these batteries.
Answer:
well, it depends on the type of battery and how you are using it. Some batteries have a limited lifespan and others can last for years. So we cannot evaluate the lifespans of these batteries by using the range.

Question 19.
Find the median, the lower quartile, and the upper quartile of the lifespans.
Answer:
Median: The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Now calculate the cumulative frequency
Cumulative frequency is defined as a running total of frequencies. The frequency of an element in a set refers to how many of that element there are in the set. Cumulative frequency can also be defined as the sum of all previous frequencies up to the current point.

n=29 (given)
Therefore, n/2=29/2=14.5
Median=1/2[(n/2)th term + (n/2+1)th term]
Median=1/2[ 14.5 + 15.5]
Median=1/2[30]
Median=15
Now check for the number, if the number was not there then check for the 15 above number and write the certain battery lifespan.
15 is not there, the next nearest number is 18 and the battery lifespan is 6 hours.
Median Q2=6 hours.
Q1 lower quartile=(n/4)th term
Q1=29/4
Q1=7.25
Q1=3
the corresponding battery lifespan is 3.5 which is nothing but 4.
Therefore, Q1 is 4 hours.
Q3 upper quartile=(3n/4)th term
Q3=(3*29/4)
Q3=87/4
Q3=21.75
21.75 (22) is not there, the next nearest number is 24 and the battery lifespan is 8 hours.
Q3=8 hours.

Question 20.
Calculate the interquartile range.
Answer:
Upper quartile Q3=8
Lower quartile Q1=4
Interquartile range=upper quartile-lower quartile
Interquartile range=8-4
Interquartile range=4 hours.

Question 21.
Math Journal Compare the lifespans less than the lower quartile to the lifespans greater than the upper quartile. Based on this comparison, do you consider the difference in battery lifespans significant? Explain your reasoning.
Answer:
Q1 lower quartile=(n/4)th term
Q1=29/4
Q1=7.25
Q1=3
the corresponding battery lifespan is 3.5 which is nothing but 4.
Therefore, Q1 is 4 hours.
Q3 upper quartile=(3n/4)th term
Q3=(3*29/4)
Q3=87/4
Q3=21.75
21.75 (22) is not there, the next nearest number is 24 and the battery lifespan is 8 hours.
Q3=8 hours.
The difference is significant. The battery lifespans above the upper quartile are more than double the battery lifespans below the lower quartile.

Question 22.
Math Journal Another statistic used to analyze data is the midrange.

Is the midrange the same as the median of a set of data values? Explain your reasoning by using an example.
Answer:
Midrange in layman terms is the middle of any data set or simply the average, mean of the data. A midrange is a statistical tool that is also known as the measure of centre in statistics. Along with the existence of the midrange formula means, medium, average, mode, and range are also known as the measure of central tendency. The midrange of the data set is simply the value between the biggest value and the lowest value. In order to find the midrange of the data set the value is then divided by 2 after summing the lowest value present in the data set with the highest value present in the data set.
No, the midrange and median are not the same.
Example: The daily temperature recorded in the city of Colombia Bogata is 55, 65, 67, 69, 70, 80, 81, 87, 90. We need to calculate the mid-temperature in Bogata during this period.
According to the midrange formula:
midrange=greatest value-lowest value/2
midrange=90-55/2
midrange=35/2
midrange=17.5
midrange and median are not the same.

Solve.

Question 23.
The finishing times of 40 people in a 100-meter freestyle swimming event were collected. The quartiles for the data are shown:
Q₁ = 62.05 seconds, Q₂ = 69.16 seconds, and Q₃ = 71.43 seconds.

List the letters of all the following statements that are correct.
a) 50% of the swimmers managed to complete the 100-meter event in between 69.16 and 71.43 seconds.
Answer: wrong statement.
The finishing times have already been taken. But in this question the event is not completed so the statement is wrong.

b) 75% of the swimmers took longer than 62.05 seconds to complete the event.
Answer:
The event is completed so the statement is correct.
75%of swimmers mean:
75% of 40 people
75*40/100
=30
30 people took longer than 62.05 seconds to complete the event.

c) Swimmers with times greater than the upper quartile are fast swimmers.
Answer:
The statement is wrong.
The given time is 3 h 35 min 49 sec
The upper quartile time is 71.43 sec is which is nothing but 0.01 hours
swimmers take time greater than the upper quartile might be slower
For example, if a swimmer has taken 75 seconds means he has taken more time to complete the event then he cant be a fast swimmer. So definitely, the statement is correct.

d) Swimmers with times between the first and the second quartiles are faster than the first 10 swimmers who finished the swimming event.
Answer:
the above-given quartiles:
Q1=62.05 seconds,
Q₂ = 69.16 seconds
This we cannot say about the faster swimmers. Because we know quartile timings and we don’t know about the first 10 swimmers timings.
In how much time they finished the event. Until we know the finish timings of the first 10 swimmers we cannot tell the Swimmers with times between the first and the second quartiles are faster than the first 10 swimmers who finished the swimming event.

e) The interquartile range shows that the time differences of the middle 50% of the swimmers is not more than 9.38 seconds.
Answer:
interquartile range=Q3-Q1
Interquartile range= 71.43 – 62.05
Interquartile range=9.38
Yes, it is correct.
It is not more than 9.38
therefore, the statement is correct.

f) 50% of the swimmers completed the 100-meter event between 62.05 and 71.43 seconds.
Answer: yes
Nearly 20 people completed the event between 62.05 and 71.43
In the above question, we got the timings of the swimmer that are more than 62.05
definitely, this statement is correct.
g) 25% of the swimmers took 71.43 or more seconds to complete the event.
Answer:
25% means the remaining people completed the event.
25*40/100
=10
10 people took 71.43 or more seconds to complete the event.
h) 10 swimmers completed the event in less than 62.05 seconds.
Answer:
total 40 people have recorded the finishing timings.
20 people took time between 62.05 and 71.43
10 people took time 71.43 or more seconds.
and the other 10 remaining people might take a chance of 62.05 seconds or less than that.
So the statement is correct.

i) The number of swimmers with times between the first and the second quartiles is more than the number of swimmers with times between the second and the third quartiles.
Answer:
The above-given quartiles:
Q₁ = 62.05 seconds, Q₂ = 69.16 seconds, and Q₃ = 71.43 seconds.
This is the record:
total 40 people have recorded the finishing timings.
20 people took time between 62.05 and 71.43
10 people took time 71.43 or more seconds.
and the other 10 remaining people might take a chance of 62.05 seconds or less than that.
By observing the above statements, the given statement is wrong.

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 9 Statistics detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 9 Answer Key Statistics

Math in Focus Grade 7 Chapter 9 Quick Check Answer Key

Find the mean of each set of data. Round your answer to 2 decimal places if it is not exact.

Question 1.
9, 11, 6, 29, 5
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 9, 11, 6, 29, 5
Mean=9+11+6+29+5/5
Mean=60/5
Mean=12.

Question 2.
43, 88, 39, 10, 26, 17, 35
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 43, 88, 39, 10, 26, 17, 35
Mean=43+88+39+10+26+17+35/7
Mean=258/7
Mean=36.85 (round to nearest number)
Mean=37.

Question 3.
53.6, 36.7, 88.5, 90.6
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 53.6, 36.7, 88.5, 90.6
Mean=53.6+36.7+88.5+90.6/4
Mean=269.4/4
Mean=67.35
Mean=67 (rounded to nearest number)

Question 4.
0.14, 1.05, 3.1, 7.18, 4.3, 8
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

The above-given data: 0.14, 1.05, 3.1, 7.18, 4.3, 8
Mean=0.14+1.05+3.1+7.18+4.3+8/6
Mean=23.77/6
Mean=3.96
Mean=4

Solve. Show your work.

Question 5.
The heights, in inches, of 10 children are
54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Find the mean height of the children.
Answer:
Definition of mean: The most common measure of central tendency is the arithmetic mean. In layman’s terms, the mean of data indicates an average of the given collection of data.  It is equal to the sum of all the values in the group of data divided by the total number of values.
For n values in a set of data namely as x₁, x_2, x_{3, …} x_n, the mean of data is given as:

It can also be denoted as:

the above-given data: 54, 66, 52, 60.5, 61.25, 55, 58.75, 51.5, 53, 50.
Mean=54+66+52+60.5+61.25+55+58.75+51.5+53+50/10
Mean=562/10
Mean height of children=56.2

Find the median of each set of numbers.

Question 6.
41,29, 78, 12, 56, 30, 22
Answer: 30
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
The formula to calculate the median of the data set is given as follow.
An odd number of observations:
If the total number of observations given is odd, then the formula to calculate the median is:
Median = {(n+1)/2}^thterm
where n is the number of observations
Even number of observations:
If the total number of observations is even, then the median formula is:
Median  = [(n/2)^th term + {(n/2)+1}^th]/2
where n is the number of observations
How to  calculate Median:
To find the median, place all the numbers in the ascending order and find the middle.
Put them in ascending order: 12, 22, 29, 30, 41, 56, 78
The middle number is 30, so the median is 30.

Question 7.
193, 121.5, 162.3, 125, 103.8, 149.6
Answer:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 103.8, 121.5, 125, 149.6, 162.3, 193
Meadian=125+149.6/2
Median=274.6/2
Median=137.3
Thus, the median is 137.3

Question 8.
9, 2, 2, 4, 4, 4, 1, 3, 6, 5, 3, 6
Answer: 4
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 6, 9
Median=4+4/2
Median=8/2
Median=4
Thus, the median is 4.

Question 9.
1,011, 1,100, 1,001, 1,010, 1,110, 1,000, 1,011, 100
Answer:
Explanation:
The median of a set of data is the middlemost number or centre value in the set. The median is also the number that is halfway into the set.
To find the median, the data should be arranged, first, in order of least to greatest or greatest to the least value. A median is a number that is separated by the higher half of a data sample, a population or a probability distribution, from the lower half. The median is different for different types of distribution.
Median formula:
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 100, 1000, 1001, 1010, 1011, 1011, 1100, 1110
Median=2021/2
Median=1010.5
Thus, the median is 1010.5

Solve. Show your work.

Question 10.
The daily low temperatures for the past 10 days were
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
What was the median daily low temperature?
Answer:
The statistical concept of the median is a value that divides a data sample, population, or probability distribution into two halves. Finding the median essentially involves finding the value in a data sample that has a physical location between the rest of the numbers. Note that when calculating the median of a finite list of numbers, the order of the data samples is important. Conventionally, the values are listed in ascending order, but there is no real reason that listing the values in descending order would provide different results. In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values. While this can be confusing, simply remember that even though the median sometimes involves the computation of a mean, when this case arises, it will involve only the two middle values, while a mean involves all the values in the data sample. In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same. Given the same data set as before, the median would be acquired in the following manner:
30.6°F, 32.1 °F, 29.5°F, 30.2°F, 26.4°F, 34.3°F, 31.6°F, 32°F, 25.9°F, and 26.4°F.
The formula to calculate the median of the finite number of data set is given here. The median formula is different for even and odd numbers of observations. Therefore, it is necessary to recognise first if we have an odd number of values or an even number of values in a given data set.
Sorted data set: 25.9, 26.4, 26.4, 29.5, 30.2, 30.6, 31.6, 32, 32.1, 34.3
Median=30.2+30.6/2
Median=60.8/2
Median=30.4
Thus, the median is 30.4

Summarize each of the following data sets in a frequency table and draw a dot plot.

Question 11.
The data show the number of misspelled words found in 15 essays.

Answer:
Definition of frequency: Frequency means the number of times a value appears in the data. A table can quickly show us how many times each value appears. If the data has many different values, it is easier to use intervals of values to present them in a table.

A dot chart or dot plot is a statistical chart consisting of data points plotted on a fairly simple scale, typically using filled in circles.

Question 12.
In a game, each child is given 10 balls to hit at moving targets. The data show the number of hits scored by 20 children.

Answer:
Definition of frequency: Frequency means the number of times a value appears in the data. A table can quickly show us how many times each value appears. If the data has many different values, it is easier to use intervals of values to present them in a table.

A dot chart or dot plot is a statistical chart consisting of data points plotted on a fairly simple scale, typically using filled in circles.

This handy Math in Focus Grade 7 Workbook Answer Key Cumulative Review Chapters 6-8 detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Cumulative Review Chapters 6-8 Answer Key

Concepts and Skills

Tell whether each pair of angles is supplementary, complementary, or neither. (Lesson 6.1)

Question 1.
m∠1 = 18° and m∠2 = 82°
Answer:
Given each pair of angles are complementary angles.

Explanation:
The sum of two angles is equal to 90° is called complementary angles. In the given figure if we sum up both the angles that is 18°+ 82°we get a 90° angle.

Question 2.
m∠3 = 103° and m∠4 = 77°
Answer:
Given each pair of angles are supplementary angles.

Explanation:
Supplementary angles are those angles that sum up to 180°.

Question 3.
m∠5 = 95° and m∠6 = 85°
Answer:
Given each pair of angles are supplementary angles.

Explanation:
Supplementary angles are those angles that sum up to 180°. Therefore add 95° and 85° then we get 180°.

Question 4.
m∠7 = 21° and m∠8 = 69°
Answer:
Given each pair of angles are complementary angles.

Explanation:
The sum of two angles is equal to 90° is called complementary angles. In the given figure if we sum up both the angles that are 21°+ 69°we get a 90° angle.

Find the measure of each numbered angle. (Lessons 6.1, 6.2)

Question 5.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
54°

Explanation:
Given, ∠SQR = 126°
∠QPR = 180°
∠PQS = 1
Here we need to calculate the numbered angle.
To find that we need to subtract ∠QPR from ∠SQR
∠1 = 180°-126°
∠1   = 54°
Hence the numbered angle is 54°

Question 6.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
88°

Explanation:
The straight-line QPR has 180°.
Given, ∠PQS = 54°
∠QTR = 38°
∠SQT is numbered angle that is ∠2
Now we need to calculate the numbered angle
Therefore QPR = ∠PQS + ∠QTR + ∠2
180°= 54°+38°+ ∠2
180°= 92°+ ∠2
180°-92° = ∠2
∠2 = 88°

Question 7.

Answer:
The numbered angle 3 is 94°

Explanation:
The sum of angles around a point is 360°
40°+125°+101°+3=360°
3= 360°-40°-125°-101°
3= 360° – 266°
3 = 94°
Hence the numbered angle 3 is 94°

Question 8.

Answer:
36°

Explanation:

The sum of angles around a point is 360°
When two lines cross the angles are opposite to each other. Therefore opposite angles are equal to each other.
The angle ∠ABC and ∠DBE are opposite to each other. So ∠ABC = ∠DBE
Given, ∠ABC = 162°, ∠DBE means the numbered angle 4 is also 162°
Now we need to calculate the numbered angle 5
To calculate the angle 5 we need to add ∠ABC + ∠DBE+ 5 = 360°
162°+162°+∠5 = 360°
324°+ ∠5 =360°
∠5 = 360°-324°
∠5 = 36°

\(\overleftrightarrow{M N}\) is parallel to \(\overleftrightarrow{P Q}\). Find the measure of each numbered angle. (Lesson 6.3)

Question 9.

Answer:
m∠1 = 81°
Vertical ∠s

Question 10.

Answer:
m∠3 = 49°
Alternate interior ∠s

Find the measures of ∠1 and ∠2 in each diagram. (lesson 6.4)

Question 11.
Triangle ABC is an isosceles triangle.

Answer:
32°

Explanation:

1+32°+x =180°
1+x° = 180°-32°
1+x°= 148°
1+x°+2 = 180°
148° +2 = 180°
2 = 180°-148°
2 = 32°
1+x°= 148°
32°+x°= 148°
x°=148°-32°
x°=  116°
1=148-116
1=32°

Question 12.
\(\overleftrightarrow{P R}\) is a straight line.

Answer:
110°+1 = 180°
1 = 180°-110°
1 = 70°

Use the given Information to construct each polygon.

Construct triangle ABC with sides AB = 5.8 cm, BC = 6.6 cm, and AC = 4.5 cm. (Lessons 7.1, 7.2, 7.3)

Step 1:
we are given
AB = 5.8
BC = 6.6
AC = 4.5

Step 2:
Use a ruler to draw the segment 66 cm long. LabeL its endpoints by B and C.

Step 3:
Because AB = 5.8 cm, use the ruLer to set the compass to a radius of 5.8 cm Then using A as center draw an arc of radius 5.8 cm above \(\overline{B C}\).

Step 4:
Because AC = 4.5 cm, use the ruler to set the compass to a radius of 4.5 cm. Then using G as center draw an arc of radius 4.5 cm that intersects the first arc. Label the intersection as A.

Step 5:
Use the ruler to draw \(\overline{A B}\) and \(\overline{A C}\)

a) Measure the angles of the triangle.
Answer:
m∠A ≈ 79°
m∠B ≈ 42°
m∠C ≈ 59°

b) Use a compass and straightedge to bisect angle B.
Answer:

c) Use a compass and straightedge to construct the perpendicular bisector of AB.
Answer:

d) The two bisectors intersect at a point D, label D in your construction.
Answer:

e) Measure the distances BD and CD.
Answer:
BD ≈ 3.2 cm
CD ≈ 3.8 cm

Question 14.
Construct quadrilateral PQRS with PQ = PR = 7.8 cm, PQ ||SR, m∠PQR = 73°, and RS = 4 cm. (Lessons 7.1, 7.2, 7.4)

Step 1:
We are given
PQ = PR = 7.8
PQ || SR
m∠PQR = 73°
RS = 4

Step 2:
Sketch the quadrilateral

Step 3:
First we construct △PRQ using the side-side-angle method.

Step 4:
Determine m∠RPQ
m∠RPQ + 2.73° = 180°
m∠RPQ = 180° – 146°
m∠RPQ = 34°

Step 5:
As PQ || SR, we have
m∠SRP = m∠RPQ = 34°

Step 6:
We construct △PRS, using the side-angle-side method.

a) Measure the length of PS.
Answer:
PS ≈ 5 cm

b) Use a compass and straightedge to construct the angle bisector of ∠S.
Answer:

c) Use a compass and straightedge to construct the perpendicular bisector of \(\overline{P Q}\).
Answer:

d) Label the point where the angle bisector and perpendicular bisector intersect as W.
Answer:

Solve. Show your work. (Lesson 7.5)

Question 15.
Jacob builds a model plane that is 800 millimeters long. If the actual plane is 40 meters, what is the scale of the model? (10 mm = 1 cm)
Answer:
We are given
Length on the plane: 800 mm
Actual length: 40 m

Determine the scale of the model:
\(\frac{\text { Length on the plane }}{\text { Actual length }}=\frac{800 \mathrm{~mm}}{40 \mathrm{~m}}=\frac{80 \mathrm{~cm}}{4,000 \mathrm{~cm}}=\frac{1}{50}\)
= \(\frac{1}{50}\)

Question 16.
The scale of a map is 1 : 1,500. If a square piece of land measures 4 inches on each side on the map, find the actual area of this piece of land to the nearest tenth of an acre. (1 acre = 43,560 ft²)
Answer:

We have:
Step 1:
Map length : Actual length = 1 in. : 1,500 in.
Map area: ActuaL area = 1 in.² : 1, 5002 in²

Step 2:
Let y represent the actual area of the piece of land in square inches.

Step 3:
Write a proportion: \(\frac{\text { Piece area on map }}{\text { Actual area of piece }}=\frac{1}{2,250,000}\)

Step 4:
Substitute:
\(\frac{4^{2}}{y}=\frac{1}{2,250,000}\)

Step 5:
Write the cross products:
y = 16.2,250,000

Step 6:
Simplify
y = 36,000,000 in²

Step 7:
Convert to acres:
1 acre = 43,560 ft = 43. 560. 12² = 6. 272, 640 in²
\(\frac{36,000,000}{6,272,640}\) ≈ 5.7 acres
= 5.7 acres

Match each solid with its net. (Lesson 8.1)

Question 17.

Answer:
The net of a cone consists of a circle and a sector of a circle.
The net b) does not consists a sector of circle, so we eliminate it
The net a) also doesn’t contain a sector of circle, therefore we eliminate it
The net d) is an ellipse sector, not a circle sector, so we eliminate it

The correct answer is (c)

Question 18.

Answer:

The net of a trianguLar prism contains 4 triangles.

The net b) contains 3 triangles and a rectangle, not 4 triangles, so we eliminate it
The net c) contains 4 triangles and a rectangle, not 4 triangles, so we eliminate it
The net d) contains 3 triangles, not 4 triangles, so we eliminate it

The correct answer is (a)

For each solid, sketch and describe the shape of the cross section formed when a plane slices the solid in the direction indicated. (Lesson 8.1)

Question 19.
A cube of length 6 centimeters perpendicular to a base and parallel to two opposite faces.

Answer:

A cross-section perpendicular to a base and parallel to two opposite faces can be drawn starting with a point for example on one of the edges of the top face. We draw a parallel line to two of the edges on the top face other than the one opposite to the one containing the initial point From the ends of the line we draw two vertical lines to the vertical edges. We join the ends of these two paralLels.

Question 20.
A cylinder of radius 3 centimeters parallel to its bases.

Answer:

A cross section parallel to its bases has the shape of a circle of radius 3:

Question 21.
A right triangle base prism parallel to its bases.

Answer:

A cross-section paralleL to its bases can be drawn starting with a point for example on one of the edges other than the bases We draw a parallel line to each of the corresponding edges on the bases. We join the ends of these two parallels.

Question 22.
Copy and complete the table. Use 3.14 as an approximation for π. (Lesson 8.2)

Answer:
Step 1:
We are given the cylinder:
V = 942
h = 12

Step 2:
Determine the radius r:
V = πr²h
942 ≈ 3.14 · r² · 12
942 = 37.68 r²
r² = \(\frac{942}{36.68}\)
r² = 25
r = \(\sqrt{25}\)
r = 5 ft

Step 3:
Determine the diameter:
d = 2 · r = 2 · 5 = 10 ft

Step 4:
Determine the total surface area:
S = 2πr² + 2πrh
= 2π · 5² + 2π · 5 · 12
= 50π + 120π
= 170π
≈170 · 3.14
= 533.8 ft²

Step 5:
We are given the cylinder:
V = 879.2
d = 10

Step 6:
Determine the radius:
r= \(\frac{d}{2}\) = \(\frac{10}{2}\) = 5 cm

Step 7:
Determine the height:
V = πr²h
879.2 ≈ 3.14 · 5² · h 
879.2 = 78.5h
h = \(\frac{879.2}{78.5}\) = 11.2cm

Step 8:
Determine the total surface area:
S = 2πr² -1- 2πrh
= 2π · 5² + 2π · 5 · 11.2
= 50π + 112π
= 162π
≈ 162 · 3.14
= 508.7cm²

Step 9:
We are given the cylinder:
r = 4
S = 251.2

Step 10:
Determine the height:
S = 2πr² + 2πrh
251.2 ≈ 2 · 3.14 · 4² + 2 · 3.14 · 4 · h
251.2 = 100.48 + 25.12h
251.2 – 100.18 = 100.48 + 25.12h – 100.48
150.72 = 25.12h
h = \(\frac{150.72}{25.12}\)
h = 6 m

Step 11:
Determine the volume:
V = πr²h ≈ 3.14 · 4^{2 ·} 6 ≈ 301.4 m³

Step 12:
Determine the diameter:
d = 2r = 2 · 4 = 8 m

Step 13:
Complete the table

Find the exact surface area of each solid cone. (Lesson 8.3)

Question 23.
A cone with radius of 8.7 meters and a slant height of 12.8 meters.
Answer:
We are given the cone:
r = 8.7
l = 12.8
Use the formula for the cone’s surface area:
S = πr² + πrl
Substitute r, l to find the exact surface area:
= π · 8.7² + π · 8.7 · 12.8
= 75.69π + 111.36π
= 187.05 π m²

Question 24.
A cone of diameter 16.8 centimeters and slant height 20.6 centimeters.
Answer:
We are given the cone:
d = 16.8
l = 20.6
First we determine the radius:
r = \(\frac{d}{2}\) = \(\frac{16.8}{2}\) = 8.4
Use the formuLa for the cones surface area:
S = πr² + πrl
Substitute r, ito find the exact surface area:
= π · 8.42 + π · 8.4 · 20.6
= 70.56π + 173.04π
= 243.6π cm²

Find the volume of each solid. Use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can. (Lessons 8.3, 8.4)

Question 25.
A square pyramid with a height of 13 centimeters and a base that is 8 centimeters on each edge.
Answer:
Use the formula:
V = \(\frac{1}{3}\) Bh
Substitute for B and h.
= \(\frac{1}{3}\) · (8²) · 13
Multiply and round:
≈ 277.3 cm³

= 277.3 cm³

Question 26.
An octagonal pyramid with a height of 8.2 inches and a base area of 34.5 square inches.
Answer:
Use the formula:
V = \(\frac{1}{3}\) Bh
Substitute for B and h.
= \(\frac{1}{3}\) · 34.5 · 8.2
Multiply and round:
= 94.3 in³

Question 27.
A cone with a radius of 6.5 centimeters and a height of 14 centimeters.
Answer:
We are given the cone:
r = 6.5
h = 14
Use the formula for the cone’s volume to find the volume:
V = \(\frac{1}{3}\)πr²h
≈ \(\frac{1}{3}\) · 3.14 · 6.5² · 14
≈ 619.1 cm³
= 619.1 cm³

Question 28.
A cone with a diameter of 24.6 feet and a height of 18.5 feet.
Answer:
We are given the cone:
d = 24.6
h = 18.5
Determine the radius:
r = \(\frac{d}{2\) = \(\frac{24.6}{2}\) = 12.3 ft
Use the formula for the cone’s volume to find the volume:
V = \(\frac{1}{3}\)πr²h
≈ \(\frac{1}{3}\) · 3.14 · 12.3² · 18.5
≈ 2,929.5 ft³
= 2,929.5 ft³

Question 29.
A sphere with a radius of 9.6 feet.
Answer:
We are given the sphere:
r = 9.6
Use the formula for the volume of a sphere:
V = \(\frac{4}{3}\)πr³ ≈ \(\frac{4}{3}\) · 3.14 · 9.6³
Multiply:
= 3,704.094…
Round to the nearest tenth:
≈ 3,704.1 ft³
= 3,704.1 ft³

Question 30.
A sphere with a diameter of 39.8 centimeters.
Answer:
We are given the sphere:
d = 39.8
Determine the radius of the sphere:
r = \(\frac{d}{2}\) = \(\frac{39.8}{2}\) = 19.9 cm
Use the formula for the volume of a sphere:
V = \(\frac{4}{3}\)πr³ ≈ \(\frac{4}{3}\) · 3.14 · 19.9³
Multiply:
= 32,993.4411….
Round to the nearest tenth:
≈ 32,993.4 cm³
= 32,993.4 cm³

Problem Solving

Find the value of each variable. (Chapter 6)

Question 31.

Answer:
x°=37°
2x=74°

Explanation:

A 360-degree angle is a complete angle of a point.
∠ABC = 360°-291°
∠ABC = 69°
∠ABC+BCA+CAB = 180°
69°+x°+2x°=180°
3x°+69=180°
3x°=180-69
3x°=111
x°=111÷3
x°=37°
2x°=74°

Question 32.

Answer:

Explanation:

∠DFE = 180°-112°
∠DFE = 68°
y°+3y°+ 68°= 180°
4y°+68° = 180°
4y°= 180°- 68°
4y°=112
y°=112÷4
y°= 28°
3y°= 84°

Question 33.

Answer:
Alternate exterior ∠s:
m∠HIN = 132°
VerticaL ∠
m∠HIN = 3b° + b°
Substitute and simplify:
132 = 4b
Divide by 4:
\(\frac{132}{4}\) = \(\frac{4b}{4}\)
Simplify
b = 33
Straight ∠:
3a° + 3b° + b° = 180°
Substitute and simplify:
3a ° + 4b° = 180°
3a + 4 · 33 = 180
3a + 132 = 180
Subtract 132 from both sides:
3a + 132 – 132 = 180 – 132
Simplify
3a = 48
Divide by 3
\(\frac{3a}{3}\) = \(\frac{48}{3}\)
Simplify
a = 16

a = 16; b = 33

Question 34.

Answer:
35°

Explanation:

∠PTU = c°
∠PRS = 80°
∠PUR = 2c°
∠TPS = 5c°
The sum of angles around the point is 360°
Sum of ∠PTU+ ∠PRS + ∠PUR + ∠TPS = 360°
c°+80°+2c°+5c° = 360°
8c° + 80° = 360°
8c° = 360°- 80°
8c° = 280°
c° = 280÷8
c°= 35°
2c°= 70°
5c°= 175°

Solve. Show your Work.

Question 35.
Segment AB is parallel to segments CD and EF. Segment BC is parallel to segment ED. Find the measure of ∠ABC. (Chapter 6)
Answer:

Question 36.
The diagram shows a triangular pin. Triangle ABC is an isosceles triangle. Line BD is parallel to line EF. m∠AEF = 40° and m∠DBC = 27°.
Find the measures of ∠EBD and ∠BAC. (Chapter 6)

Answer:
We are given:

Corresponding ∠s:
m∠EBD = 40°
Determine m∠ABC:
m∠ABC – m∠EBD + m∠DBC
= 40° – 27°
= 67°
△ABC isosceles:
m∠ACB = m∠ABC = 67°
Sum of ∠s in a triangle:
m∠BAC + m∠ACB + m∠ABC = 180°
Substitute:
m∠BAC + 67° + 67° = 180°
Simplify:
m∠BAC + 134° = 180°
We subtract 134° from both sides:
m∠BAC + 134° – 134° = 180° – 134°
Simplify:
m∠BAC = 46°

m∠EBD = 40°
m∠BAC = 46°

Question 37.
X, Y, and Zare the location of three manufacturing factories linked by three straight roads as shown in the diagram. A water pipe will go from Factory Y, and will be equidistant from the two roads linking Factory X and Factory Z to Factory Y. (Chapter 7)
a) Construct triangle XYZ to represent the locations of the three factories using a scale of 1 centimeter to 100 meters.
Answer:

b) Using a compass and straightedge only, construct the path of the water pipe.
Answer:
The Line whose points are equidistant from \(\overline{Y^{\prime} X^{\prime}}\) and \(\overline{Y^{\prime} Z^{\prime}}\) is the angle bisector of ∠Y’. We construct it:

c) The water department needs to build a temporary construction worksite that must be equidistant from Factory Y and Factory Z and at a distance of 400 meters from Factory X. Label the position of the construction worksite to be set up in your diagram as point W.

Answer:
The tine whose points are equidistant from Y’ and Z’ is the perpendicular bisector of \(\overline{Y^{\prime} Z^{\prime}}\). We construct it and label by W its intersection with the angle bisector from part b):

Question 38.
A model of a gym uses a scale of 1 : 12. The actual gym will be a rectangular prism. It will be 69.6 feet long, 54 feet wide, and 22.8 feet high. Find, in cubic feet, the volume of the model of the gym. (Chapters 7, 8)
Answer:
85691 cubic feet

Explanation:
The actual gym is a rectangular prism.
Given the length of a rectangular prism is 69.6 feet
Width = 54 feet
Height = 22.8 feet
The voulme of a rectangular prism is = Length × Width × Height
Volume = 69.6 × 54 × 22.8
Volume = 85691 cubic feet

Question 39.
Mrs. Sullivan buys a 5-liter bottle of apple juice. She can either use cylindrical glass A or cylindrical glass B to serve the apple juice. Use 3.14 as an approximation for π. (1,000 cm³ = 1 L) (Chapter 8)

We are given:
d_A = 6
h_A = 9
d_B = 5
h_B = 12

a) What is the volume of glass A?
Answer:
Determine the radius of glass A:
r_A = \(\frac{d_{A}}{2}\) = \(\frac{6}{2}\) = 3 cm
Determine the volume of glass A
V_A = πr_A²h_A ≈ 3.14 · 3² · 9 = 254.34 cm³

b) How many glasses of apple juice can she serve if she uses glass A?
Answer:
Determine the number of glasses A:
\(\frac{5 \cdot 1,000}{254.34}\) = \(\frac{5,000}{254.34}\) ≈ 19 glasses

c) What is the volume of glass B?
Answer:
Determine the radius of glass B:
r_B = \(\frac{d_{B}}{2}\) = \(\frac{5}{2}\) = 2.5 cm
Determine the volume of glass B
V_B = πr_B²h_B ≈ 3.14 · 2.5² · 12 = 235.5 cm³

d) How many glasses of apple juice can she serve if she uses glass B?
Answer:
Determine the number of glasses B:
\(\frac{5 \cdot 1,000}{235.5}\) = \(\frac{5,000}{235.5}\) ≈ 21 glasses

e) If Mrs. Sullivan sells one glass of apple juice for $1.80, what is the maximum amount of money she can collect?
Answer:
21 · 1.80 = 37.80

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Review Test Answer Key

Concepts and Skills

For this review, r represents radius and h represents height. You may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth unless otherwise stated.

Question 1.
Find the volume of each cylinder to the nearest unit. Use the given dimensions,
a) r = 4.2 inches; h = 14 inches
Answer:
775 cubic units

Explanation:
The volume of the cylinder is =π r² h
π = 3.14, r = 4.2, h= 14 inches
volume of the cylinder = 3.14×4.2×4.2×14 = 775 cubic units

b) r = 7 centimeters; h = 12 centimeters
Answer:
1846 cubic units

Explanation:|
The volume of the cylinder is =π r² h
π = 3.14, r = 7, h= 12 centimeters
volume of the cylinder = 3.14×7×7×12 = 1846 cubic units

Question 2.
Find the volume of each cone to the nearest unit. Use the given dimensions.
a) r = 3 centimeters; h = 8 centimeters
Answer:
75 cm³.

Explanation:
The volume of each cone is 1/3hπr²
r = 3 centimeters; h = 8 centimeters
Therefore the volume of each cone is = 1/3×8×3.14×3×3 = 75 cm³.

b) r = 8 inches; h = 15 inches
Answer:
1004 cm³.

Explanation:
The volume of each cone is 1/3hπr².
Given r = 8 inches; h = 15 inches and π=3.14
Therefore the volume of each cone is = 1/3×8×3.14×15 ×15 = 1004 cm³.

Question 3.
Find the volume of each pyramid. Use the given dimensions.

a) A square base with an edge length of 6 centimeters; h = 4 centimeters.
Answer:
We are given the cone:
l = 6
h = 4
Use the formula for the volume of a pyramid:
V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)l²h
Substitute for l, h:
= \(\frac{1}{3}\) · 6² · 4
Multiply:
= 48 cm³

b) A rectangular base with length of 6 inches and width = 3.3 inches; h = 7 inches.
Answer:
We are given the cone:
l = 6
w = 3.3
h = 7
Use the formula for the volume of a pyramid:
V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)lwh
Substitute for l, w, h:
= \(\frac{1}{3}\) · 6 · 3.3 · 7
Multiply:
= 46.2 in³

Question 4.
Find the volume of each sphere to the nearest unit. Use the given dimensions,
a) r = 9.6 centimeters
Answer:
3705 cm³

Explanation:
The volume of each sphere is = 4/3 πr³
Given r = 9.6 centimeters
Volume of the sphere is = 4/3×3.14×9.6×9.6×9.6 = 3705 cm³.

b) d = 26 centimeters
Answer:
9202 cm³.

Explanation:
The volume of each sphere is = 4/3 πr³
Given diameter is = 13 cm
Here we need a radius to calculate the volume of the sphere hence r = d/2 = 26/2 = 13 cm
Volume of the sphere is = 4/3×3.14×13×13×13=9202 cm³.

Question 5.
Find the exact surface area of each solid.

Answer:
a) We are given the cylinder:
r = 3
h = 6
Use the formula for the surface area of a cylinder.
S = 2πr² + 2πrh
Substitute for r, h:
= 2π · 3² + 2π · 3 · 6
Multiply:
= 18π + 36π
Add:
= 54π ft²

b) We are given the sphere:
d = 28
Determine the radius of the sphere:
r = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14
Use the formula for the surface area of a sphere:
S = 4πr²
Substitute for r:
= 4π · 14²
Multiply:
= 784π m²

c) We are given the cone:
r = 6
l = 10
Use the formula for the surface area of a cone:
S = πr² + πrl
Substitute for r, l:
= π · 6² + π · 6 · 10
Multiply:
= 36π + 60π
Add:
= 96π in²

Question 6.
Find the volume and surface area of each solid. Round to the nearest tenth.
a) A solid cone with a diameter of 5 feet, a slant height of 7 feet, and a height of 6.5 feet.
Answer:
We are given the cone.
d = 5
l = 7
h = 6.5
Find the radius:
r = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5
Use the formula for the volume of the cone:
V = \(\frac{1}{3}\) πr²h
≈ \(\frac{1}{3}\) · 3.14 · 2.5² · 6.5
≈ 42.5 ft³
Use the formula for the surface area of the cone:
S = πr² + πrl
= π · 2.5² + π · 2.5 · 7
= 6.25π + 17.5π
= 23.75π
≈ 23.75 · 3.14
≈ 74.6 ft²

b) A sphere with a radius of 28 millimeters.
Answer:
We are given the sphere:
r = 28
Use the formula for the volume of the sphere:
V = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) · 3.14 · 28³
≈ 91,905.7 mm³
Use the formula for the surface area of the sphere:
S = 4πr² ≈ 4 · 3.14 · 28²
≈ 9,487 mm²

c) A solid cylinder with a radius of 1.4 inches and a height of 4.2 inches.
Answer:
We are given the cylinder:
r = 1.4
h = 4.2
Use the formula for the volume of the cylinder:
V = πr²h ≈ 3.14 · 1.4² · 4.2
≈ 25.8 in³
Use the formula for the surface area of the cylinder:
S = 2πr² + 2πrh
= 2π · 1.4² + 2π · 1.4 · 4.2
= 3.92π + 11.76π
= 15.68π
≈ 15.68 · 3.14
≈ 49.2 in²

Problem Solving

Solve. Show your work.

Question 7.
The volume of a cone is 450 cubic centimeters and the radius of the base is 5 centimeters. What is the height of the cone to the nearest tenth?
Answer:
We are given:
V = 450
r = 5
Use the formula for the volume of the sphere:
V = \(\frac{1}{3}\) · πr² · h
Substitute for V, πn r:
450 = \(\frac{1}{3}\) · 3.14 · 5² · h
Multiply:
450 = \(\frac{78.5}{3}\) · h
Multiply:
450 = \(\frac{78.5}{3}\) · h
Multiply both sides by 3:
3 · 450 = 3 · \(\frac{78.5}{3}\) · h
Multiply:
1,350 = 78.5h
Divide both sides by 78.5
\(\frac{1,350}{78.5}\) = \(\frac{78.5h}{78.5}\)
Simplify
h ≈ 17.2 cm

Question 8.
The surface area of a sphere is 498.96 square centimeters. What is the radius of the sphere to the nearest tenth?
Answer:
We are given the sphere:
S = 498.96
Use the formula for the surface area of a sphere:
S = 4πr²
Substitute S and π:
498.96 = 4 · 3.14 · r²
Multiply:
498.96 = 12.56 · r²
Divide each side by 12.56
\(\frac{498.96}{12.56}=\frac{12.56 r^{2}}{12.56}{78.5}\)
Evaluate:
39.7261 ≈ r²
Find the square root of each side:
\(\sqrt[2]{39.7261}\) ≈ r²
Round to the nearest tenth:
r ≈ 6.3 cm

Question 9.
A cone with a height of 6 inches and a slant height of 7.5 inches has a lateral surface with an area of approximately 106 square inches.

a) What is the radius? Round to the nearest tenth.
Answer:
We are given:
h = 6
l = 7.5
A = 106
Use the formula for the cone’s lateral surface area:
A = πrl
Substitute A, l, π to find r:
106 = 3.14 · r · 7.5
Multiply:
106 = 23.55r
Divide both sides by 23.55
\(\frac{106}{23.55}=\frac{23.55 r}{23.55}\)
r ≈ 4.5 in

b) What is the volume of the cone? Round to the nearest tenth.
Answer:
Use the formula for the cone’s volume:
V = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) · 3.14 · 4.5² · 6
≈ 127.2 in³

Question 10.
A cylinder, a cone, and a sphere are shown below. Each solid has a radius of 1 inch and a height of 2 inches. Which of them has the greatest volume? Justify your answer.

Answer:
We are given:
r = 1
h = 2
Determine the cylinder’s volume:
V_cylinder = πr²h = π · 1² · 2 = 2π
Determine the cone’s volume:
V_cone = \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) · π · 1² · 2
= \(\frac{2 \pi}{3}\)
Determine the sphere’s volume:
V_sphere = \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) · π · 1³
= \(\frac{4 \pi}{3}\)
We have:
\(\frac{2 \pi}{3}\) < \(\frac{4 \pi}{3}\) < 2π
V_cone < V_sphere < V_cylinder
The greatest volume is the cylinder’s

Question 11.
Two metal cubes each have an edge length of 4 centimeters. They are melted and recast into a square pyramid with a height of 5 centimeters. Find the area of the base of the pyramid.
Answer:
Determine the volume of the two cubes:
V = 2 · 4³ = 2 · 64 = 128 cm³
Use the formula of a pyramids volume:
V = \(\frac{1}{3}\) Bh
Substitute for V, h:
128 = \(\frac{1}{3}\) · B · 5
Multiply both sides by 3:
3 · 128 = 3 · \(\frac{5B}{3}\)
384 = 5B
Divide by 5:
\(\frac{384}{5}\) = \(\frac{5B}{5}\)
B = 76.8 cm²

Question 12.
A conical party hat has a diameter of 7 centimeters and a slant height of 14.5 centimeters. Paul wants to wrap the lateral surface of the party hat with plastic wrap. How much plastic wrap will he use? Round to the nearest square centimeter.
Answer:
We are given the cone:
d = 7
l = 14.5
Determine the radius of the cone:
r = \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5
Determine the lateral surface of the cone:
A = πrl ≈ 3.14 · 3.5 · 14.5 ≈ 159 cm²
= 159 cm²

Question 13.
The composite solid shown is made up of a hemisphere attached to the top of a cube. The diameter of the hemisphere is the same as the edge length of the cube. Find the volume of the composite solid. Round to the nearest tenth.
Answer:

Determine the radius of the hemisphere:
r = \(\frac{9}{2}\) = 4.5
Determine the volume of the hemisphere:
V_hemisphere = \(\frac{1}{2}\) · \(\frac{4}{3}\) πr³
≈ \(\frac{2}{3}\) · 3.14 · 4.5³
≈ 190.8 cm³
Determine the volume of the cube:
V_cube = 9³ = 729 cm³
Determine the volume of the composite solid:
V_solid = V_hemisphere + V_cube
= 190.8 + 729
= 919.8 cm³

Question 14.
The circumference of a fully inflated beach ball is 12π, or about 37.68 inches. What is the radius of the beach ball? What is the volume of
the beach ball? Round to the nearest tenth if necessary.
Answer:
The radius of the beach ball is 6 inches.
The volume of the beach ball is 151 inches.

Explanation:
The circumference of a fully inflated beach ball is 12π.
Circumference of a circle = 2πr
12π=2πr
r=6
Hence radius of the beach ball is 6 inches.
Now we need to calculate the volume of a beach ball
The volume of the sphere is 4/3 πr³
Volume = 4/3×3.14×6×6×6
The volume of the beach ball = 904 inches.

Question 15.
A pyramid made of clay has a square base of length 8 inches and a height of 12 inches. The pyramid is reshaped into a cylinder with a radius of 8 inches. What is the height of the cylinder? Round to the nearest inch
Answer:
The height of the cylinder is 1.27 inches

Explanation:
Given: The base(b) of the square pyramid as 8 inches length and height(h) of the square pyramid as 12 inches.
Now we need to calculate Volume of square pyramid to know the height of cylinder.
The volume of the square pyramid is given by = 1/3(b)²h
Volume = 1/3 (8×8)²×12
Volume = 1/3 (64)×12
Volume = 256 inches
Now pyramid is reshaped into a cylinder with a radius = 8 inches
Volume of cylinder is π r² h
Volume = 3.14×8×8×h
Volume is 256 inches
256 = 3.14×8×8×h
h = 1.27 inches