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Chapter 13, Lesson 3: Measures of Variation

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Go Math Answer Key

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions

Students who are searching for 6th Grade chapter 13 solutions can get them here. Download Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions for free. We have provided the answers with clear cut explanations on HMH Go Math 6th Grade Chapter 13 Variability and Data Distributions.

Before you start your preparation checkout the topic provided below. The Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions helps the students to score good marks in exams. Click on the below links and download the solutions.

Chapter 13 – Lesson: 1

Share and Show – Page No. 709

Big cats – page no. 710, patterns in data – page no. 711, lesson check – page no. 712.

Chapter 13 – Lesson: 2

Share and Show – Page No. 715

  • Problem Solving + Applications – Page No. 716

Box Plots – Page No. 717

Lesson check – page no. 718.

Chapter 13 – Lesson: 3

Share and Show – Page No. 721

Problem solving + applications – page no. 722, mean absolute deviation – page no. 723, lesson check – page no. 724.

Chapter 13 – Lesson: 4

Share and Show – Page No. 727

Problem solving + applications – page no. 728, measures of variability – page no. 729, lesson check – page no. 730.

Chapter – 13 – Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 731

  • Mid-Chapter Checkpoint – Page No. 732

Chapter 13 – Lesson: 5

Share and Show – Page No. 735

Unlock the problem – page no. 736, choose appropriate measures of center and variability – page no. 737, lesson check – page no. 738, share and show – page no. 741, problem solving + applications – page no. 742, apply measures of center and variability – page no. 743, lesson check – page no. 744.

Chapter 13 – Lesson: 6

Share and Show – Page No. 747

Problem solving + applications – page no. 748, describe distributions – page no. 749, lesson check – page no. 750.

Chapter 13 – Lesson: 7

Share and Show – Page No. 753

On your own – page no. 754, problem solving misleading statistics – page no. 755, lesson check – page no. 756.

Chapter 13 – Review/Test

Chapter 13 Review/Test – Page No. 757

Chapter 13 review/test – page no. 758, chapter 13 review/test – page no. 759, chapter 13 review/test – page no. 760, chapter 13 review/test – page no. 761, chapter 13 review/test – page no. 762.

For 1–3, use the dot plot.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 1

Answer: Between the intervals of 4 – 7 excluding 4 and 7

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 1

Question 2. Identify any clusters in the data. Type below: _________________

Answer: 1-4

Explanation: A group of dots is called a cluster The dots form a cluster at 1 – 4

Question 3. Summarize the information in the dot plot. Type below: _________________

Answer: It says about the number of paintings done by each student in the art club.

Explanation: The number of paintings is represented by the number line. The dots represent the students. Therefore we can say that It says about the number of paintings done by each student in the art club.

On Your Own

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 2

Explanation: STEP 1 Identify any peaks in the data. The histogram has 6 peaks. The interval representing the greatest number of visitors is for ages between 60 and 69 age group. STEP 2 The data changes across the intervals. The number of visitors increases from 0 to 29 age group and from 40 to 69 age group. So, the data values increase to one peak in the interval from 0 to 9 age group and then decrease. The visitors of the age group 30 – 39 did not visit the zoo.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 3

Answer: 5a. Yes 5b. Yes 5c. No 5d. No 5e. No

Explanation: 5a. There are dots between 4-5 so we can say that there is a gap from 4 to 5. 5b. The number of dots is more at the interval 0 So we can say that there is a peak at 0. 5c. The symmetrical view is nothing but having the same number of dots on both sides of the figure but we cannot observe it in the above figure. Therefore we can say that the dot plot has no symmetry. 5d. The most frequently occurring observation is known as a mode. One dot repeats in all the intervals so we can say that the mode is 1. 5e. A group of observations form a cluster, there are more than 1 group of dots in the figure given above.

There are 41 species of cats living in the world today. Wild cats live in places as different as deserts and the cold forests of Siberia, and they come in many sizes. Siberian tigers may be as long as 9 feet and weigh over 2,000 pounds, while bobcats are often just 2 to 3 feet long and weigh between 15 and 30 pounds.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 4

Question 6. Look for a Pattern Describe the overall shape of the histogram. Type below: _________________

Answer: The graph starts from a small interval and increases to the highest and then decreases to the smallest interval. The histogram has rectangles that are closely packed.

Explanation: STEP 1 Identify any peaks in the data. The histogram has 1 peak(s). The interval representing the greatest number of bobcats is for weights between 18 and 20 pounds. STEP 2 Describe how the data changes across the intervals. The number of bobcats increases from 12 to 17 pounds and from 21 to 29 pounds. STEP 3 Describe any symmetry the graph has. If I draw a vertical line through the interval for 18 to 20 pounds, the left and right parts of the histogram are very close to being mirror images. The histogram has line symmetry.

So, the data values increase to one peak in the interval for 18 to 20 pounds and then decrease. The data set has a vertical line symmetry about the peak.

Question 7. Sense or Nonsense? Sunny says that the graph might have a different shape if it was redrawn as a bar graph with one bar for each number of pounds. Is Sunny’s statement sense or nonsense? Explain. Type below: _________________

Answer: Sense

Explanation: Bar graph also contains rectangles but they are not closely packed hence the statement is correct which is said by Sunny as a bar graph with one bar for each number of pounds.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 5

Question 1. The dot plot shows the number of omelets ordered at Paul’s Restaurant each day. Does the dot plot contain any gaps? Type below: _________________

Answer: Yes, the dot plot contains gaps

Explanation: The dots are filled from 10 – 11, from 14 – 16, and from 18 – 19 but the region between these two intervals is left unfilled so the region containing gaps is 12-13 including 12 and 13, and 17 is also left unfilled.

Answer: 14 – 16 and 18 – 19

Explanation: A group of dots is called a cluster. The dots which form a cluster are 14 – 16 and from 18 – 19.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 6

Question 3. The histogram shows the number of people that visited a local shop each day in January. How many peaks does the histogram have? Type below: _________________

Answer: The histogram has only one peak.

Explanation: The rectangle with tall length represents the highest peak in the graph given above. The number of people who visited a local shop each day in January was 0 – 9  visitors and this was the highest frequency having 14 days. The highest peak is in the interval of 0 – 9.

Question 4. Describe how the data values change across the intervals. Type below: _________________

Answer: They decrease from highest to lowest values in the given picture above.

Explanation: The graph represents the number of visitors in the month of January Visitors of number 0 – 9 have the highest frequency, 10 – 19 are the second-highest among the visitors who went to the local shop in the month of January, followed by 20 – 29, 30 – 39

Problem Solving

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 7

Answer: Yes, the graph has a symmetry

Explanation: If I draw a vertical line through the interval for _ to_ pounds, the left and right parts of the histogram are very close to being mirror images. The histogram __ line symmetry. A geometric figure has line symmetry if you can draw a line through it so that the two parts are mirror images of each other. So, the data values increase to one peak in the interval for _ to _ pounds and then decrease. The data set __ line symmetry about the peak.

Go Math 6th Grade Homework Answers Question 6. A histogram that shows the ages of students at a library has intervals 1–5, 6–10, 11–15, 16–20, and 21–25. There is a peak at 11–15 years and the graph is symmetric. Sketch what the histogram could look like and describe the patterns you see in the data. Type below: _________________

lesson 3 homework practice measures of spread answer key

The histogram shows the ages of students at a library has intervals 1–5, 6–10, 11–15, 16–20, and 21–25. There is a peak at 11–15 years and the graph is symmetric.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 8

Answer: 10 – 14 interval has the highest frequency of 6

Explanation: The rectangle with a peak can be said it have the greatest frequency. The interval with a peak is 11 – 15 and the frequency of the peak is 6

Question 2. Meg makes a dot plot for the data 9, 9, 4, 5, 5, 3, 4, 5, 3, 8, 8, 5. Where does a gap occur? Type below: _________________

Answer: 6 – 7 including 6 and 7

Explanation: Let us consider an axis with 3 to 9 numbers on it and plot the dots as given in the question at points 3,4,5,8 and 9 the gap occurs between 6 and 7 including 6 and 7.

Spiral Review

Question 3. A rectangular fish tank is 20 inches long, 12 inches wide, and 20 inches tall. If the tank is filled halfway with water, how much water is in the tank? ________ in.

Answer: 37500 cubic centimeter

Explanation: The length of the rectangle of the rectangular fish tank = 20 inches x 2.5 cm = 50 cm (since 1 inch = 2.5 cm) The breadth of the rectangle of the rectangular fish tank = 12 inches x 2.5 cm = 30 cm (since 1 inch = 2.5 cm) The height of the rectangle of the rectangular fish tank = 20 inches x 2.5 cm = 50 cm (since 1 inch = 2.5 cm) Water filled in the tank = Volume of the tank = 50 x 50 x 30 = 75000 cubic centimeter If the tank is filled halfway = the volume of the tank / 2 = 37500 cubic centimeters

Question 4. Look at the histogram below. How many students scored an 81 or higher on the math test? ________ students

Explanation: The interval 81 – 90 has 10 frequencies and the interval 91 – 100 has 4 frequencies. So the total number of students = 14

Lesson 3 Homework Practice Mean Absolute Deviation Answer Key Question 5. The Little League coach uses a radar gun to measure the speed of several of Kyle’s baseball pitches. The speeds, in miles per hour, are 52, 48, 63, 47, 47. What is the median of Kyle’s pitch speeds? The median is ________ miles.

Answer: Median is 48

Explanation: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for odd number of observations n = 5 (odd) Median = (5 + 1 / 2) = (6/2) = 3rd observation = 48 Therefore the median is 48.

Find the median, lower quartile, and upper quartile of the data.

Question 1. The scores of 11 students on a geography quiz: 87, 72, 80, 95, 86, 80, 78, 92, 88, 76, 90 Type below: _________________

Answer: Median: 86, lower quartile: 72, upper quartile: 95

Explanation: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for an odd number of observations n = 11 (odd) Median = (11 + 1 / 2) = (12/2) = 6th observation = 86 Therefore the median is 86.

Lower quartile: 72  Upper quartile: 95

Question 2. The lengths, in seconds, of 9 videos posted online: 50, 46, 51, 60, 62, 50, 65, 48, 53 Type below: _________________

Answer: Median: 51 Lower quartile: 46 Upper quartile: 65

Explanation: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for an odd number of observations n = 9 (odd) Median = (9 + 1 / 2) = (10/2) = 5th observation = 51 Therefore the median is 51.

Lower quartile: 46 Upper quartile: 65

Question 3. Make a box plot to display the data set in Exercise 2. Type below: _________________

lesson 3 homework practice measures of spread answer key

Explanation: The box is drawn to understand the clear view of the raw data, in a precise manner. This box gives us information about the lengths of the videos posted online. We can directly say the median, lower quartile, upper quartile seeing the box plot.

Question 4. 13, 24, 37, 25, 56, 49, 43, 20, 24 Type below: _________________

Explanation: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for an odd number of observations n = 9 (odd) Median = (9 + 1 / 2) = (10/2) = 5th observation =25 Therefore the median is 25.

Question 5. 61, 23, 49, 60, 83, 56, 51, 64, 84, 27 Type below: _________________

Explanation: First, write the observations in ascending order or descending order. The formula to calculate the median is the mean of (n/2) and (n/2+1)th observations, for an even number of observations n = 10 (even) Median = Mean of (5)th and (6)th observations = 56 + 60 divided by 2 = 116/2 = 58 Therefore the median is 58.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 9

Explanation: Lower limit: 8 Upper limit: 30 Median: First, write the observations in ascending order or descending order. The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observations n = 12 (even) Median = Mean of (6)th and (7)th observations = 18 + 20 divided by 2 = 38/2 = 19 Therefore the median is 19.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 10

Answer: The lower and upper limits are marked wrong.

Explanation: The box drew above the number line is wrong. It does not show the correct upper and lower limits. The lower limit is 5 and the upper limit is 35.

Problem Solving + Applcations – Page No. 716

Pose a Problem

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 12

Answer: Airline A: greatest value − least value = 8 Airline B: greatest value − least value = 10 The data for airline B is more spread out.

lesson 3 homework practice measures of spread answer key

Explanation: The greatest value and lowest value can be identified by seeing the box drew above the number line. The ends represent the lower and upper limits in both the box plots.

The solution to the question framed: The start and end of the rectangle represent the lower and upper limits. And the middle line represents the median. The lower limit is 5 Upper limit is 35 Median: First, write the observations in ascending order or descending order. The formula to calculate the median is the mean of (n/2) and (n/2+1)th observations, for even number of observations n = 6 (even) Median = Mean of (3)th and (4)th observations = 15 + 25 divided by 2 = 40/2 = 20 Therefore the median is 20.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 13

Answer: Median: 24 The lower quartile is 16. The upper quartile is 30.

Explanation: Seeing the data in the box we can identify the lower and upper quartiles. Median: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for odd number of observations n = 11 (odd) Median = (11 + 1 / 2) = (12/2) = 6th observation =24 Therefore the median is 24.

Question 1. The amounts of juice in 12 glasses, in fluid ounces: 11, 8, 4, 9, 12, 14, 9, 16, 15, 11, 10, 7 Type below: _________________

Answer: Median: 10.5 Lower quartile: 4 Upper quartile: 16

Explanation: Median: First, write the observations in ascending order or descending order. The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observations n = 12 (even) Median = Mean of (6)th and (7)th observations = 10 + 11 divided by 2 = 21/2 = 10.5 Therefore the median is 10.5. After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively. They are: Lower quartile: 4 Upper quartile: 16

Question 2. the lengths of 10 pencils, in centimeters: 18, 15, 4, 9, 14, 17, 16, 6, 8, 10 Type below: _________________

Answer: Median: 12 Lower quartile: 4 Upper quartile: 18

Explanation: Median: First, write the observations in ascending order or descending order. The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observations n = 10 (even) Median = Mean of (5)th and (6)th observations = 10 + 14 divided by 2 = 24/2 = 12 Therefore the median is 12. After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively. They are: Lower quartile: 4 Upper quartile: 18

lesson 3 homework practice measures of spread answer key

Explanation: Box plot is drawn using the number line and the rectangle which is drawn above it. The ends of the rectangles say about the lower and upper limits and the middle line indicates the median.

Question 4. The numbers of students on several teams are 9, 4, 5, 10, 11, 9, 8, and 6. Make a box plot for the data. Type below: _________________

lesson 3 homework practice measures of spread answer key

Explanation: Box plot is drawn using the number line and the rectangle which is drawn above it. The ends of the rectangles say about the lower and upper limits and the middle line indicates the median. Therefore the lower and upper quartiles are 4 and 11 respectively. Median: First, write the observations in ascending order or descending order. The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observations n = 8 (even) Median = Mean of (4)th and (5)th observations = 8 + 9 divided by 2 = 17/2 = 8.5 Therefore the median is 8.5.

Question 5. The amounts spent at a gift shop today are $19, $30, $28, $22, $20, $26, and $26. What is the median? What is the lower quartile? Type below: _________________

Answer: Median: $26 Lower quartile: $19 Upper quartile: $30

Explanation: Median: First, write the observations in ascending order or descending order. The formula to calculate the median is (n+1/2)th observation, for odd number of observations n = 7 (odd) Median = (7 + 1 / 2) = (8/2) = 4th observation =26 Therefore the median is 26. After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively. They are: Lower quartile: $19 Upper quartile: $30

Question 6. The weights of six puppies in ounces are 8, 5, 7, 5, 6, and 9. What is the upper quartile of the data? Type below: _________________

Answer: Upper quartile: 9

Explanation: The highest value in the data is defined as the upper quartile. The highest value in the raw data given is 9

Question 7. Draw a box plot to display this data: 81, 22, 34, 55, 76, 20, 56. Type below: _________________

lesson 3 homework practice measures of spread answer key

Explanation: A box plot gives information about the lower and upper quartiles and about the median. The box plot is drawn using a rectangle and the number line.

Question 1. The values in a data set are 15, 7, 11, 12, 6, 3, 10, and 6. Where would you draw the box in a box plot for the data? Type below: _________________

Answer: The box is drawn above the number line.

lesson 3 homework practice measures of spread answer key

Question 2. What is the lower quartile of the following data set? 22, 27, 14, 21, 22, 26, 18 Type below: _________________

Explanation: The value which is lowest in the given data is called the lowest quartile. Therefore the lowest quartile in the given data is 14.

Question 3. Jenn says that “What is the average number of school lunches bought per day?” is a statistical question. Lisa says that “How many lunches did Mark buy this week?” is a statistical question. Who is NOT correct? Type below: _________________

Answer: Lisa’s statement is wrong.

Explanation:

Question 4. The prices of several chairs are $89, $76, $81, $91, $88, and $70. What is the mean of the chair prices? The mean is $ _________

Answer: $82.5

Explanation: Number of observations= 6 Mean of the observations= $89 + $76 + $81+ $91+$88+ $70/ 6= 495/6 = $82.5

Question 5. By how much does the mean of the following data set change if the outlier is removed? 13, 19, 16, 40, 12 Type below: _________________

Answer: The mean shows a difference if the lower limit is removed the mean increases and if the upper limit is removed the mean decreases.

Explanation: Outliers are nothing but both upper and lower limits. The actual mean is 20 But when the lower limit is removed the mean increases to 22 while when the upper limit is removed the mean decreases to 15. Therefore, we can say that the mean shows a difference when the outliers are removed.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 14

Answer: 52 – 54

Explanation: A cluster is nothing but a group of dots. In the intervals 52 – 54 a cluster has occurred.

Use counters, a dot plot, or iTools to find the mean absolute deviation of the data.

Question 1. Find the mean absolute deviation for both data sets. Explain which data set is more spread out. the number of laps Shawna swam on 5 different days: 5, 6, 6, 8, 10 mean = 7

the number of laps Lara swam on 5 different days: 1, 3, 7, 11, 13 mean = 7 Type below: _________________

Answer: Case 2 is more spread out.

Explanation: CASE1 The number of laps Shawna swam on 5 different days: 5,6,6,8,10 Mean = 7 Deviations: 7 – 5 = 2 7 – 6 = 1 7 – 6 = 1 7 -8 = -1 7 -10=-3 Mean of deviations = 2+1+1+1+3/5 = 8/5 = 1.6

CASE2 The number of laps Lara swam on 5 different days: 1, 3, 7, 11, 13 Mean = 7 Deviations: 7 – 1 = 6 7 – 3 = 4 7 – 7 = 0 7 -11= -4 7 -13= -6 Mean of deviations = 6+ 4 + 0 + 4 + 6 / 5 = 20/5 = 4

Use the dot plot to find the mean absolute deviation of the data.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 15

Answer: Mean absolute deviation is 2.4

Explanation: STEP 1 Label each dot with its distance from the mean. Starting from left to right: 4: 7-4=3 5: 7-5=2 6: 7-6=1 9: 7-9=-2 10: 7-10=-3 11: 7-11=-4

STEP 2 Find the mean of the distances. (3) + (2) +(2) +(2) +(2) +(1) + (2) +(3) +(3) +(4) / 10 = 24/10 = 2.4

So, the mean absolute deviation of the data is 2.4

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 17

Answer: Mean Absolute deviation is 3.2

Explanation: STEP 1 Label each dot with its distance from the mean. Starting from left to right: 26: 29-26=03 27: 29-27=02 32: 29-32=-3 33: 29-33=-4 35: 29-35=-6

STEP 2 Find the mean of the distances. (3) + (2) +(3) +(4) +(6) +(3) + (3) +(2) / 8 = 26/8 = 3.2

So, the mean absolute deviation of the data is 3.2

Lesson 13.3 Practice A Data Distributions Answer Key Question 4. The mean absolute deviation of the number of daily visits to Scott’s website for February is 167.7. In March, the absolute mean deviation is 235.9. In which month did the number of visits to Scott’s website vary more? Explain how you know. Type below: _________________

Answer: As the mean absolute deviation is more in the month of February we can say that there are more visitors in this month.

Explanation: As the mean of the month of February is less it means that the number of observations is more. Similarly, as the mean of the month of March is more it means that the number of observations is less. Therefore we can say that the number of visitors was higher in the month of February compared to March.

Question 5. Write an Inequality Algebra In April, the data for Scott’s website visits are less spread out than they were in February. Use a to represent the mean absolute deviation for April. Write an inequality to describe the possible values of a. Type below: _________________

Answer: a < February

Explanation: Since the data is more spread out in the month of April than they were in February. Therefore the inequality represents a “less than” sign.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 18

Explanation: STEP 1 Label each observation with its distance from the mean. Starting from left to right: 10: 11-10= 1 12: 11-12=-1 13: 11-13=-2 18: 11-18=-7 10: 11-10= 1 08: 11-08= 3 07: 11-07= 4 06: 11-06= 5 16: 11-16=-5 14: 11-14=-3 08: 11-08= 3 10: 11-10= 1

STEP 2 Find the mean of the distances. 1+1 +2 +7 +1 +3 +4 +5+5+3+3+1/ 12 = 36/12 = 3

So, the mean absolute deviation of the data is 3.

Question 7. Suppose all of the players on a basketball team had the same height. Explain how you could use reasoning to find the mean absolute deviation of the players’ heights. Type below: _________________

Explanation: If the players on a basketball team had the same height. The mean deviation will be equal to 0 because the difference between the mean and the observations is 0. Let the observations be 2,2,2,2,2 Mean = 10/5 = 2 Mean deviation =  (2-2)+(2-2)+(2-2)+(2-2)+(2-2)/5 = 0/5 = 0

Question 8. Explain Tell how an outlier that is much greater than the mean would affect the mean absolute deviation of the data set. Explain your reasoning. Type below: _________________

Answer: An outlier increases the mean absolute deviation of the data set.

Explanation: The difference between the outlier and the mean is a greater number when added to the sum of observations the mean absolute deviation increases.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 19

Answer: 9a. No 9b. Yes 9c. No

Explanation: Player A Mean = 1+2+3/3 = 6/3 = 2 Mean absolute deviation = 1+0+1/3 = 2/3 = 0.6

Player B Mean =2+2+2/3 = 6?3 = 2 Mean absolute deviation = 0/3 = 0

Player C Mean = 1+2+1/3 = 4/3 = 1.3 Mean absolute deviation = 0.3+0+0.3/3 = 0.2

Use counters and a dot plot to find the mean absolute deviation of the data.

Question 1. the number of hours Maggie spent practicing soccer for 4 different weeks: 9, 6, 6, 7 mean = 7 hours _______ hour

Answer: The mean absolute deviation of the data is 1.

Explanation: STEP 1 Label each observation with its distance from the mean. Starting from left to right: 9: 7-9=-2 6: 7-6=-1 6: 7-6=-1 7: 7-7= 0

STEP 2 Find the mean of the distances. 2 +1 +1+0/ 4 = 4 /4 = 1

So, the mean absolute deviation of the data is 1.

Question 2. the heights of 7 people in inches: 60, 64, 58, 60, 70, 71, 65 mean = 64 inches _______ inches

Answer: The mean absolute deviation of the data is 4.

Explanation: STEP 1 Label each observation with its distance from the mean. Starting from left to right: 60: 64-60= 4 64: 64-64= 0 58: 64-58= 6 60: 64-60= 4 70: 64-70=-6 71: 64-71=-7 65: 64-65=-1

STEP 2 Find the mean of the distances. 4+0+6+ 4+6+7+1/7 = 28/7 = 4

So, the mean absolute deviation of the data is 4.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 20

Answer: The mean absolute deviation of the data is 1

Explanation: STEP 1 Label each dot with its distance from the mean. Starting from left to right: 08: 10-08=02 09: 10-09=01 10: 10-10= 0 11: 10-11=-1 12: 10-12=-2

STEP 2 Find the mean of the distances. (2) + (1) +(0) +(1) +(2) +(2)+(1)+(0)+(0)+(0)+(1)+(2) /12 = 12/12= 1

So, the mean absolute deviation of the data is 1

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 21

Answer: The mean absolute deviation of the data is 2.4

Explanation: STEP 1 Label each dot with its distance from the mean. Starting from left to right: 03: 8-03=05 04: 8-04=04 05: 8-05=03 07: 8-07=01 08: 8-08= 0 09: 8-09=-1 10: 8-10=-2 11: 8-11=-3 12: 8-12=-4

STEP 2 Find the mean of the distances. (5) + (4) +(3) +(1) +(0) +(1) + (2) +(3) +(4)+(5)+(0)+(1)+(1)+(2)+(4)/ 15 = 36/15 = 2.4

Question 5. In science class, Troy found the mass, in grams, of 6 samples to be 10, 12, 7, 8, 5, and 6. What is the mean absolute deviation? _______ grams

Answer: The mean absolute deviation of the data is 2.

Explanation: Mean = 10+12+7+8+5+6/6 = 48/6 = 8

Mean absolute deviation:

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 10: 8-10= -2 12: 8-12= -4 07: 8-07= 01 08: 8-08= 0 05: 8-05=03 06: 8-06=02

STEP 2 Find the mean of the distances. 2+4+1+0+3+2/6 = 12/6 = 2

So, the mean absolute deviation of the data is 2.

Question 6. Five recorded temperatures are 71°F, 64°F, 72°F, 81°F, and 67°F. What is the mean absolute deviation? _______ °F

Answer: The mean absolute deviation of the data is 4.4.

Explanation: Mean = 71+64+72+81+67/5 = 355/5 = 71

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 71: 71-71= 0 64: 71-64= 07 72: 71-72= -1 81: 71-81=-10 67: 71-67= 04

STEP 2 Find the mean of the distances. 0+7+1+10+4/5 = 22/5 = 4.4

So, the mean absolute deviation of the data is 4.4.

Question 7. Make a dot plot of the following data: 10, 10, 11, 12, 12, 13, 13, 15. Use the dot plot to find the mean absolute deviation. Type below: _________________

Answer: The mean absolute deviation of the data is 1.25

Explanation: Mean = 10+10+11+12+12+13+13+15/8 = 96/8 = 12

lesson 3 homework practice measures of spread answer key

STEP 1 Label each dot with its distance from the mean. Starting from left to right: 10: 12-10=02 11: 12-11=01 12: 12-12=0 13: 12-13=-1 15: 12-15=-3

STEP 2 Find the mean of the distances. (2) + (2) +(1) +(0) +(0) +(1) + (1) +(3) / 8 = 10/8 = 1.25

So, the mean absolute deviation of the data is 1.25

Question 1. The six test grades are 86, 88, 92, 90, 82, and 84. The mean of the data is 87. What is the mean absolute deviation? _______

Answer: The mean absolute deviation of the data is 3.5

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 86: 87-86= 01 88: 87-88= -1 92: 87-92= -5 90: 87-81= 06 82: 87-82= 05 84: 87-84= 03

STEP 2 Find the mean of the distances. 1+5+1+6+5+3/6 = 21/6 = 3.5

So, the mean absolute deviation of the data is 3.5

Question 2. Eight heights in inches are 42, 36, 44, 46, 48, 42, 48, and 46. The mean of the data is 44. What is the mean absolute deviation? _______ inches

Explanation: The mean absolute deviation of the data is 3

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 42: 44-42= 02 36: 44-36= 08 44: 44-44= 0 46: 44-46= -2 48: 44-48= -4 42: 44-42= 02 48: 44-48= -4 46: 44-46= -2

STEP 2 Find the mean of the distances. 2+8+2+4+0+2+4+2/8 = 24/8 = 3

So, the mean absolute deviation of the data is 3

Question 3. What is the volume of a rectangular prism with dimensions 4 meters, 1 \(\frac{1}{2}\) meters, and 5 meters? _______ m 3

Answer: 30m 3

Explanation: Dimensions: 4 meters, 1 1/2 meters, 5 meters Change the mixed fraction into improper fraction = 3/2 Volume of the rectangle = 4 x 3/2 x 5 = 30m 3

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 22

Explanation: Total number of days in the month of September = 30 Number of days given in the frequency table = 17+8+4 = 29 Frequency in the interval 9 – 11 = 30 – 29 = 1 day

Question 5. The following data shows the number of laps each student completed. What number of laps is the mode? 9, 6, 7, 8, 5, 1, 8, 10 The mode is _______ laps.

Answer: The mode is 8 laps.

Explanation: The most frequently occurring observation is known as mode. 8 is the mode in the above raw data given.

Question 6. What is the upper quartile of the following data? 43, 48, 55, 50, 58, 49, 38, 42, 50 The upper quartile is _______

Answer: The upper quartile is 58

Explanation: The highest observation in the data given is known as the upper quartile. The upper quartile is 58

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 23

Answer: $12, $3

Explanation: The difference between the highest observation and the lowest observation is called a range. Range = 19 – 7 = $12 The difference and the highest and the lowest dots of the dot plot is called as interquartile range. Interquartile range = 15 – 12 = $3

Practice: Copy and Solve Find the mean absolute deviation for the data set.

Question 2. heights in inches of several tomato plants: 16, 18, 18, 20, 17, 20, 18, 17 _______ inch

Explanation: Mean: Mean = 16+18+18 +20+17+20+18+17/8 = 144/8 = 18

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 16: 18-16= 02 18: 18-18= 0 18: 18-18= 0 20: 18-20= -2 17: 18-17= 01 20: 18-20= -2 18: 18-18= 0 17: 18-17= 01

STEP 2 Find the mean of the distances. 2+0+0+2+1+2+0+1/8 = 8/8 = 1

Question 3. times in seconds for students to run one lap: 68, 60, 52, 40, 64, 40 _______ seconds

Answer: The mean absolute deviation of the data is 10

Explanation: Mean: Mean = 68+60+52+40+64+40/6 = 54

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 68: 54-68= -14 60: 54-60= -6 52: 54-52= 02 40: 54-40= 14 64: 54-64= -10 40: 54-40= 14

STEP 2 Find the mean of the distances. 14+6+2+14+10+14/6 = 60/6 = 10

So, the mean absolute deviation of the data is 10

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 24

Question 4. What is the range of the data? $ _______

Answer: $24

Explanation: The difference between the highest observation and the lowest observation is called a range. Range = $56 – $32 = $24

Question 5. What is the interquartile range of the data? $ _______

Answer: $16

Explanation: The difference and the highest and the lowest dots of the dot plot is called as interquartile range. Interquartile range = $52 – $36 = $16

Question 6. Times in minutes spent on a history quiz: 35, 35, 32, 34, 34, 32, 34, 36 _______ minute

Explanation: Mean: Mean = 35+ 35+ 32+ 34+34+ 32+ 34+36/8 = 272/8 = 34

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 35: 34-35= -1 35: 34-35= -1 32: 34-32= 02 34: 34-34= 0 34: 34-34= 0 32: 34-32=02 34: 34-34=0 36: 34-36=-2

STEP 2 Find the mean of the distances. 1+1+2+0+0+2+0+2/8 = 8/8 = 1

Question 7. number of excused absences for one semester: 1, 2, 1, 10, 9, 9, 10, 6, 1, 1 _______

Answer: The mean absolute deviation of the data is 3.8

Mean: Mean =1+2+1+10+9+9+10+6+1+1 /10 = 50/10 = 5

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 1: 5-1= 4 2: 5-2= 3 1: 5-1= 4 10: 5-10= -5 9: 5-9= -4 9: 5-9=-4 10: 5-10=-5 6: 5-6=-1 1: 5-1=4 1: 5-1=4

STEP 2 Find the mean of the distances. 4+3+4+5+4+4+5+1+4+4/10 =38/10 = 3.8

So, the mean absolute deviation of the data is 3.8

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 25

The mean absolute deviation of the data is 3.6 Range = 32-16 = 16 Interquartile range = 24 – 20 = 4

Mean: Mean =18+24+20+26+24+20+32+20+16+20 /10 = 220/10 = 22

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 18: 22-18= 4 24: 22-24= -2 20: 22-20= 2 26: 22-26= -4 24: 22-24= -2 20: 22-20= 2 32: 22-32=-10 20: 22-20= 2 16: 22-16= 6 20: 22-20= 2

STEP 2 Find the mean of the distances. 4+2+2+4+2+2+10+2+6+2/10 =36/10 = 3.6

So, the mean absolute deviation of the data is 3.6 The difference between the highest observation and the lowest observation is called a range. Range = 32-16 = 16 The difference and the highest and the lowest dots of the dot plot is called as interquartile range. Interquartile range = 24 – 20 = 4

Question 9. Hyato’s family began a walking program. They walked 30, 45, 25, 35, 40, 30, and 40 minutes each day during one week. At the right, make a box plot of the data. Then find the interquartile range. _______ minutes

Answer: 35 minutes

Explanation: Ascending order: 25,30,30,35,40,40,45 n=7 (odd) Median= Number of (n+1/2) = 8/2 = 4th observation = 35 Median:(four terms of the data) Median = 30+30/2 = 60/2 = 30 Median:(last 3 terms of the data) Median = (n+1/2) = 2nd observation = 40 Interquartile range = 30+40/2 = 70/2 = 35

Question 10. Compare Jack recorded the number of minutes his family walked each day for a month. The range of the data is 15. How does this compare to the data for Hyato’s family? Type below: _________________

Answer: Jack’s family walked less number of minutes each day compared to Hyato’s family.

Explanation: The range of Hyato’s family is 20 while the range of Jack’s family is 15. Therefore we can say that Hyato’s family walked more minutes compared to Jack’s family in a day. The range can define the data with large observations and the data with the least observations.

Question 11. Sense or Nonsense? Nathan claims that the interquartile range of a data set can never be greater than its range. Is Nathan’s claim sense or nonsense? Explain. Type below: _________________

Answer: Nonsense, The Interquartile range of a data set can be less than or greater than the range.

Explanation: The interquartile range is the difference between the medians of the observations. Nathan’s claim is nonsense as he said, ” The interquartile range can never be greater than its range.” The range is the difference between the highest observation and the lowest observation. The interquartile range can be less than greater than the range.

Example: Ascending order: 25,30,30,35,40,40,45 n=7 (odd) Median= Number of (n+1/2) = 8/2 = 4th observation = 35 Median:(four terms of the data) Median = 30+30/2 = 60/2 = 30 Median:(last 3 terms of the data) Median = (n+1/2) = 2nd observation = 40 Interquartile range = 30+40/2 = 70/2 = 35

Range= 45-25 = 20

In the above case, the interquartile range is more than the range proving that the given statement is nonsense.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 26

Answer: greater than

Explanation: The range is the difference between the highest and the lowest observations. Range of Farm A: 72-58 = 14 Range of Farm B: 70-55 = 15

Therefore, The range of Farm A’s heights is greater than the range of Farm B’s heights.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 27

Answer: 16, 8

Explanation: The difference between the highest and the lowest observations is range. Range = 17 – 1 = 16 The difference between the highest and lowest observations of the box is the interquartile range. Interquartile range = 12 – 4 = 8

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 28

Question 2. What is the range of the data? _____

Explanation: The difference between the highest and the lowest observations is range. Range = 95 – 60 = 35

Question 3. What is the interquartile range of the data? _____

Explanation: The difference between the highest and lowest observations of the box is the interquartile range. Interquartile range = 90 – 70 = 20

Find the mean absolute deviation for the set.

Question 4. heights in centimeters of several flowers: 14, 7, 6, 5, 13 _____ cm

Answer: The mean absolute deviation of the data is 3.6

Mean: Mean =14+7+ 6+5+13/5= 45/5 = 9

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 14: 9-14= -5 07: 9-07= 02 06: 9-06= 03 05: 9-05= 04 13: 9-13= -4

STEP 2 Find the mean of the distances. 5+2+3+4+4/5 = 18/5 = 3.6

So, the mean absolute deviation of the data is 3.6

Question 5. ages of several children: 5, 7, 4, 6, 3, 5, 3, 7 _____ years

Explanation: Mean: Mean = 5+7+4+6+ 3+5+3+7/8 = 40/8 = 5

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 5: 5-5= 0 7: 5-7= -2 4: 5-4= 01 6: 5-6= -1 3: 5-3= 02 5: 5-5= 0 3: 5-3= 02 7: 5-7=-2

STEP 2 Find the mean of the distances. 0+2+1+1+2+0+2+2/8 = 10/8 = 1.25

Question 6. The following data set gives the amount of time, in minutes, it took five people to cook a recipe. What is the mean absolute deviation for the data? 33, 38, 31, 36, 37 _____ minutes

Explanation: Mean: Mean = 33+38+31+36+37/5 = 175/5 = 35

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 33: 35-33= 02 38: 35-38= -3 31: 35-31= 04 36: 35-36= -1 37: 35-37= -2

STEP 2 Find the mean of the distances. 2+3+4+1+2/5 = 12/5 = 2.4

Question 7. The prices of six food processors are $63, $59, $72, $68, $61, and $67. What are the range, interquartile range, and mean absolute deviation for the data? Type below: _________________

Answer: Range = $9 The mean absolute deviation of the data is 4

Explanation: The difference between the highest and the lowest observations is range. Range = $68 – $59 = $9 The difference between the highest and lowest observations of the box is the interquartile range. Interquartile range = 12 – 4 = 8

Mean: Mean = $63+$59+$72+$68+$61+$67/6 = 390/6 = 65

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 63: 65-63= 02 59: 65-59= 06 72: 65-72= -7 68: 65-68= -3 61: 65-61= -4 67: 65-67= -2

STEP 2 Find the mean of the distances. 2+6+7+3+4+2/6 = 24/6 = 4

So, the mean absolute deviation of the data is 4

Question 8. Find the range, interquartile range, and mean absolute deviation for this data set: 41, 45, 60, 61, 61, 72, 80. Type below: _________________

Answer: The mean absolute deviation of the data is 9.7

Mean: Mean = 41+45+60+61+61+72+80 /7 = 420/7 = 60

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 41: 60-41= 19 45: 60-45= 15 60: 60-60= 0 61: 60-61= -1 61: 60-61= -1 72: 60-72= -12 80: 60-80= -20

STEP 2 Find the mean of the distances. 19+15+0+1+1+12+20/7 = 68/7 = 9.7

So, the mean absolute deviation of the data is 9.7

Question 1. Daily high temperatures recorded in a certain city are 65°F, 66°F, 70°F, 58°F, and 61°F. What is the mean absolute deviation for the data? _____ °F

Mean: Mean = 65+66+70+58+61 /5 = 320/5 = 64

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 65: 64-65=-1 66: 64-66=-2 70: 64-70=-6 58: 64-58=06 61: 64-61=03

STEP 2 Find the mean of the distances. 1+2+6+6+3/5 = 18/5 = 3.6

Question 2. Eight different cereals have 120, 160, 135, 144, 153, 122, 118, and 134 calories per serving. What is the interquartile range for the data? _____ calories

Answer: 42cereals

Explanation: Ascending order of the data: 118,120,122,134,135,144,153,160 Median:(for first 4 terms) Median= 120+122/2 = 242/2 = 121 Median:(for first 4 terms) Median= 144+153/2 = 297/2 = 148.5 The difference and the highest and the lowest dots of the dot plot is called as interquartile range. Interquartile range = 148.5 – 121 = 27.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 29

Explanation: After 59 there is 1 interval from 60-79 Number of days the restaurant sells more than 59 pizzas = 20

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 30

Answer: 20 – 39

Explanation: Number of days the restaurant sold the maximum pizzas = 30 Number of pizzas sold each day = 20 – 39

Question 5. What is the mode of the data set? 14, 14, 18, 20 The mode is ________

Explanation: The most frequently occurring observation is known as a mode. In the above data mode is 14.

Question 6. The data set below lists the ages of people on a soccer team. The mean of the data is 23. What is the mean absolute deviation? 24, 22, 19, 19, 23, 23, 26, 27, 24 ________

Answer: The mean absolute deviation of the data is 2

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 24: 23-24=-1 22: 23-22= 1 19: 23-19= 4 19: 23-19= 4 23: 23-23=0 23: 23-23=0 26: 23-26=-3 27: 23-27=-4 24: 23-24=-1

STEP 2 Find the mean of the distances. 1+1+4+4+3+4+1+0+0/9 = 18/9 = 2

So, the mean absolute deviation of the data is 2

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 31

Question 1. The _____ is the difference between the upper quartile and the lower quartile of a data set. Type below: _________________

Answer: Range

Explanation: The difference between the upper and lower quartiles of the data is known as the range.

Question 2. A graph that shows the median, quartiles, and least and greatest values of a data set is called a(n) _____. Type below: _________________

Answer: Box plot

Explanation: The figure that shows the median, quartiles, and least and greatest values of a data set is called a box plot, A box plot is a figure that represents the median with a horizontal line, and the starting and ending line represents the upper and lower quartiles and the end dots represent the upper limit and the lower limit.

Question 3. The difference between the greatest value and the least value in a data set is the _____. Type below: _________________

Explanation: Each data set consists of upper and lower limits the difference between these limits is called range.

Question 4. The _____ is the mean of the distances between the values of a data set and the mean of the data set. Type below: _________________

Answer: Mean absolute

Explanation: Mean absolute deviation is calculated by subtracting each observation from the mean and then the mean is calculated for these observations. Therefore we can say that the mean absolute is the mean of the distances between the values of a data set and the mean of the data set.

Concepts and Skills

Question 5. Make a box plot for this data set: 73, 65, 68, 72, 70, 74. Type below: _________________

Answer: Median = 71

Explanation: Median: Ascending order: 65, 68, 70, 72, 73,74 n = even = 6 Median = Mean of 3rd and 4th terms = 70+72/2 = 142/2 = 71

Find the mean absolute deviation of the data.

Question 6. 43, 46, 48, 40, 38 ________

Answer: The mean absolute deviation of the data is 3.2

Explanation: Mean:

Mean = 43+46+48+40+38/5 = 215/5 = 43

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 43: 43-43=0 43: 43-46=-3 43: 43-48=-5 43: 43-40= 3 43: 43-38= 5

STEP 2 Find the mean of the distances. 0+3+5+3+5/5 = 16/5 = 3.2

Question 7. 26, 20, 25, 21, 24, 27, 26, 23 ________

Answer: The mean absolute deviation of the data is 2.125

Mean = 26+20+25+21+24+27+26+23/8 = 192/8 = 24

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 26: 24-26=-3 20: 24-20= 4 25: 24-25= -1 21: 24-21= 3 24: 24-24=0 27: 24-27=-3 26: 24-26=-2 23: 24-23=1

STEP 2 Find the mean of the distances. 3+4+1+3+0+3+2+1/8 = 17/8 = 2.125

So, the mean absolute deviation of the data is 2.125

Question 8. 99, 70, 78, 85, 76, 81 ________

Mean = 99+70+78+85+76+81/6 = 489/6 = 81.5

Find the range and interquartile range of the data.

Question 9. 2, 4, 8, 3, 2 The range is _________ . The interquartile range is _________ .

Explanation: The difference between the upper quartile and lower quartile. Range = 8 – 2 =6

Question 10. 84, 82, 86, 87, 88, 83, 84 The range is _________ . The interquartile range is _________ .

Explanation: The difference between the upper quartile and lower quartile. Range = 88 – 82 = 6

Question 11. 39, 22, 33, 45, 42, 40, 28 The range is _________ . The interquartile range is _________ .

Explanation: The difference between the upper quartile and lower quartile. Range = 45 – 22 = 23

Page No. 732

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 32

Answer: There is a gap in the data in the intervals: between 1 and 2

Explanation: As shown in the dot plot there is a gap between 1 and 2. This means that Yasmine’s team did not score only one goal when they played the game.

Question 13. What is the interquartile range of the data shown in the dot plot with Question 12? The interquartile range is _________ .

Explanation: The interquartile range is the difference between the lower and upper quartiles. Interquartile range = 3-1 = 2

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 33

Answer: The graph has only one peak

Explanation: In the given graph there are rectangles out of which one has a tall rectangle which can be addressed as the peak.

Question 15. In a box plot of the data below, where would the box be drawn? 55, 37, 41, 62, 50, 49, 64 Type below: _________________

Answer: The box is drawn above the number line and this represents the median and the lower and upper limits.

Explanation: A box is drawn to represent the median and the upper and lower limits in a box plot.

Question 1. The distances in miles students travel to get to school are 7, 1, 5, 9, 9, and 8. Decide which measure(s) of center best describes the data set. Explain your reasoning. Type below: _________________

Answer: Mean: 6.5 Median: 7.5 Mode: 9

Explanation: Mean: 7+1+5+9+9+8/6 = 39/6 = 6.5 Median: Ascending order: 1,5,7,8,9,9 Median = Mean of 7 and 8 = 7+8/2 = 15/2 = 7.5 Mode: The most frequently occurring observation is known as the mode. The mode is 9.

Question 2. Use Graphs The numbers of different brands of orange juice carried in several stores are 2, 1, 3, 1, 12, 1, 2, 2, and 5. Make a box plot of the data and find the range and interquartile range. Decide which measure better describes the data set and explain your reasoning. Type below: _________________

Answer: Range: 11      Interquartile: 3 Interquartile range is the best way to represent the data.

Explanation: Range = 12 – 1 = 11 Interquartile range : Median(of first 4 terms): Median = 1+1/2 = 2/2 = 1 Median (of last 4 terms): Median = 3+5/2 = 8/2 = 4 Interquartile range = 4 – 1 = 3

Question 3. Use Reasoning The ages of students in a computer class are 14, 13, 14, 15, 14, 35, 14. Decide which measure of center(s) best describes the data set. Explain your reasoning. Type below: _________________

Answer: Mean:17  Median:14   Mode:14  Median and mode is the best ways to represent the data.

Explanation: Mean: 14+13+14+15+14+35+14/7 = 119/7 = 17 Median: Ascending order: 13,14,14,14,14,15,35 Median = 14 Mode: The most frequently occurring observation is known as the mode. The mode is 14.

Question 4. Mateo scored 98, 85, 84, 80, 81, and 82 on six math tests. When a seventh math test score is added, the measure of center that best describes his scores is the median. What could the seventh test score be? Explain your reasoning. Type below: _________________

Answer: Median is the best way to represent the data.

Explanation: Median: Ascending order: 80,81,82,84,85,98 Median = Mean of 3 and 4 = 82+84/2 = 166/2 = 83 The seventh score can be 83

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 34

Answer: Mean, median,mode

Explanation: Mean= 1+2+2+3+3+1+18+2/8 = 32/8 = 4 Median: Ascending order: 1,1,2,2,2,3,3,18 Median = 2+2/2 = 2 Mode: The most frequently occurring observation is called a mode. Mode=2

Question 5. b. What information do you need to solve the problem? Type below: _________________

Answer: We need to have the data to find the centre of tendencies.

Explanation: The given data can be used to find the mean, median, and mode.

Question 5. c. What are the measures of center? Type below: _________________

Answer: Mean = 4 Median = 2 Mode = 2

Explanation: There are three measures to calculate their approximate values.

Question 5. d. Which measure of center should Jaime use? Explain. Type below: _________________

Answer: Median or mode

Explanation: Median or mode are nearer to the solution, therefore, they can be used.

Question 6. The number of sit-ups students completed in one minute are 10, 42, 46, 50, 43, and 49. The mean of the data values is 40 and the median is 44.5. Which measure of center better describes the data, the mean or median? Use words and numbers to support your answer. Type below: _________________

Answer: Median is the better way to represent the data.

Explanation: 44.5 is closer and represents the more number of observations compared to the mean.

Question 1. The distances, in miles, that 6 people travel to get to work are 14, 12, 2, 16, 16, and 18. Decide which measure(s) of center best describes the data set. Explain your reasoning. Type below: _________________

Answer: Mean= 13 miles Median= 15 miles Mode= 16 miles

Explanation: Mean is less than the data points. The Median describes the data in the best way compared to the mean and mode.

Question 2. The numbers of pets that several children have are 2, 1, 2, 3, 4, 3, 10, 0, 1, and 0. Make a box plot of the data and find the range and interquartile range. Decide which measure better describes the data set and explain your reasoning. Type below: _________________

Answer: Range = 10-0 = 10 Interquartile range = 3.5 – 0.5 = 3 The interquartile range is the best way to represent the data.

Explanation: Ascending order: 0,0,1,1,2,2,3,3,4,10 Median = 2+2/2 = 2 Lower quartile = 0.5 Upper quartile= 7/2 = 3.5 Highest observation= 10 Lowest observation = 0 Range = 10-0 = 10 Interquartile range = 3.5 – 0.5 = 3 The interquartile range is the best way to represent the data.

Question 3. Brett’s history quiz scores are 84, 78, 92, 90, 85, 91, and 0. Decide which measure(s) of center best describes the data set. Explain your reasoning. Type below: _________________

Answer: Mean is the best measure of centre to describe the data set.

Explanation: Mean: Mean= 84+78+92+90+85+91+0/7 = 74.2 Median: Ascending order: 0,78,84,85,90,91,92 Median = 4th observation = 85

Question 4. Eight students were absent the following number of days in a year: 4, 8, 0, 1, 7, 2, 6, and 3. Decide if the range or interquartile range better describes the data set, and explain your reasoning. Type below: _________________

Answer: 8 represents all the terms range is preferable compared to the interquartile range.

Explanation: Ascending order: 0,1,2,3,4,6,7,8 Range = 8-0 = 8 Median = Mean of 3 and 4 = 3+4/2 = 7/2 = 3.5 Median of first 3 terms = 1 Median of last 3 terms = 7 Interquartile range = 7-1 = 6 Since 8 represents all the terms range is preferable compared to the interquartile range.

Question 5. Create two sets of data that would be best described by two different measures of centre. Type below: _________________

Answer: The given below are the examples of two sets of data that would be best described by two different measures of centre.

Explanation: Example 1:

The numbers of pets that several children have are 2, 1, 2, 3, 4, 3, 10, 0, 1, and 0. Make a box plot of the data and find the range and interquartile range. Decide which measure better describes the data set and explain your reasoning. Type below: _________________

Brett’s history quiz scores are 5,6,7,8,9,10. Decide which measure(s) of centre best describes the data set. Explain your reasoning. Type below: _________________

Answer: Mean and median is the best measure of centre to describe the data set.

Explanation: Mean: Mean= 5+6+7+8+9+10/6 = 7.5 Median: Ascending order: 5,6,7,8,9,10 Median = Mean of 7 and 8 = 15/2 = 7.5

Question 1. Chloe used two box plots to display some data. The box in the plot for the first data set is wider than the box for the second data set. What does this say about the data? Type below: _________________

Answer: The graphs say that the interquartile range is more for the second graph compared to the first.

Explanation: The interquartile range is the difference between the lower and upper quartiles. It is more for wider data compared to compact data.

Question 2. Hector recorded the temperature at noon for 7 days in a row. The temperatures are 20°F, 20°F, 20°F, 23°F, 23°F, 23°F, and 55°F. Which measure of center would best describe the data? Type below: _________________

Answer: Mode

Explanation: The most frequently occurring observation is known as mode. The mode of the above data describes the data well, the mode of the data is 20°F

Question 3. By how much does the median of the following data set change if the outlier is removed? 13, 20, 15, 19, 22, 26, 42 Type below: _________________

Answer: 0.5

Explanation: Median: Ascending order: 13,15,19,20,22,26,42 Median = 20 If the outlier is removed then the median= 19+20/2 = 39/2 = 19.5 The difference in the medians = 0.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 35

Answer: 8 people

Explanation: Total number of people = 40 Percentage = 8/40 x 100 = 20%

Question 5. What is the lower quartile of the following data? 12, 9, 10, 8, 7, 12 The lower quartile is _______ .

Explanation: Ascending order: 7,8,9,10,12,12 Median = 9+10/2 = 9.5 Lower quartile = 8

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 36

Explanation: The difference between the upper and lower quartiles is called as interquartile range. Interquartile range = 14 – 9 = 5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 37

Answer: less than ; less than ; more ; more

Explanation: The mean of Player 1’s points is less than the mean of Player 2’s points. The interquartile range of Player 1’s points is less than the interquartile range of Player 2’s points. So, Player 2 typically scores more points than Player 1, but Player 2’s scores typically vary more Player 1’s scores

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 38

Answer: They differ slightly but on an average, we can say that shelter B is more as compared to shelter A

Explanation: Shelter A Median: Ascending order: 5,7,7,7,10,12,15 Median = 4th observation = 7 Range = 15-5 = 10 Shelter B Median: Ascending order: 4,5,5,11,11,13,15 Median = 4th observation = 11 Range = 15-4 = 11

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 39

Question 3. Which two students typically practiced the same amount each week, with about the same variation in practice times? Type below: _________________

Answer: Sally and Jennifer

Explanation: They are slightly different but on the whole, the average shows no difference and we can say that Sally and Jennifer practiced for the same amount each week

Question 4. Which two students typically practised the same number of hours, but had very different variations in their practice times? Type below: _________________

Answer: Tim and Sally

Explanation: They are different in range but on the whole, the average shows no difference and we can say that Sally and Tim practiced for the same number of hours, but had very different variations in their practice times.

Question 5. Which two students had the same variation in practice times, but typically practiced a different number of hours per week? Type below: _________________

Answer: Matthew and Tim

Explanation: Matthew and Tim practiced for the same number of hours but they had a high variation in the range.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 40

Answer: The interquartile range is the best way to compare the data in week 1 While the median is the best way to compare the data in week 2

Explanation: Week 1 Median: Ascending order: 1,2,2,3,3,4,5 Median = 4th observation = 3 Lower quartile range= 2 Upper quartile range= 4 Interquartile range = 4-2 = 2

Week 2 Median: Ascending order: 1,1,1,3,3,8,8 Median = 4th observation = 3 Lower quartile range= 1 Upper quartile range= 8 Interquartile range = 8-1 = 7

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 41

Explanation: Yashi said that thinks that the variation in bean plant growth was about the same as the variation in tomato plant growth. It is a true statement because the range of both bean and tomato plants’ growth is the same and they have the same medians.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 42

Answer: more than; more than

Explanation: The mean of boys height: Mean = 72+68+70+56+58+62+64/7 = 64.2 Range= 72-56 = 16

The mean of girls height: Mean = 55+60+56+51+60+63+65/7 = 58.5 Range= 65-51 = 14 The mean of the boys’ heights is _more than____ the mean of the girls’ heights. The range of the boys’ heights is _more than____ the range of the girls’ heights.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 43

Answer: less than; interquartile range; less than; interquartile range; warmer than; more than

Explanation: The mean of City 1’s temperatures is the —-less than———- mean of City 2’s temperatures. The –interquartile range——— of City 1’s temperatures is—less than—– the —-interquartile range—–of City 2’s temperatures. So, City 2 is typically —warmer than—–City 1, but City 2’s temperatures vary —–more than——-City 1’s temperatures.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 44

Answer: Lake A’s average is greater but varies more.

Explanation: Lake A Median: Ascending order: 4,6,7,9,10,12 Median = Mean 3rd and 4th observation = 7+9/2 = 8 Range = 12 – 4 = 8

Lake B Median: Ascending order: 4,4,5,6,6,7 Median = Mean 3rd and 4th observation = 5+6/2 = 5.5 Range = 7 – 3 = 4

Question 3. Mrs. Mack measured the heights of her students in two classes. Class 1 has a median height of 130 cm and an interquartile range of 5 cm. Class 2 has a median height of 134 cm and an interquartile range of 8 cm. Write a statement that compares the data. Type below: _________________

Answer: Class 2 is greater but varies more.

Explanation: The interquartile range is the difference between the lower and upper quartiles. Since the interquartile range is more for class 2 we can say that the extremes are greater while the interquartile range is less for class 1 which means that the data is compact.

Question 4. Richard’s science test scores are 76, 80, 78, 84, and 80. His math test scores are 100, 80, 73, 94, and 71. Compare the medians and interquartile ranges. Type below: _________________

Answer: Medians are equal but the interquartile range varies a large, math test scores are more spread out compared to science test scores.

Explanation: Science test scores: Median: Ascending order: 76,78,80,80,84 Median = 3rd observation = 80 Interquartile range = 84 – 76 = 10

Math test scores: Median: Ascending order: 71,73,80,94,100 Median = 3rd observation = 80 Interquartile range = 100 – 71 = 29

Medians are equal but the interquartile range varies a large, math test scores are more spread out compared to science test scores.

Question 5. Write a short paragraph to a new student that explains how you can compare data sets by examining the mean and the interquartile range. Type below: _________________

Answer: average and consistency

Explanation: If the mean is more it means that the data has more observations or observations with more value. Interquartile range and median range say about the consistency.

Question 1. Team A has a mean of 35 points and a range of 8 points. Team B has a mean of 30 points and a range of 7 points. Write a statement that compares the data. Type below: _________________

Answer: Similar variation but team A average is more than team B

Explanation: The range has only a difference of 1 point which can be said as a slight variation but while the average/ mean has a large variation.

Question 2. Jean’s test scores have a mean of 83 and an interquartile range of 4. Ben’s test scores have a mean of 87 and an interquartile range of 9. Compare the students’ scores. Type below: _________________

Answer: Ben’s average is more than Jean’s but Ben is less consistent compared to Jean.

Explanation: Ben’s average score is more than the average scores of Jean while the interquartile range

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 45

Answer: 2 students

Explanation: Median of students in a class of school A = 24 Median of students in a class of school B = 26 Difference between the medians of the schools = 26-24 = 2 students

Question 4. The distances in miles that 6 people drive to get to work are 10, 11, 9, 12, 9, and 27. What measure of center best describes the data set? Type below: _________________

Answer: Median is the centre best describes the data set

Explanation: Median: AScending order: 9,9,10,11,12,27 Median= 3rd and 4th observations = 10+11/2 = 21/2 = 10.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 46

Answer: Team A and C

Explanation: Team A and C have medians which nare only slightly different but the consistency varies a lot that is range.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 47

Question 1. What statistical question could Connie ask about her data? Type below: _________________

Answer: Mean, Median, Mode

Explanation: The graph shows the age and number of people. The questions which can be asked can be of mean, median, mode.

Question 2. Describe any peak or gap in the data. Type below: _________________

Answer: Peak : 21-30   Gap : 61-70

Explanation: There is a peak in the graph at the interval 21-30 There is a gap in between the bars of the histogram the bar which had a gap before it was 61-70

Question 3. Does the graph have symmetry? Explain your reasoning. Type below: _________________

Answer: No the graph doesn’t have symmetry

Explanation: The symmetry of the graph means there must be equal parts of the graph on both the sides of the line of the graph. This is not possible in the above situation.

Question 4. The lower quartile of the data set is 16.5 years, and the upper quartile is 51.5 years. Find the interquartile range. Is it a better description of the data than the range? Explain your reasoning. Type below: _________________

Answer: Interquartile range = 35; The interquartile range is better than the range.

Explanation: The interquartile range is the difference between the upper quartile and the lower quartile. Interquartile range = 51.5-16.5 = 35 The interquartile range is better than the range because if we take the example of the above graph we can see thatthe most of the data fall in the range of the interquartile range ie. 35. Therefore we can say that the interquartile range is better than the range.

Question 5. Make Arguments The mode of the data is 16 years old. Is the mode a good description of the center of the data? Explain Type below: _________________

Answer: No mode is not a good description of the data.

Explanation: The mode is just a frequently occurring observation. It cannot be the best way to describe the data.

Use the dot plot for 6–8.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 48

Answer: No, the data has no symmetry.

Explanation: No, the data has no symmetry. Because there are gaps between the dots drawn.

Question 7. Jason claims that the median is a good description of his data set, but the mode is not. Does his statement make sense? Explain. Type below: _________________

Answer: Median can be a better centre of description. Therefore his statement makes a sense.

Explanation: Median = 7+8/2 = 15/2 = 7.5 The number 7.5 represents more number of observations.

Question 8. Trinni surveyed her classmates about how many siblings they have. A dot plot of her data increases from 0 siblings to a peak at 1 sibling and then decreases steadily as the graph goes to 6 siblings. How is Trinni’s dot plot similar to Jason’s? How is it different? Type below: _________________

Answer: Trinni graph represents a part of Jason’s graph

Explanation: In Jason’s graph, there is a peak in the middle and then it decreases on both sides. But according to Trinni graph, there is no peak in the middle.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 49

Answer: Range = 12  Interquartile range = 2

Explanation: The range is the difference between the highest and lowest observations. Range = 12-0 = 12 The interquartile range is the difference between the upper and lower quartiles. Ascending order: 0,0,1,1,2,2,2,2,3,3,3,5,12 Lower quartile = 1+1/2 = 2/2 = 1 Upper quartile = 3+3/2 = 6/2 = 3 Interquartile range = 3-1 = 2

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 50

Question 1. What statistical question could Chase ask about the data? Type below: _________________

Answer: What is the median number of songs purchased?

Explanation: Many questions can be formed from the data given to us We can ask about the mean, median, mode.

Question 2. Describe any peaks in the data. Type below: _________________

Answer: Peak : 0-4

Explanation: The peak is a bar in the histogram which has the highest value. The peak of the given graph is 0-4

Question 3. Describe any gaps in the data. Type below: _________________

Answer: There are no gaps in the graph.

Explanation: The gap is something between a bar of the histogram and all the other adjacent bars. There no such case in the graph. Therefore there are no gaps in the graph.

Question 4. Does the graph have symmetry? Explain your reasoning. Type below: _________________

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 51

Answer: Peaks: At 6 and 8 Gaps: Between 3 and 5 , 6 and 8 Clusters: Between 0-3 ; 9-11

Explanation: The highest points in the graph are known as peaks. They are at 6 and 8 in this graph. The space between the dots in the dot plot graph is known as gaps. Gaps are between 3 and 5; 6 and 8 The group of the dots in the dot plot are known as clusters. The clusters are at the 0-3; 9-11

Question 6. Describe how a graph of a data set can be used to understand the distribution of the data. Type below: _________________

Answer: Mean, median, mode

Explanation: There are three measures of centre which can be used to describe the data given in the form of a graph. The three measures of centre are mean, median, mode.

Question 1. The ages of people in a restaurant are 28, 10, 44, 25, 18, 8, 47, and 30. What is the median age of the people in the restaurant? _______ years old

Answer: Median age of the people in the restaurant is 26.5 approximately 27

Explanation: Median: Ascending order: 8,10,18,25,28,30,44,47 Median = Mean of 4th and 5th observations = 25+28/2 = 53/2 = 26.5

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Explanation: Median is the middlemost value and it is 11 in the above graph. We need to consider the middle value by neglecting the same number on both the sides.

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Answer: 30-33

Explanation: The gap is a space between the intervals. The intervals are 30-33.

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Explanation: The highest point in the graph is known as the peak. The peak in the dot plot is 37.

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Answer: Red and blue

Explanation: The difference between the upper and lowest observations is called a range. The range (consistency) in the data given is the same but they vary in the mean. But we can say that Red and Blue teams typically earned a different number of points per game.

Question 6. Manny’s monthly electric bills for the past 6 months are $140, $165, $145, $32, $125, and $135. What measure of center best represents the data? Type below: _________________

Answer: Median is the best way to represent the data. Median= 137.5

Explanation: Median: Ascending order: 32,125,135,140,145,165 Median = 135+140/2 = 275/2 = 137.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 56

Answer: middlemost value; 20; 15

Explanation: First, look at the median. The median is 20 points. Next, work backwards from the statistics. The median is the middlemost value of the data. So, at least 20 of the values are scores less than or equal to 15. Finally, use the statistics to draw a conclusion.

Question 2. What if a score of 15 or greater resulted in a prize? How would that affect Josh’s decision? Explain. Type below: _________________

Answer: It doesn’t affect his decision.

Explanation: Josh wanted to play only when most of the prizes were awarded. Therefore if the minimum score was 15 or greater than it then they would get the prize. So there will be no difference in his decision.

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Answer: No the manager is not correct.

Explanation: The range is given to be 62. The range is correct when we check it with the help of the given data, therefore, the data is not inconsistent.

Question 4. Gerard is fencing in a yard that is 21 feet by 18 feet. How many yards of fencing material does Gerard need? Explain how you found your answer. _______ yards

Answer: 78 yards of the fencing material is required

Explanation: Length = 21 feet  Breadth = 18 feet The perimeter of the rectangle = The number of yards of fencing material required = 2(l+b) = 2(21+18) = 2(39) = 78 yards

Question 5. Susanna wants to buy a fish that grows to be about 4 in. long. Mark suggests she buys the same type of fish he has. He has five of these fish with lengths of 1 in., 1 in., 6 in., 6 in., and 6 in., with a mean length of 4 in. Should Susanna buy the type of fish that Mark suggests? Explain. Type below: _________________

Answer: Yes Susanna buy the type of fish that Mark suggests

Explanation: The length of fish suggested by Mark is 4 in. long. Mark has 5 fishes with a mean length of 4 in. To buy a fish of that grows to be about 4 in. long. Susanna should buy the fishes suggested by Mark.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 58

Answer: 7 stamps

Explanation: In week 4 and 6, the number of stamps are 4, 5. Therefore in week 6 and 8, the number of stamps are 5, 7

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Answer: According to the question he should spend 3 hours per week. His statement is correct.

Explanation: Sum of the data = 1+2+1+3+2+10+2 = 21 Number of days in a week = 7 Mean = 21/7 = 3 hours

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Question 1. What is the minimum price and the maximum price? minimum: $ _________ maximum: $ _________

Answer: minimum: $6 maximum: $19

Explanation: The minimum value is the most minimum price in the given data. The maximum value is the most maximum price in the given data.

Question 2. What is the mean of the prices? $ ________

Answer: $15

Explanation: Mean = sum of all the observations/ total number of observations = 6+16+18+16+18+19/6 = 93/6 = 15.2 Approximately therefore the mean of the observations is $15

Question 3. Construct a box plot for the data. Type below: _________________

Answer: The box plot is a diagram which signifies the information about the data.

lesson 3 homework practice measures of spread answer key

Question 4. What is the range of the prices? $ ________

Answer: the range is $13

Explanation: The difference between the upper and lower observations is known as the range. Range = 19-6 = $13

Question 5. What is the interquartile range of the prices? $ ________

Answer: $7.5

Explanation: Ascending order: $6, $16, $16, $18, $18, $19 Median = 16+18/2 = 34/2 = 17 Lower quartile = 6+16/2 = 11 Upper quartile = 18+19/2 = 18.5 Interquartile range = 18.5-11 = 7.5

Question 6. Does the menu match Mr. Jackson’s requirements? Explain your reasoning. Type below: _________________

Answer: Yes the menu matches Mr Jackson’s requirements.

Explanation: Mr Jackson wants to make dinner arrangements with cost less than $16. The mean of the items in the menu: Mean = $6+$16+$16+$18+$18+$19/6 = 93/6 = $15.5 Therefore the requirements of Mr Jackson is satisfied.

Question 7. Give an example of a misleading statistic. Explain why it is misleading. Type below: _________________

Answer: The Waterside Inn advertises the misleading statement.

Explanation: According to the information given in the question, The Waterside Inn advertises that they have meals that average $15. But it is more than that, so this is the misleading statement. The mean of the items in the menu: Mean = $6+$16+$16+$18+$18+$19/6 = 93/6 = $15.5 Therefore the requirements of Mr Jackson is satisfied.

Question 1. Mary’s science test scores are 66, 94, 73, 81, 70, 84, and 88. What is the range of Mary’s science test scores? ________

Explanation: The difference between the highest and the lowest observations is called a range. Range = 94 – 66 = 28

Question 2. The heights in inches of students on a team are 64, 66, 60, 68, 69, 59, 60, and 70. What is the interquartile range? ________

Answer: Interquartile range = 9

Explanation: Ascending order: 59,60,60,64,66,68,69,70 Median = Mean of 64 and 66 = 64+66/2 = 130/2 = 65 Lower quartile = 60 Upper quartile = 69 Interquartile range = 69 – 60 = 9

Question 3. By how much does the median of the following data set change if the outlier is removed? 26, 21, 25, 18, 0, 28 Type below: _________________

Answer: The median changes by 3.5

Explanation: Ascending order: 0,18,21,25,26,28 Median = 21+28/2 = 49/2 = 24.5 If the outlier is removed then the Median = 21 Difference between the 1st and 2nd median = 24.5 – 21 = 3.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 61

Answer: Interquartile range = 6

Explanation: The difference between the lower and upper quartiles is known as the interquartile range. Interquartile range = 50 – 44 = 6

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 62

Answer: Mean and median are the best centre of tendencies to compare the data.

Explanation: Mean = 20+20+18+19+23+40+22+19/8 = 181/8 = 22.6 Median: Ascending order: 18,19,19,20,20,22,23,40 Median = 40/2 = 20

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 63

Answer: 1a. Yes 1b. No 1c. Yes 1d. No 1e. Yes

Explanation: 1a. The highest point in the dot plot is called the peak. The peak in the given dot plot is at 5 and 11 the value of the peak is 3 1b. The group of dots form a cluster with 3 or more intervals. 1c. There is a gap between the intervals 6-8 1d. The maximum number of people did 11 chin-ups while only a single person did 15 chin-ups. 1e. The most frequently occurring observation is known as mode. The mode of the given data is at the intervals 3,4 and 9.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 64

Answer: The histogram has 1 peak(s). The histogram is symmetry.

Explanation: The is one and only one peak at the interval 41 – 50 We can say that the graph is symmetrical because if we draw a line between the graph we can observe that the graph has two parts symmetric to each other.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 65

Answer:Median = 6 Lower quartile = 1 Upper quartile =19.5

Explanation: Ascending order: 0,0,1,1,4,5,6,9,13,17,22,30,47 Median = 6 Lower quartile = Mean of 1 and 1 = 1+1/2 = 2/2 = 1 Upper quartile = Mean of 17 and 22 = 17+22/2 = 39/2 = 19.5

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 66

Answer: A= 16 B=17 C= 20 D= 21 E=24

Explanation: Ascending order: 16,17,17,18,19,20,20,21,21,21,22,24 Median = 20+20/2 = 20 Lower quartile = 17 Upper quartile = 21

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 68

Answer: 5a.  false 5b. true 5c. true 5d. true

Explanation: Median is the middlemost value of the given data. Median of the data is 14 The range is the difference between the upper and lower observations. Range = 21-6 = 15 The interquartile range is the difference between the upper and lower observations. Upper quartile range: 18 Interquartile range = 18-9 = 9

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 69

Answer: The mean number of glasses of water Dalia drinks each day is 8 glasses.

Explanation: Mean = sum of all the observations/ total number of observations= 6+7+9+9+8+7+10/7 = 8

Question 6. Part B What is the mean absolute deviation of the number of glasses of water Dalia drinks each day? Round your answer to the nearest tenth. Use words and numbers to support your answer. _______

Answer: Mean absolute deviation is 1.14

Mean: Mean = sum of all the observations/ total number of observations= 6+7+9+9+8+7+10/7 = 8

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 06: 8-6  = 02 07: 8-7  = 01 09: 8-9  = -1 09: 8-9  = -1 08: 8-8  =  0 07: 8-7  = 01 10: 8-10= -2

STEP 2 Find the mean of the distances. 2+1+1+1+0+1+2/7 = 8/7 = 1.14

So, the mean absolute deviation of the data is 1.14

Question 7. The numbers of emails Megan received each hour are 9, 10, 9, 8, 7, and 2. The mean of the data values is 7.5 and the median is 8.5. Which measure of center better describes the data, the mean or median? Use words and numbers to support your answer. Type below: _________________

Answer: Mean is the best centre of tendency to represent the data given in the question

Explanation: Ascending order of the data: 2,7,8,9,9,10 Mean = 7.5 Mean represents the observations 8,9,9,10 which come after 7.5 Therefore mean is the best way to represent the data.

Question 8. The number of miles Madelyn drove between stops was 182, 180, 181, 184, 228, and 185. Which measure of center best describes the data? Options: a. mean b. median c. mode

Answer: b. Median

Explanation: Ascending order: 180,181,182,184,185,228 Median = 182+184/2 = 183 183 represents all the observations after 182 So the median is the best way to represent the data.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 70

Answer: 321-330 ; 341-350

Explanation: The histogram has 2 intervals which show equal heights which means that the monthly earnings of these intervals is the same.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 71

Answer: b. There is a gap from 38 to 40.

Explanation: The dot plot represents a gap between 38-40. So we can say that there is a gap between the intervals 38 to 40.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 72

Answer: There are 2 peaks, Yes there is an increase across the intervals, Yes there is a decrease across the intervals

Explanation: The highest point in the histogram is called is as a peak. There is a peak near the month’s May and August.

There is an increase between the bars in the bar graph. At the months February, March, November there is an increase in the graph.

There is a decrease between the bars in the bar graph. At the months September, October, December.

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 74

Answer: Range = 8-0 = 8 Interquartile range = 3-1 = 2

lesson 3 homework practice measures of spread answer key

Answer: Yes I agree that it is a good place to move.

Explanation: After analyzing the data we can say that this country would be a good place to move for Gavin. Gavin wants to move to a place which has an average of 5 cm rainfall. So this country is the best choice because it has a mean equal to 5 and median equal to 4.35

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 76

Answer: No Peyton’s statement is not misleading because the mean of the data is 4. Therefore Peyton says she usually reads 4 books per month.

Explanation: Mean = 2+3+2+4+3+11+3/7 = 28/7 = 4

Go Math Grade 6 Answer Key Chapter 13 Variability and Data Distributions img 77

Answer:15a. No 15b. Yes 15c.  Yes

Explanation: Mean of player A = 90+90+90/3 = 90

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 90: 90-90  = 0 90: 90-90  = 0 90: 90-90  = 0

STEP 2 Find the mean of the distances. 0+0+0/3 = 0 So, the mean absolute deviation of player A is 0

Mean of player B = 110+100+90/3 = 100

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 100: 100-110  = -10 100: 100-100  = 0 100: 100-90    = 10

STEP 2 Find the mean of the distances. 10+0+10/3 = 20/3 = 6.67 So, the mean absolute deviation of player B is 6.67

Mean of player C = 95+100+95/3 = 96.67

STEP 1 Label each observation with its distance from the mean. Starting from left to right: 96.67: 96.67-95    = 1.67 96.67: 96.67-100  = -3.33 96.67: 96.67-95    = 1.67

STEP 2 Find the mean of the distances. 1.67+3.33+1.67/3 = 6.67/3 = 2.22 So, the mean absolute deviation of player C is 2.22

Conclusion:

Hope the solutions provided in this Go Math Grade 6 Answer Key Chapter 13 are helpful for all the students. Download the Answer Key of Go Math Grade 6 Chapter 13 Variability and Data Distributions and start your preparation now. Stay with us to get the solutions of all Grade 6 chapters.

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Measures of Spread (Lesson with Homework)

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Algebra I - Measures of Spread (Dispersion)

Common Core Aligned Lesson Plan with Homework

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Lesson Objectives:

-Find the different measure of spread (Mean Absolute Deviation, Variance, and Standard Deviation).

-Analyze data sets using different measures of spread, and describe how they are different.

Common Core Standards:

S.ID.2 - Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.

S.ID.3 - Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).

----Chapter: Introduction to Probability and Statistics----

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Introduction to Probability and Statistics Lesson Plan Bundle

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Lesson 3 Homework Practice Measures Of Variation Answer Key

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Lesson 3 homework practice measures of variation answer key – Embark on a journey into the realm of data analysis with Lesson 3 Homework Practice: Measures of Variation Answer Key. This comprehensive guide delves into the significance of measures of variation, their applications, and their interpretation, providing a solid foundation for understanding data distribution and variability.

Measures of variation, including range, variance, and standard deviation, are essential tools for statisticians and researchers. They quantify the spread or dispersion of data, offering valuable insights into the underlying patterns and relationships within datasets. This guide will equip you with the knowledge and skills to calculate and interpret these measures effectively.

Measures of Variation: Lesson 3 Homework Practice Measures Of Variation Answer Key

Lesson 3 homework practice measures of variation answer key

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  1. Measures of Spread

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  4. Results for measures of spread

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  6. Lesson 3 Reteach Measures Of Variation Answer Key

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  1. PDF NAME DATE PERIOD Lesson 3 Homework Practice

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    Course 2 • Chapter 8 Measure ... cGraw-Hill Comp anies, Inc. Perm ission is granted to repr oduce for c lassr oom use. Lesson 3 Homework Practice Area of Composite Figures Find the area of each figure. Round to the nearest tenth if necessary. 1. 5 mi 12 mi 8 mi 18 mi 2. 5.9 cm 3.6 cm 1.1 cm 4.8 cm 3. 5 ft 4 ft 4. 8 m 6 m 10 m 6 m 20 m 5. 8 yd ...

  3. PDF NAME DATE PERIOD Lesson 3 Homework Practice

    Lesson 3 Homework Practice Measures of Variation 1. Use the data in the table. Weights of Black Bears (lb) 277 448 279 334 132 599 237 251 183 191 a. Find the range of the data. b. Find the median and the first and third quartiles. c. Find the interquartile range. d. Name any outliers in the data.

  4. PDF NAME DATE PERIOD Lesson 3 Homework Practice

    Lesson 3 Homework Practice Misleading Graphs and Statistics ANTIQUES For Exercises 1-3, use the table. 1. Find the mean, median, and mode of the data. 2. Which measure might be misleading in describing the value of each item? Explain. 3. Which measure would best describe the value of each item? Explain. MOUNTAINS For Exercises 4 and 5, use ...

  5. PDF 3.3 Statistical Studies Variations (Spread) Name: A company is shooting

    7. What does the difference in the measures of variability (spread) suggest? 8. Using your measures, explain which student you think the teacher should choose and why. 9. Matching: Has a standard deviation of 0. A. 1, 7, 1, 7, 7,1 Has a standard deviation of 1. B. 1, 3, 1, 3, 1, 3 Has a standard deviation of 2. C. 4, -2, 8, 2, 4, 2

  6. Statistics 3.2: Measures of Variations Flashcards

    These measures determine the spread of a data set. What is variance? Variance is denoted as s^2, and it is the average square difference between each data value and the mean. The variance formula is: s^2 = nEx^2 - (Ex)2 / n (n-1) How do you find variance? Ex. 8, 12, 10, 10, 20, 6, 4, 8. 1. You usually find the range first, which in this case ...

  7. PDF Lesson 3 Homework Practice Measures Of Variation Answer Key

    Lesson 3 Homework Practice Measures Of Variation Answer Key. Lesson 3 Homework Practice Measures Of Variation Answer Key. I The answer is 6. II. the answer is 28. THINGS TO REMEMBER: Print the chart with the question numbers on the left, the data sets on the top and. The measure of the central tendency for a sample set of.

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    Lesson 3 Homework Practice Box Plots Draw a box plot for each set of data. 1. ages of children taking dance classes: 10, 8, 9, 7, 10, 12, 14, 14, 10, 16 2. prices, in dollars, of bicycles: 150, 134, 136, 120, 145, 170, 125, 130, 145, 190, 140 3. PRODUCTS Use the box plot that shows the average prices in cents per pound farmers received for eggs ...

  10. Section 3.2: Measures of Spread Flashcards

    Study with Quizlet and memorize flashcards containing terms like Measure of Spread #1:, Measure of Spread #2:, Population Variance: and more.

  11. *STAT 1401

    Step 2: S^2 = 10 / 5. Step 3: S^2 = 2. Compute the Standard Deviation of a Population and a Sample. because the variance is computed using standard deviations, the units of the variance are the squared units of the data. in most situations, it is better to use a measure of spread that has the same units as the data.

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    Lesson 3 Skills Practice Angles of Triangles Find the value of x in each triangle with the given angle measures. 1. 81 x 84 2. 24 36 x 3. 49 ...

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    Lesson 3 Skills Practice Measures of Variation Find the range, median, first and third quartiles, and interquartile range for each data set. Name any outliers. 1. Number of Boxes of Popcorn Sold 52 72 96 21 58 40 75 2. Number of Text Messages Sent 20 23 18 4 17 21 15 56 ...

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    About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...

  15. Student Answer Keys

    Answer Key - Chapter 25 (31.0K) Answer Key - Chapter 26 (36.0K) To learn more about the book this website supports, please visit its Information Center .

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    Standardized Test Practice Vocabulary Review Lesson Resources Extra Examples Personal Tutor Self-Check Quizzes. Common Core State Standards Supplement, SE Hotmath Homework Help Multilingual eGlossary Math Review Math Tools Study to Go Online Calculators. Mathematics. Home > Chapter 13 > Lesson 3. Oklahoma Pre-Algebra. Chapter 13, Lesson 3 ...

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    Lesson 3 Homework Practice Mean Absolute Deviation Answer Key Question 5. ... Answer: Case 2 is more spread out. Explanation: CASE1 The number of laps Shawna swam on 5 different days: 5,6,6,8,10 Mean = 7 ... Lesson 13.3 Practice A Data Distributions Answer Key Question 4.

  19. PDF Lesson 3 Homework Practice

    Find the range of the data. Find the median and the first and third quartiles. Find the interquartile range. Find any outliers in the data and name them. 3. TRAIN The table shows the number of riders on the train each day for two weeks. Compare and contrast the measures of variation for both weeks.

  20. Section 3.3 Measures of Variation Flashcards

    The measure of variation equal to the square of the standard deviation. The units are the squares of the units of the original data values. This value is never negative. Unlike the Empirical Rule, this applies to ANY Data set. It states: the proportion (or fraction) of any set of data lying within K standard deviations of the mean is always at ...

  21. Measures of Spread (Lesson with Homework) by Infinitely Pi Learning

    Algebra I - Introduction to Probability and Statistics Lesson Plan BundleThis bundle includes:Unit Outline (Including Links to Extra Resources)Measures of Central TendencyMeasures of SpreadRepresenting DataDistributions of DataNormal Distribution and Z-ScoreSimple Probability and OddsPermutations an. 8. Products. $17.00 $20.00 Save $3.00.

  22. PDF NAME DATE PERIOD Lesson 5 Homework Practice

    Lesson 5 Homework Practice Measures of Variation For Exercises 1 and 2, find the mean absolute deviation of each set of data. Round to the nearest tenth if necessary. Describe what the mean absolute deviation represents. 1. Exercise Time (min) 45 100 60 75 75 90 90 105 2. Food Items Donated 38 44 46 47 52 55 3. Refer to the table in Exercise 1.

  23. Lesson 3 Homework Practice Measures Of Variation Answer Key

    Measures of Variation: Lesson 3 Homework Practice Measures Of Variation Answer Key. Measures of variation are statistical tools used to quantify the spread or dispersion of data. They provide valuable insights into the variability and distribution of data, enabling researchers and analysts to make informed decisions and draw meaningful conclusions.