DISTRIBUTIONS AND VARIABILITY - PDF Free Download

LESSON 2: DATA ABOUT US 1. Check your work. Did you answer and enter all the questions correctly? 2. a. Answers will vary. Ask your partner if your question makes sense. b. Answers will vary. Ask your partner if your data collection method seems reasonable. c. Answers will vary. Ask your partner if he agrees with your prediction. 3. Answers will vary. Possible answer: Typical data will be where the majority of the data points are. Typical data could also be the middle value, or the average of the values. 4. a. How many teeth lost is a numerical answer given as a whole number. b. The time it takes to run a mile is a numerical answer given in minutes and seconds. c. Which type of bagel a typical sixth grade student likes best is not a numerical answer. The answer is a type of bagel. 5. Answers will vary. Possible answer: The answers to these questions would change if answered by a third grader: How many text messages do you send each day? Third graders would probably send fewer text messages; they may not have a cell phone. What time do you get up on a school day? Third graders may start school at a different time than sixth graders or take less time to get ready for school. How high can you reach? The reach of third graders is definitely different than that of sixth graders, since third graders are generally shorter. How much time do you spend on homework each week? Third graders probably spend less time on homework than sixth graders. How many hours a week do you watch television? Third graders responses may be different than those of sixth graders. Copyright 2015 Pearson Education, Inc. 32

LESSON 4: MEAN 1. A 7 (3 + 4 + 6 + 8 + 9 + 9 + 10) 7 = 49 7 = 7 2. C 9 3, 4, 6, 8, 9, 9, 10 There are two 9s. 3. B 8 3, 4, 6, 8, 9, 9, 10 The middle value is 8. 4. The mean high temperature is 65 F. (62 + 65 + 63 + 64 + 67 + 66 + 68) 7 = 455 7 = 65 5. a. The mean is 5. (2 + 4 + 7 + 3 + 8 + 6) 6 = 30 6 = 5 b. Answers will vary. Possible answer: The mean is different than the other values in the data set; there is no data value of 5. 6. a. The mean is 5.5. (5 + 3 + 9 + 6 + 4 + 6) 6 = 33 6 = 5.5 b. Answers will vary. Possible answer: The mean is different than the other values in the data set; there is no data value of 5.5. The mean is also not a whole number, while the values in the data set are all whole numbers. 7. Answers will vary. Possible answer: 0, 7, 7, 8, 8 (0 + 7 + 7 + 8 + 8) 5 = 30 5 = 6 The mean is 6. 8 0 = 8 The range is 8. Copyright 2015 Pearson Education, Inc. 33

LESSON 5: MEAN ABSOLUTE DEVIATION 1. B 8 (2 + 4 + 8 + 8 + 8 + 8 + 12 + 14) 8 = 64 8 = 8 2. B 2.5 (2 + 4 + 8 + 8 + 8 + 8 + 12 + 14) 8 = 64 8 = 8 The mean is 8. MAD: 8 2 = 6 8 4 = 4 8 8 = 0 8 8 = 0 8 8 = 0 8 8 = 0 12 8 = 4 14 8 = 6 (6 + 4 + 0 + 0 + 0 + 0 + 4 + 6) 8 = 20 8 = 2.5 3. A 1, 3, 9, 14, 27 Data set A has the largest range, which means it is the most spread out and has more data points farther away from the mean. 4. Answers will vary; all of the data values must be the same. 5. The mean is 4.4, not 4. Therefore, the calculation of the MAD is incorrect. (3 + 3 + 3 + 4 + 4 + 5 + 5 + 5 + 6 + 6) 10 = 44 10 = 4.4 6. Project a. Ask a classmate to read your revised question. b. Schedules will vary. Have a classmate comment on your schedule. 7. No. Possible answer: Since the values must be different, the narrowest range will be 4 for a set of 5 values: 0, 1, 2, 3, 4. This gives a mean of 2. The MAD is greater than 1 because the sum of the distances from the mean is 6, which is more than the number of data values. As the number of data values increases by 1, the differences from the mean will increase by more than 1. So the total will always be greater than the number of data values, resulting in a MAD greater than 1. Copyright 2015 Pearson Education, Inc. 34

LESSON 6: MEAN, MEDIAN, AND MODE 1. Mean = (0 + 6 + 8 + 9 + 9 + 11 + 11 + 11 + 12 + 13) 10 = 90 10 = 9 Median = 10; there are five values greater than 10 and five values less than 10. Mode = 11 2. A Half of the students scored 9 or 10 on the quiz. If the median is 9, half the scores were 9 and 10. B More than one student scored below 6. If the mean is 7, then some students scored very low. 3. A Math Quiz Class 1 0 1 2 3 4 5 6 Only in figure A are the mean, mode, and median the same value, which in this case is 3. 4. If one data point is moved from 15 to 3, the mean, median, mode, and range will change. The mode will change from 15 to 15 and 16. The mean will decrease because the total of the data will be 12 less (15 3 = 12). The range will change from 12 (6 to 18) to 15 (3 to 18). The median will change from 14.5 to 14 because the data point at 15 is moving from above the median to below it. 5. The data set has an equal number of values that are greater than 8 and less than 8. 6. Answers will vary. Possible answer: 0, 0, 0, 1, 3, 5, 12 Mean: (0 + 0 + 0 + 1 + 3 + 5 + 12) 7 = 21 7 = 3 Median: 1 Copyright 2015 Pearson Education, Inc. 35

LESSON 6: MEAN, MEDIAN, AND MODE 7. Answers will vary. Possible answer: Move one point from 1 to 2, and the point at 4 to 3. Mean: (0 + 1 + 2 + 3 + 3 + 3 + 5 + 5 + 6) 9 = 28 9 = 3.1 Median: 3 Math Quiz Class 1 0 1 2 3 4 5 6 Copyright 2015 Pearson Education, Inc. 36

LESSON 7: DESCRIBE A SITH GRADE STUDENT 1. C 60 in. A typical height for a sixth grader would be about 60 in. 2. A The range is fairly narrow. C The mode, median, and mean are close to 9.5 in. 3. A There is an outlier. B The measures of center are all different. 4. A typical value for this set of data is 14 or 15. Although the data is spread out, you can see that 1 is an outlier and that there is a cluster of data. The mode is 15 and the median is 14. The mean is close to 14 as well (about 13.77). If you look at the middle 50% of the values (15 out of 30), you can see they are clustered around 14 and 15. Math Test 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 mean mode median 5. Answers will vary. Possible answer: The line plot probably has a fairly wide range since none of the measures are the same. Since the mean is greater than the median, there may be a small cluster of data, or an outlier, toward the higher end of the range. The three measures are fairly close though, so there may be a cluster of data around 34 to 37. 6. Project Answers will vary. Discuss any changes to your data collection method with your partner. Copyright 2015 Pearson Education, Inc. 37

LESSON 7: DESCRIBE A SITH GRADE STUDENT 7. Answers will vary. Possible answer: Since 3 is not the typical score, it is possible that there are no scores of 3. The data could be clustered at either end of the range. Math Quiz Class 1 0 1 2 3 4 5 6 The mean and median are both 3, but there are no data values at 3. The mode is 0 and 6, indicating two clusters of data. So 3 is clearly not the typical score. Copyright 2015 Pearson Education, Inc. 38

LESSON 8: MATCHING LINE PLOTS WITH DATA 1. The median is incorrect. The median is 5; it is the middle of the 11 values with 5 values at or below 5 and 5 values above 5. The mean is 5: (2 + 4 + 4 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 6) 11 = 55 11 = 5. The mode is 6, since there are the most data points at 6. The range is 4: 6 2. 2. C Line Plot C 1 2 3 4 5 6 The range of A is 4; the median of D is 3. Therefore, neither A nor D can be the correct answer. So the answer is either B or C. The range of B and C is 5; the mode is 2. The mean of B is (1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 4 + 4 + 6) 11 = 29 11 3. The answer cannot be B. The mean of C is (1 + 2 + 2 + 2 + 2 + 2 + 3 + 4 + 4 + 5 + 6 ) 11 = 33 11 = 3. The median of C is 2. Therefore, C is the correct answer. 3. D Mean: 3 Median: 3 Mode: 2 Range: 5 The mean is (1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 6) 11 = 33 11 = 3. The median is 3: 3 is the middle value. The mode is 2. There are three data points of 2, which is the most. The range is 5: 6 1 = 5. Copyright 2015 Pearson Education, Inc. 39

LESSON 8: MATCHING LINE PLOTS WITH DATA 4. Answers will vary. Possible answer: 1 2 3 4 5 6 5. The mean is not correct. If the median is 4, half of the data values are 4 or less. So, there is no way that the average of the data is 5. 6. Answers may vary. Possible answer: 1 2 3 4 5 6 There must be at least one data point at 1, 3, and 6 in order to meet the median and range parameters. The sum of the data values must be 13 4 = 52 in order for the mean to equal 4. Since the median is 3, there must be six points between 1 and 3, and six points between 3 and 6. Starting with this information, data points can be added and adjusted to create an appropriate line plot. Copyright 2015 Pearson Education, Inc. 40

Grade 6 Unit 8: Distributions and Variability LESSON 9: BO PLOTS 1. B 6 The median is 8; the lower quartile is halfway between 4 and 8. 2. A 4 The median is 7.5; the lower quartile is 6; the upper quartile is 10; the IQR is 4. 3. The whiskers would be very short because a lot of data would be close to either extreme. However, the box would be long in comparison because the data in the box is pushed to either end. Here is an example: 0 1 2 3 4 5 6 7 8 9 10 4. There is probably an outlier at the upper extreme. Half of the data is tightly clustered (within the box) and 25% more is in an even narrower range next to it. So, 75% of the data is near the lower extreme. Only 1 in the long right whisker. 4 of the data is contained 5. A The median is 14. The lower extreme is 8; the upper extreme is 20. Based on the line plot the lower quartile is 12 and the upper quartile is 16. The values are clustered about the median. The graph that shows this is A. 6. Project Answers will vary. Share your successes and challenges with a classmate. 7. The median is 15. The lower quartile is 7.5 and the upper quartile is 22.5. The interquartile range is 15. Copyright 2015 Pearson Education, Inc. 41

LESSON 10: EPLORING BO PLOTS 1. B 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 The median is 13.5, the lower quartile is 12; the upper quartile is 16. 2. B 9 If there are 36 data points, then 1 4 or 9 points are in each quartile. 3. Many of the data points can be moved. There are 24 data points so each quartile contains 6: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Data within each quartile can be moved, as long as they stay in that quartile and at least one data point remains at each of the five-number summary values. Here is one possibility: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 4. C 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 For the data 0, 18, 19, 20, the median is 18.5. The lower quartile is 9 and the upper quartile is 19.5. Box plot C shows this. Copyright 2015 Pearson Education, Inc. 42

LESSON 10: EPLORING BO PLOTS 5. Answers will vary. Possible answer: The box plot shows the upper and lower extremes, so 0 and 20 must be data values. The upper quartile is 16.5, so it could be a data value. If not, there are two data values that add to 33, such as 16 and 17, since 16.5 is halfway between (33 2 = 16.5). Similarly, 5 and 6 could be data values, or two numbers that add to 11, since the lower quartile is 5.5 (11 2 = 5.5). Copyright 2015 Pearson Education, Inc. 43

LESSON 11: HISTOGRAMS 1. C A data set of the ages of 10,000 people living in a town. The histogram would be the easiest method of showing this information. 2. D There is an outlier between 0 and 10. Measures of center and spread cannot be found by looking at a histogram. 3. A 55 people Add the frequencies to find the number of people. 4. Answers will vary. Possible answer: A typical person who came to Back to School Night was around 30 to 40 years old. While there was a range from under 10 to over 60, most of the data were clustered around the middle interval of 30 to 40, with the amount of data tapering off to either side. 5. The histogram would be flat, except for possibly the last bin. Since the data is evenly distributed across the range, each bin will have the same amount of data, except possibly the last bin. If the bin width were 9, there would be 9 bins of 11 data points, and a 10th bin with 1 data point. If the bin width were 10, there would be 10 bins of 10 data points each. 6. Project Answers will vary. Share your advice with a classmate. 7. The graphs would be similar because they would each have four sections. However, the four sections in the box plot, the quartiles, each contain the same amount of data, whereas the four sections of the histogram, the bins, each have the same range but not necessarily the same amount of data. Copyright 2015 Pearson Education, Inc. 44

LESSON 12: WHICH BIN IS BEST? 1. C 10 2. As long as the number of data values stays the same for each bin, the data can be moved in many ways. How Long Can Sixth Grade Students Hold Their Breath? 5 10 15 20 25 30 35 40 45 50 Seconds Here is one possibility, making the line plot even flatter: How Long Can Sixth Grade Students Hold Their Breath? 5 10 15 20 25 30 35 40 45 50 Seconds 3. D The mode is 84. The mode cannot possibly be 84 since there is a frequency of 1 for a value between 80 and 90. There are much greater frequencies for values between 40 and 50. Copyright 2015 Pearson Education, Inc. 45

LESSON 12: WHICH BIN IS BEST? 4. Answers will vary. Possible answer: A bin width of 5 would be best to show the shape of the data. If a bin width of 10 was used, the bins would have about the same frequency and the clusters of data would not show. A bin width of 5 shows the clusters of data because every other bin is empty or almost empty. Frequency 8 7 6 5 4 3 2 1 10 15 20 25 30 35 40 45 50 10 15 20 25 30 35 40 45 50 5. B Data with a range of 200 and with 1,000 data points 6. Histograms are useful for large sets of numerical data, especially when the shape of the data is the important aspect to see, even if the range is narrow. Copyright 2015 Pearson Education, Inc. 46