Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability

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HMH Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability

I Can find and use range, IQR, and MAD to summarize a data set.

Step It Out

One measure of variability is the mean absolute deviation.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 1
Question 1.
The ages, in years, of two teams of cyclists at the Tour de France are shown.
Team 1: 29, 32, 33, 26, 22, 32, 22, 23, 24
Team 2: 27, 33, 31, 29, 26, 30, 29, 33, 32

A. Remember that a measure of center of a data set summarizes its values with a single number. Find the mean age of each team.
Team 1 mean: Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 years
Team 2 mean: Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 years
Answer:
Team 1 Mean = 27 years
Team 2 Mean = 30 years

Explanation:
Given,
Mean = Sum of terms ÷ no of terms
Given terms in ages are 29, 32, 33, 26, 22, 32, 22, 23, 24
Now arrange them in ascending to descending order
22, 22, 23, 24, 26, 29, 32, 32, 33
Sum of terms = 22 + 22 + 23 + 24 + 26 + 29 + 32 + 32 + 33
= 243
No of terms = 9
Mean = 243 ÷ 9 = 27
Mean of Team 1 is 27.
Now find the mean for Team 2
Given terms are 27, 33, 31, 29, 26, 30, 29, 33, 32
Arranage them in order
26, 27, 29, 29, 30, 31, 32, 33, 33,
Sum of terms = 26 + 27 + 29 + 29 + 30 + 31 + 32 + 33 + 33 = 270
No of terms = 9
Mean = Sum of terms ÷ No of terms
Mean = 270 ÷ 9
= 30
Team 2 Mean is 30 years.

B. Find the mean absolute deviation (MAD) for each team.
Team 1 MAD: Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 years
Team 2 mean: Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 years
Answer:
Team 1 MAD is 4 years.
Team 2 MAD is 2 years.

Explanation:
The mean absolute deviation is the average distance of all the terms in a data set from the mean of the same data set.
First, find the mean of the data set.
Add all the terms in the data set for Team 1
The mean for Team 1 is 27.
Now calculate how each term is away from the mean. And use the absolute value because the distance is always positive.
Team 1: 29, 32, 33, 26, 22, 32, 22, 23, 24
29 – 27 = |2| = 2
32 – 27 = |5| = 5
33 -27 = |6| = 6
26 – 27 = |1| = 1
22 – 27 = |-5| = 5
32 – 27 = |5| = 5
22 – 27 = |-5| = 5
23 – 27 = |-4| = 4
24 – 27 = | -3| = 3
2 + 5 + 6 + 1 + 5 + 5 + 5 + 4 + 3 = 36
And 9 is the number of terms in the data set.
Mean absolute Deviation is = 36 ÷ 9
MAD = 4
Next Find MAD for Team 2
Team 2: 27, 33, 31, 29, 26, 30, 29, 33, 32
Mean for Team 2 is 30 years
27 – 30 = |-3| = 3
33 – 30 = |-3| = 3
31 – 30 = |-1| = 1
29 – 30 = |-1| = 1
26 – 30 = |-4| = 4
30 – 30 = |0| = 0
29 – 30 = |-1| = 1
33 – 30 = |-3| = 3
32 – 30 = |-2| = 2
Total = 18
MAD = 18 ÷ 9 =2

C. Compare the MAD of Team 1’s ages to the MAD of Team 2’s ages. What does the MADs tell you about the two team’s ages? Explain.
Answer:
Team 1’s age is 4 years and Team 2’s age is 2 years. By comparing the two team’s ages Team 1’s age is greater than Team 2’s age.

D. If the distance between a cyclist’s age and the team’s mean age is less than or equal to the MAD, that cyclist’s age falls within the MAD. For Team 1, list the ages that fall within the MAD.
Answer:
The ages that fall within the MAD are 29, 26, 23, and 24.

Turn and Talk What is the difference between a measure of center, such as the mean, and a measure of variability, such as the mean absolute deviation?
Answer:
Mean is the sum of all the given values divided by the total number of values given.|
The mean absolute deviation is the average distance between each data value and the mean.

Other measures of variability are the range and interquartile range.
Connect to Vocabulary
The range is the difference between the greatest and least values in a data set.
The interquartile range (IQR) is the difference between the upper and lower quartiles in a data set.

Question 2.
The prices for Wi-Fi routers at two stores are shown in the box plots. Find and compare the ranges and interquartile ranges of the prices at each store.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 3

A. To find the range for each box plot, find the difference between the greatest and least values.
Store A
300 – 120 = Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2
The range for store A is $ Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2.
Answer:
$180

Explanation:
The range for store A is $180.
300 – 120 = $180.

Store B
250 – Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 = Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2
The range for store B is $ Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2.
Answer:
Store B
250 – 100 = 150
The range for store B is $150.

B. To find the IQR for each box plot, find the difference between the upper and lower quartiles.
Store A
230 – 180 = Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2
The IQR for store A is $ Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2.

Store B
200 – Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2 = Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2
The IQR for store B is $ Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 2.
Answer:
The IQR for store A is $50.
The IQR for store B is $70.

Explanation:
Given,
The Upper Quartile for Store A is 230 and the Lower quartile for store A is $180.
Then 230 – 180 = $50.
The upper quartile for store B is $200 and the lower quartile for store B is $ 130.
Then IQR 200 – 130 = $70

C. Compare the range of the prices at each store and the interquartile range of the prices at each store.
Store A has a greater ____________, but Store B has a greater _____________.
Answer:
Store A has a greater Range, but Store B has a greater Interquartile range.

Turn and Talk Can you tell the exact price of any of the routers at the two stores? If so, explain how.
Answer:

Question 3.
The number of visitors each day last week at a train exhibit hosted by a local museum were: 29, 3, 45, 33, 30, 38, and 25 visitors.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 4
A. The range of the data is ___________ visitors.
Answer:
42

Explanation:
Range = Difference between greatest and lowest values
The greatest value is 45 and the lowest value is 3
Range = 45 -3
= 42

B. The interquartile range (IQR) of the data is __________ visitors.
Answer:
IQR = 38 -25
= 13

Explanation:
Interquartile range = Difference between upper and lower quartile.
Interquartile range = 38 – 25
= 13

C. Looking at the shape of the data, which measure of variability, range or IQR, best represents this data? Explain.
Answer:
Interquartile range is the best measure of variability for data sets with outliers because it is based on the values that come from the middle half of the distribution set.

Check Understanding

Question 1.
A new swimming club has just started. The ages, in years, of its members, are 39, 27, 51, 42, 33, 73, 49, and 46. What is the mean absolute deviation of the ages of the club members? Which ages fall within the MAD?
Answer:
MAD = 9.75.
The ages that fall within the MAD is 27, 33, and 73.

Explanation:
Given,
The ages, in years, of its members, are 39, 27, 51, 42, 33, 73, 49, and 46
Sum of terms = 39 + 27 + 51 +42 + 33 + 73 + 49 + 46
= 360
No of terms = 8
Mean = 360 ÷ 8
= 45
39 – 45 = |-6| = 6
27 – 45 = |-18| = 18
51 – 45 = |6| = 6
42 – 45 = |-3| = 3
33 – 45 = |-12| = 12
73 – 45 = |28| = 28
49 – 45 = |4| = 4
46 – 45 = |1| = 1
Total terms = 6 + 18 + 6 + 3 + 12 + 28 + 4 + 1
= 78
MAD = 78 ÷ 8
= 9.75

Question 2.
What are the range and interquartile range of the data set in the box plot? Why might the range be wide, but the IQR narrow?
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 5
Answer:
Range = 85
IQR = 15

Explanation:
The difference between upper and lower values is called range.
Upper value = 95 and lowe value = 10
Range = 95 -10 = 85
Interquartile range = Upper quartile – lower quartile
Upper Quartile = 50 lower Quartile = 35
IQR = 50 – 35 = 15

Question 3.
Looking at the shape of the data, compare the range and IQR of the data set shown in the box plot with the data set in Problem 2.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 6
Answer:
Range = 55
IQR = 30

Explanation:
Upper value = 100 and lower value = 45
Range = 100 – 45
= 55
IQR = 85 – 55
= 30
By comparing the range and IQR with the problem no 2, the range of problem 2 is greater but IQR is greater than problem 2.

On Your Own

For Problems 4-7, use the box plot. It summarizes the number of goals per game for one team in a field hockey tournament.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 7

Question 4.
The range of the data is __________ goals.
The interquartile range of the data is ___________ goals.
Answer:
Range = 10
Interquartile range = 3

Explanation:
The range of the data is 10 – 0 = 10 goals.
The interquartile range of the data is 4 – 1 = 3 goals.

Question 5.
Looking at the shape of the data, which measure of variability, range or IQR, best represents this data? Explain.
Answer:
Interquartile range is the best measure of variability than range.

Question 6.
Name one measure of variability for the data and give its value. What does this measure of variability tell you about the goals per game?
Answer:

Question 7.
Reason Can you find the mean absolute deviation of the data set using the box plot? Explain.
Answer:
The mean absolute deviation of the data set using the box plot is 2.88

Explanation:
Given data set is 0, 1, 2, 4, 10.
Mean = (0 + 1 + 2 + 4 +10) ÷  5
Mean  = 17 ÷  5
= 3.4
0 – 3.4 = |-3.4| = 3.4
1 – 3.4 = |-2.4| = 2.4
2 – 3.4 = |-1.4| = 1.4
4 – 3.4 = |0.6| = 0.6
10 – 3.4 = |6.6| = 6.6
3.4 + 2.4 +1.4 + 0.6 + 6.6 = 14.4
MAD = 14.4 ÷  5
MAD = 2.88

Find the range and interquartile range (IQR) for each box plot.

Question 8.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 8
Answer:
Range = 17
IQR = 16

Explanation:
Range = 32 – 5 = 17
IQR = 30 – 14 = 16

Question 9.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 9
Answer:
Range = 470
IQR = 150

Explanation:
Given,
Upper value = 790 and lower value = 320
Range = 790 – 320
= 470
Interquartile range = upper quartile  – lower quartile
= 630 – 480
= 150

Question 10.
The box plots summarize data about the ages of the people who live in two counties.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 10
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 11
A. What are the range and interquartile range of the population in County A?
Answer:
Range for county A = 105
Interquartile range for county A = 44

Explanation:
Range = 105 – 0 = 105
Interquartile range = 58 – 14 = 44

B. What are the range and interquartile range of the population in County B?
Answer:
Range of the population in County B is 109.
Interquartile range in county B is 40 – 13 = 27

Explanation:
Range = 109 – 0 =109
Interquartile range = 40 – 13 = 27.

C. How do the range and interquartile range of the population in the two counties compare?
Answer:
The range of the population in County A is less than the range in county B.
The interquartile range in County A is greater than the interquartile range in county B.

D. Reason Which data set is likely to have the greater mean absolute deviation? Explain.
Answer:
The data set of County A have the greater mean absolute deviation.

Explanation:
County A = 0, 14, 32, 58 and 105
County B = 0, 13, 26, 40 and 109.
Mean of county A = 209 ÷ 5
Mean = 41.8
Mean of county B = 188 ÷  5
Mean = 37.6
The data set of County A have the greater mean absolute deviation.

For Problems 11-12, find and compare the range and IQR between the two box plots.

Question 11.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 12
Answer:
Range for store A = 1.5
IQR for store A = 0.9
Range for store B = 0.8
IQR for store B = 0.35

Explanation:
Given for store A
Upper value = 3.09 and lower value = 1.59
Range = 3.09 – 1.59
= 1.5
Upper quartile = 2.79 and lower quartile = 1.89
IQR = 2.79 – 1.89
= 0.9
Given for store B
Upper value = 2.79 and lower value = 1.99
Range = 2.79 – 1.99
= 0.8
Upper quartile = 2.49 and lower quartile = 2.14
IQR = 2.49 – 2.14
= 0.35
So by comparing the two box plots the range of Store A is greater than the range of store B.
The IQR of Store A is greater than the IQR of store B.

Question 12.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 13
Answer:
Range of class A = 39
IQR of class A = 22
Range of Class B = 50
IQR of Class B = 20

Explanation:
Range of class A = 98 – 59 = 39
IQR of Class A = 94 – 72 = 22
Range of Class B = 96 – 46 = 50
IQR of class B = 89 – 69 = 20
So by comparing the two box plots the range of class B is greater than the range of Class A.
IQR of class A is greater than the IQR of class B.

Question 13.
The points scored by a youth basketball team in each of their games so far this season are 43, 30, 34, 26, 46, 47, 48, 37, 29, and 40.
A. The mean number of points per game is ___________ points.
The mean absolute deviation of points per game is __________ points.
Answer:
Given,
43, 30, 34, 26, 46, 47, 48, 37, 29, and 40.
Sum of terms = 43 + 30 + 34 + 26 + 46 + 47 + 48 + 37 + 29 + 40
= 380
No of terms = 10
Mean = Sum of terms ÷ No of terms
= 380 ÷ 10
= 38
43 – 38 = |5| = 5
30 – 38 = |-8| = 8
34 – 38 = |-4| = 4
26 – 38 = |-8| = 8
46 – 38 = |8| = 8
47 – 38 = |9| = 9
48 – 38 = |10| = 10
37 – 38 = |-1| = 1
29 – 38 = |-11| = 11
40 – 38 = |2| = 2
Total terms =  5 + 8 + 4 + 8 + 8 + 9 + 10 + 1 + 11 + 2
= 66
MAD = 66 ÷ 10
= 6.6

B. The mean is a measure of ___________. The mean absolute deviation is a measure of ____________.
Answer:
The mean is a measure of central tendancy.
And the mean absolute deviation is a measure of variability.

C. Which scores fall within the mean absolute deviation? Which scores are outside of the mean absolute deviation?
Answer:
The scores fall within the mean absolute deviation is 43, 34, 37 and 40.
The scores are outside of the mean absolute deviation is 30, 26, 46, 47, 48, and 29.

D. The team plays two more games and scores 15 points in one and 60 points in the other. How does this affect the mean absolute deviation?
Answer:
Given,
The team plays two more games and scores 15 points and 60 points.
15 – 38 = |-23| = 23
60 – 38 = |22| = 22
Total terms =  5 + 8 + 4 + 8 + 8 + 9 + 10 + 1 + 11 + 2 + 23 + 22
= 111 ÷ 12
= 9.25
The mean absolute deviation is 2.65 more than the previous game.

Question 14.
The table shows the ages of members of the city council. What is the MAD of their ages? What does this say about the age of the typical member?
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 14
Answer:
Given the age of council members
34, 49, 72, 53, 46, 58, 41, 63
Sum of terms = 34 + 49+ 72 + 53 + 46 + 58 + 41 + 63
= 416
No of terms = 8
Means = sum of terms ÷ no of terms
= 416 ÷ 8
= 52
34 – 52 = |-18| = 18
49 – 52 =|-3| = 3
72 – 52 = |20| = 20
53 – 52 =|1| = 1
46 – 52 = |-6| = 6
58 – 52 = |6| = 6
41 -52 = |11| = 11
63 – 52 = |11| = 11
18 + 3 + 20 + 1 + 6 + 6 + 11 + 11 = 76
No of terms = 8
Mean absolute deviation = 76 ÷ 8
= 9.5

Question 15.
Open-Ended Write a set of data with 10 data points that could have the box plot. Then find the range, interquartile range, and mean absolute deviation of the data set.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 15
Answer:
Range = 4
InterQuartile range = 2
Mean absolute deviation = 1.2

Explanation:
Given,
Sum of terms = 2 + 3 + 4 + 5 + 6
= 20
No of terms = 5
Mean = sum of terms ÷ no of terms
= 20 ÷ 5
= 4
The given data set is 2, 3, 4, 5, 6
2 – 4 =|-2| = 2
3 -4 = |-1| = 1
4 – 4 = |0| = 0
5 – 4 = |1| = 1
6 – 4 = |2| = 2
Now add all the terms 2 + 1 + 0 + 1 + 2 = 6
Mean absolute deviation = 6 ÷ 5
= 1.2

For Problems 16-17, find the mean and the MAD. Circle the data points that fall within the mean absolute deviation of each data set.

Question 16.
45, 85, 70, 25, 80, 95, 75, 85, 90, 80
Mean = _________
MAD = __________
Answer:
Mean = 73
MAD = 15.4

Explanation:
Given,
No of terms = 45, 85, 70, 25, 80, 95, 75, 85, 90, 80
sum of terms = 45 + 85 + 70 + 25 + 80 + 95 + 75 + 85 + 90 + 80
= 730
No of terms = 10
Mean = sum of terms ÷no of terms
= 730 ÷ 10
= 73
Now we need to find MAD
45 – 73 = |-28| = 28
85 – 73 = |12| = 12
70 – 73 = |-3| = 3
25 – 73 = |-48| = 48
80 – 73 = |3| = 3
95 – 73 = |22| = 22
75 – 73 = |2| = 2
85 – 73 = |12| = 12
90 – 73 = |17| = 17
80 – 73 = |7| = 7
28 + 12 + 3 + 48 + 3 + 22 + 2 + 12 + 17 + 7 =154
MAD = 154 ÷ 10
= 15.4

Question 17.
3, 5, 2, 6, 4, 5, 4, 6, 2, 6
Mean = _________
MAD = __________
Answer:
Mean = 4.3
MAD = 1.3

Explanation:
Given,
data set 3, 5, 2, 6, 4, 5, 4, 6, 2, 6
Sum of terms = 3 + 5 + 2 + 6 + 4 + 5 + 4 + 6 + 2 + 6
= 43
No of terms = 10
Mean = 43 ÷ 10
= 4.3
3 – 4.3 = |1.3| = 1.3
5 – 4.3 = |-0.7| = 0.7
2 – 4.3 = |-2.3| = 2.3
6 – 4.3 = |1.7| = 1.7
4 – 4.3 = |-0.3| = 0.3
5 – 4.3 = |0.7| = 0.7
4 – 4.3 = |-0.3| = 0.3
6 – 4.3 = |1.7| = 1.7
2 – 4.3 = |-2.3| = 2.3
6 – 4.3 = |1.7| = 1.7
1.3 + 0.7 + 2.3 + 1.7 + 0.3 + 0.7 + 0.3 + 1.7 + 2.3 + 1.7 = 13
MAD = 13 ÷ 10
= 1.3

Lesson 16.4 More Practice/Homework

Question 1.
Darren counts the number of devices that are plugged into the wall in each room of his house and records them in a list: 1, 2, 0, 6, 5, 3, 4, 3. Which values fall outside the mean absolute deviation?
Answer:
MAD is 1.5
The values fall outside the mean absolute deviation is 1, 0, and 6.

Explanation:
Given list is 1, 2, 0, 6, 5, 3, 4, 3.
First calculate the mean
sum of terms = 1 + 2 + 0 + 6 + 5 + 3 + 4 + 3 =24
No of terms = 8
Mean = sum of terms ÷ no of terms
Mean = 24 ÷ 8
= 3
1 – 3 = |-2| = 2
2 – 3 = |-1| = 1
0 – 3 = |-3| = 3
6 – 3 = |-3| = 3
5 – 3 = |-2| = 2
3 – 3 = |0| = 0
4 – 3 = |1| = 1
3 – 3 = |0| = 0
Sum of the terms given  = 2 + 1 + 3 + 3 + 2 + 0 + 1 + 0 = 12
MAD = 12 ÷ 8
= 1.5
The values fall outside the mean absolute deviation is 1, 0, and 6.

Question 2.
During a gray September, the amount of rainfall in millimeters per day over six days is 1.6, 1.3, 14.2, 11.7, 2.8, 0. 8. Find the range and interquartile range of the amount of rainfall. Which is the best measure of variability to describe this data? Explain.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 16
Answer:
Range 13.4 and IQR is 1.6

Explanation:
Let us arrange the given data from lowest to highest
0.8, 1.3, 1.6, 2.8, 11.7, 14.2
Range = 14.2 – 0.8
= 13.4
IQR = 1.6
The best measure of variability is IQR because it is based on the values that come from the middle half of the distribution set.

Question 3.
The ages of people who competed in a road race last weekend are shown in the box plots. Interpret the differences in the range and the IQR for both groups to come to a conclusion about the runners.
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 17
Answer:
Range for male runners is 73 – 9 = 64
IQR for male runners is 37
Range for female runners is 82 – 10 = 72
IQR for female runners is 35.

Explanation:
The range is the difference between the highest and lowest values
So the range for male runners is 73 – 9 = 64
IQR is the middle half of the given set
IQR for Male runners is 37
Same for female runners
Range = 82 – 10 = 72
IQR = 35.

Question 4.
Math on the Spot Find the interquartile range for the data set.
Data set: 11, 5, 16, 20, 31, 27, 9, 15, 26
Answer:
Let us arrange the given data set.
5, 9, 11, 15, 16, 20, 26, 27, 31
The interquartile range is called the middle half of the distribution.
The interquartile range for the data set is 16.

Question 5.
Name a measure of variability and how it is calculated.
Answer:
Variability is commonly measured with the following statistics
The first one is Range
The range is called the difference between the highest and lowest values.
The next one is the interquartile range
The interquartile range is called the middle half of the distribution.
And the third one is the standard deviation
It is called the average distance from the mean.

Test Prep

Question 6.
The box plot shows the numbers of grams of protein in several brands of protein bars. What are the range and interquartile range of the data?
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 18
Range: _________ grams
Interquartile range: __________ grams
Answer:
Range = 44 grams
Interquartile range = 29 grams

Explanation:
The range is called the difference between the highest and lowest values
The highest value is 48 and the lowest value is 4
The difference between 48 – 4 = 44 grams
The interquartile range is the range of the middle half of the distribution.
The interquartile range is 29 grams.

Question 7.
What are the range and interquartile range for the data set?
Into Math Grade 6 Module 16 Lesson 4 Answer Key Explore Measures of Variability 19
Range = __________
IQR = ____________
Answer:
Range = 9.2 – 1.9 = 7.3
IQR = 5.1

Explanation:
The range is called the difference between the highest and lowest values
The highest value is 9.2 and the lowest value is 1.9
The difference between 9.2 – 1.9 = 7.3
The interquartile range is the range of the middle half of the distribution.
The interquartile range is 5.1.

Question 8.
A data set has a mean of 5.3 and a mean absolute deviation of 2. Which of the following points fall within the mean absolute deviation? Select all that apply.
(A) 2
(B) 3
(C) 5
(D) 6
(E) 7
(F) 9
Answer:
option B 3
option C 5
option D 6

Spiral Review

Question 9.
Juana and her friends have collected shells from the beach. The number of shells collected by each person is 4, 9, 7, 5, 12, 8, 6, 5. How many shells should each person get if they want to share them evenly?
Answer:
The shells each person gets they want to share evenly is 7.

Explanation:
The number of shells collected by each person is 4, 9, 7, 5, 12, 8, 6, 5.
4 + 9 + 7 + 5 + 12 + 8 + 6 + 5 = 56
There are 8 numbers given
Each person will get if they want to share them is 56 ÷ 8 = 7

Question 10.
A college student made videos showing his dorm life each week for two months. The number of weekly videos he recorded were 3, 9, 5, 2, 14, 6, 7, 5, and 8. What is the median number of videos he recorded each week?
Answer:
2, 3, 5, 5, 6, 7, 8, 8, 9 and 14
The median number of videos he recorded each week is 7.

Explanation:
Let us arrange the given number of videos
2, 3, 5, 5, 6, 7, 8, 8, 9 and 14
The median is the middle number of the sorted list of numbers
Hence the median number of videos recorded each week is 7.

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