Chapter 1 Exploring Data
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Chapter 1 Exploring Data Chapter Review
(pp. 72–77)
1. a. Subtract the smallest data point from the largest so 20 – 4 = 16. b. There are 10 data points, so the median will be the mean of the fifth and sixth data points. 14 +2 16 = 15 c. With 10 data points, Q1 is the third data point or 12, and Q3 is the eights data point or 18. IQR = Q3 – Q1 = 18 – 12 = 6 2. a. Add all of the data points and divide by the number of data points. The sum of the data points is 140. 140 = 14 10 b. Subtract the mean from each data point, square it, add them together, and divide by the number of samples minus 1. The sum of the squared = 24 deviations is 216. 216 9 c. Take the square root of the variance. 24 = 2 6 4.9 3. a. With 14 data points the median will be the mean of the seventh and eighths data points. 24 + 29 = 26.5 years 2 b. Add all of the data points and divide by the number of data points. The sum of the data points is 396. 396 28.3 years 14 c. Find the data point(s) with the highest frequency. 9 years and 42 years d. Subtract the smallest data point from the largest, so 61 – 2 = 59 years. e. Subtract the mean from each data point, square it, add them together, and divide by the number of samples minus 1. Then take the square root. The sum of the squared deviations is about 4724.9.
4724.9 13
19.1 years
4. a. Find the smallest value. 2 years b. Find the largest value. 61 years 23
Functions, Statistics and Trigonometry Solution Manual
c. In a data set of 14, Q1 will be the fourth smallest data point. Q1 = 12 years d. In a data set of 14, Q2 will be the mean of the seventh and eighth data points. 24 +2 29 = 26.5 years e. In a data set of 14, Q3 will be the fourth largest data point. Q3 = 42 years 5. The minimum is the smallest data point. Min = 1,045 square miles. The maximum is the largest data point. Max = 571,951 square miles. Since there are twelve data points the median is the mean of the sixth and seventh data 56,804 points. Median = 55,584 + = 2 56,194 square miles. Q1 is be the mean of the third and fourth smallest data 53,927 = 49,372 points. Q1 = 44,817 + 2 square miles. Q3 is the mean of the third and fourth largest data points. Q3 = 155,959 + 145,552 2
= 150,755.5 square
miles. 6. a. The IQR is the distance between Q1 and Q3 , so Q1 + IQR = Q3 . 99 + 50 = 149 b. Outliers are data points less than Q1 1.5(IQR) or greater than Q3 + 1.5(IQR). Data points above 75 + 149 = 224 or data points below 99 – 75 = 24 are outliers. 7. Use a weighted mean with the number of months in each season as weights. 5(70) + 7(10) = 35 cm per month 5+ 7 8. Use credits as weights for a weighted 24(4) + 16(3.3) mean. = 3.72 24 + 16 9. Use a weighted average with percent of girls in each category as weights. 0.3(3) + 0.2(5) + 0.5(7) = 5.4 notebooks 10. Use a weighted average with number of classrooms as weights. 14(1) + 6(2) + 3(0) + 1(3) 1.21 computers 14 + 6 + 3 + 1 per classroom Chapter 1, Chapter Review
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11. Set the weighted average equal to the desired grade and solve for the missing variable. 0.6(82) + 0.2(95) + 0.2(x) = 86 so x = 86 0.268.2 = 89 12. Set the weighted average equal to the desired grade and solve for the missing 3(4) + 2(x) = 4.2 so 2112 = 4.5 variable. 2 5 8
gi
13. a.
i =1
8
b. The numerator is the sum of all of the data points, which is 145. This divided by 8 is 18.125. c. This is the sum of data points four through seven. 12 + 27 + 16 + 20 = 75 4
14. a.
gi = 10 + 24 + 14 + 12 = 60;
i=1 8
gi = 27 + 16 + 20 + 22 = 85;
i=5 4
8
i=1
i=5
gi – gi = 60 – 85 = -25
b. Karen scored 25 more points in the second half of the season than in the first half. 9
ns ps
15. a.
s =1 9
ns
s =1
b. The numerator is the sum of the price times the number of shares purchased at that price, or total money spent. This equals $11,650. The denominator is the sum of the number of shares purchased, or 1,900. $11,650 $6.13 per share 1,900 16. The mean is usually affected more by extreme data points. The median is always affected by the same amount by a data point above the mean, regardless of how extreme it is, but the mean is affected increasingly by more extreme values. 17. a. false; the mode is the data point with the highest frequency and the mean is the sum of all of the data points divided by the total number of data points. 24
Functions, Statistics and Trigonometry Solution Manual
b. Answers vary. Sample: {1, 1, 2, 3, 4, 5, 6, 6}; the mean is 3.5 and the modes are 1 and 6. 18. true; A percentile is a value such that the value percent of the data is less than or equal to some number. 4 out of 8 ordered points are less than or equal to the fourth lowest value and 4 out of 8 is 50%. 19. C 20. B; The variance is defined as the sum of the squared deviations from the mean value. This only uses the mean and the data points, not the median, IQR, or mode. 21. C; The standard deviation is defined as the square root of the variance. 22. Use a weighted average with number of scores as weight and solve for the 49(26) + 1(x) = 26.4 so missing score. 50 x = 1, 320 1, 274 = 46 23. a. The minimum is the smallest data point. Min = 48. The maximum is the largest data point. Max = 94. In a data set with 20 data points, Q1 will be the mean of the fifth and sixth lowest data points, the median will be the mean of the tenth and eleventh lowest data points, and Q3 will be the mean of the fifth and sixth highest data points. Q1 = 55.5, median = 61.5, Q3 = 80 b. IQR = 24.5. Data points more than 1.5(IQR) away from either Q1 or Q3 are outliers so data smaller than 55.5 – 1.5(24.5) = 18.75 or larger than 80 + 1.5(24.5) = 116.75 are outliers. None of the data points is an outlier. 24. The first class has more variability in height. The standard deviation is a measure of spread and variability, so the class with the higher standard deviation, if the means are approximately the same, will be the one with the most variability. 25. a. The minimum is the smallest data point. Min = 97.9 thousand dollars. The maximum is the largest data point. Max = 630.0 thousand dollars. In a data set with 12 data Chapter 1, Chapter Review
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points, Q1 will be the mean of the third and fourth lowest data points, the median will be the mean of the sixth and seventh lowest data points, and Q3 will be the mean of the third and fourth highest data points. Q1 = 119.35 thousand dollars, median = 165.65 thousand dollars, and Q3 = 310.6 thousand dollars. b. Answers vary. Sample: There is a large range of home prices in U.S. metropolitan areas. 26. 5, 125; Because there are no outliers all of the data must fall within 1.5(IQR) of either Q1 or Q3 . 1.5(IQR) = 45 so all data must fall between (inclusive) 5 and 125. 27. a. The mean is the total dollars divided by the number of canvassers. The total collected is $3,768 so the mean = $188.4. The median is is $3,768 20 the average of the tenth and eleventh highest amounts collected so the $163 median is $140 + = $151.5. The 2 sample standard deviation is the square root of the sum of the squared deviations from the mean divided by 19. The sum of squared deviations is 362,689. Dividing by 19 we get about 19,088.89 and taking the square root we get about $138.16. b. Q1 = 95 and Q3 = 231. The IQR = 136. Any outliers will be less than 95 – 1.5(136) = -109 which is impossible, or greater than 231 + 1.5(136) = 435 so 440, 445, and 500 are outliers. c. Using the data points minus the three outliers and the same process we get the mean to be $140.18, the median to be $140, and the standard deviation to be $77.83. 28. Answers vary. Sample: The numbers representing age group categories, or the numbers representing gender categories, or the numbers representing education categories. 29. year; voting-age population (millions), percent of voting-age population 25 Functions, Statistics and Trigonometry Solution Manual
reporting they voted, age, gender, education 30. It means 28.4% of 18 to 20 year olds reported that they voted. 31. We find that there were 111.9 million women and that 60.1% of them reported that they voted in 2004. 60.1% of 111.9 million is about 67.25 million women. 32. We find that there were 65.3 million people that had graduated high school or gotten their GED and 57.5% of them reported that they voted in 1992. 57.5% of 65.3 million people is about 37.55 million people. 33. In each reported year between 1992 and 2004 the percentage of people aged 65 and over reporting that they voted was the largest. 34. The standard deviation of the normal temperatures in Minneapolis-St. Paul is about 21.2, whereas it is 11.0 for Juneau. This indicates a much greater variability in temperatures in Minneapolis St. Paul than in Juneau. 35. Answers vary. Sample: The mean, median, IQR, and standard deviations differ very little between the males and females. These statistics, as well as the box plots and histograms indicate little to no difference in the two-mile running times of males and females. 36.
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c. 15 students scored more than 80 35.71%. points, so 15 42
37.
38. a. The largest bar for both men and women is in the age group 30-34, so that age group has the largest population. Adding them we find a population in this age group of about 22 million. b. The seventh largest bar for both men and women in India is in the age group 30-34, so that age group has the seventh largest population. Adding them we find a population in this age group of about 90 million. c. Answers vary. Sample: India has about 6 times the population of the U.S. in the age interval 0-9. In addition, age groups 0-4 and 5-9 are the largest age groups in India, while age groups 25-29 and 30-34 are the largest groups in the U.S. d. Answers vary. Sample: Due to rapid increases in the general standard of living and health care, India has a rapidly expanding population. The population of the U.S. is stable or declining with a greater percent of the population in the oldest age groups. This may indicate that the standard of living and healthcare are at a generally high level so the rate of improvement is not as dramatic as India’s. 39. a. Add the heights of the intervals in the histogram to get 42. b. False. The median score will be between the 21st and 22nd scores. Looking at the histogram one can see that these scores are in the interval of 70-80. 26
Functions, Statistics and Trigonometry Solution Manual
40. a. $83 b. $91 c. 50%; This is the interquartile range, which holds the middle 50% of the population. d. This is the group of the 25% of people who spent the least at the baseball game. They spent between $49 and $62. e. The second quartile, between Q1 and Q2 , has the greatest spread because it is the longest quartile on the box plot. 41.
42.
43. a. Month 1 2 3 4 5 6
Cumulative Wages $3,250 $7,375 $10,125 $15,500 $20,000 $23,800
b. Total sales are $23,800 so half of total sales is $11,900. Looking at the frequency table we see that the first month with more than this much in total sales was month 4. c.
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44. a. Score 1 2 3 4 5 6 7 8 9 10
Frequency 1 0 2 1 3 3 5 4 0 1
Relative Frequency 0.05 0 0.1 0.05 0.15 0.15 0.25 0.2 0 0.05
b.
27
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Chapter 1, Chapter Review
Chapter 1 Exploring Data Chapter Review
(pp. 72–77)
1. a. Subtract the smallest data point from the largest so 20 – 4 = 16. b. There are 10 data points, so the median will be the mean of the fifth and sixth data points. 14 +2 16 = 15 c. With 10 data points, Q1 is the third data point or 12, and Q3 is the eights data point or 18. IQR = Q3 – Q1 = 18 – 12 = 6 2. a. Add all of the data points and divide by the number of data points. The sum of the data points is 140. 140 = 14 10 b. Subtract the mean from each data point, square it, add them together, and divide by the number of samples minus 1. The sum of the squared = 24 deviations is 216. 216 9 c. Take the square root of the variance. 24 = 2 6 4.9 3. a. With 14 data points the median will be the mean of the seventh and eighths data points. 24 + 29 = 26.5 years 2 b. Add all of the data points and divide by the number of data points. The sum of the data points is 396. 396 28.3 years 14 c. Find the data point(s) with the highest frequency. 9 years and 42 years d. Subtract the smallest data point from the largest, so 61 – 2 = 59 years. e. Subtract the mean from each data point, square it, add them together, and divide by the number of samples minus 1. Then take the square root. The sum of the squared deviations is about 4724.9.
4724.9 13
19.1 years
4. a. Find the smallest value. 2 years b. Find the largest value. 61 years 23
Functions, Statistics and Trigonometry Solution Manual
c. In a data set of 14, Q1 will be the fourth smallest data point. Q1 = 12 years d. In a data set of 14, Q2 will be the mean of the seventh and eighth data points. 24 +2 29 = 26.5 years e. In a data set of 14, Q3 will be the fourth largest data point. Q3 = 42 years 5. The minimum is the smallest data point. Min = 1,045 square miles. The maximum is the largest data point. Max = 571,951 square miles. Since there are twelve data points the median is the mean of the sixth and seventh data 56,804 points. Median = 55,584 + = 2 56,194 square miles. Q1 is be the mean of the third and fourth smallest data 53,927 = 49,372 points. Q1 = 44,817 + 2 square miles. Q3 is the mean of the third and fourth largest data points. Q3 = 155,959 + 145,552 2
= 150,755.5 square
miles. 6. a. The IQR is the distance between Q1 and Q3 , so Q1 + IQR = Q3 . 99 + 50 = 149 b. Outliers are data points less than Q1 1.5(IQR) or greater than Q3 + 1.5(IQR). Data points above 75 + 149 = 224 or data points below 99 – 75 = 24 are outliers. 7. Use a weighted mean with the number of months in each season as weights. 5(70) + 7(10) = 35 cm per month 5+ 7 8. Use credits as weights for a weighted 24(4) + 16(3.3) mean. = 3.72 24 + 16 9. Use a weighted average with percent of girls in each category as weights. 0.3(3) + 0.2(5) + 0.5(7) = 5.4 notebooks 10. Use a weighted average with number of classrooms as weights. 14(1) + 6(2) + 3(0) + 1(3) 1.21 computers 14 + 6 + 3 + 1 per classroom Chapter 1, Chapter Review
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11. Set the weighted average equal to the desired grade and solve for the missing variable. 0.6(82) + 0.2(95) + 0.2(x) = 86 so x = 86 0.268.2 = 89 12. Set the weighted average equal to the desired grade and solve for the missing 3(4) + 2(x) = 4.2 so 2112 = 4.5 variable. 2 5 8
gi
13. a.
i =1
8
b. The numerator is the sum of all of the data points, which is 145. This divided by 8 is 18.125. c. This is the sum of data points four through seven. 12 + 27 + 16 + 20 = 75 4
14. a.
gi = 10 + 24 + 14 + 12 = 60;
i=1 8
gi = 27 + 16 + 20 + 22 = 85;
i=5 4
8
i=1
i=5
gi – gi = 60 – 85 = -25
b. Karen scored 25 more points in the second half of the season than in the first half. 9
ns ps
15. a.
s =1 9
ns
s =1
b. The numerator is the sum of the price times the number of shares purchased at that price, or total money spent. This equals $11,650. The denominator is the sum of the number of shares purchased, or 1,900. $11,650 $6.13 per share 1,900 16. The mean is usually affected more by extreme data points. The median is always affected by the same amount by a data point above the mean, regardless of how extreme it is, but the mean is affected increasingly by more extreme values. 17. a. false; the mode is the data point with the highest frequency and the mean is the sum of all of the data points divided by the total number of data points. 24
Functions, Statistics and Trigonometry Solution Manual
b. Answers vary. Sample: {1, 1, 2, 3, 4, 5, 6, 6}; the mean is 3.5 and the modes are 1 and 6. 18. true; A percentile is a value such that the value percent of the data is less than or equal to some number. 4 out of 8 ordered points are less than or equal to the fourth lowest value and 4 out of 8 is 50%. 19. C 20. B; The variance is defined as the sum of the squared deviations from the mean value. This only uses the mean and the data points, not the median, IQR, or mode. 21. C; The standard deviation is defined as the square root of the variance. 22. Use a weighted average with number of scores as weight and solve for the 49(26) + 1(x) = 26.4 so missing score. 50 x = 1, 320 1, 274 = 46 23. a. The minimum is the smallest data point. Min = 48. The maximum is the largest data point. Max = 94. In a data set with 20 data points, Q1 will be the mean of the fifth and sixth lowest data points, the median will be the mean of the tenth and eleventh lowest data points, and Q3 will be the mean of the fifth and sixth highest data points. Q1 = 55.5, median = 61.5, Q3 = 80 b. IQR = 24.5. Data points more than 1.5(IQR) away from either Q1 or Q3 are outliers so data smaller than 55.5 – 1.5(24.5) = 18.75 or larger than 80 + 1.5(24.5) = 116.75 are outliers. None of the data points is an outlier. 24. The first class has more variability in height. The standard deviation is a measure of spread and variability, so the class with the higher standard deviation, if the means are approximately the same, will be the one with the most variability. 25. a. The minimum is the smallest data point. Min = 97.9 thousand dollars. The maximum is the largest data point. Max = 630.0 thousand dollars. In a data set with 12 data Chapter 1, Chapter Review
Back to Chapter Review
points, Q1 will be the mean of the third and fourth lowest data points, the median will be the mean of the sixth and seventh lowest data points, and Q3 will be the mean of the third and fourth highest data points. Q1 = 119.35 thousand dollars, median = 165.65 thousand dollars, and Q3 = 310.6 thousand dollars. b. Answers vary. Sample: There is a large range of home prices in U.S. metropolitan areas. 26. 5, 125; Because there are no outliers all of the data must fall within 1.5(IQR) of either Q1 or Q3 . 1.5(IQR) = 45 so all data must fall between (inclusive) 5 and 125. 27. a. The mean is the total dollars divided by the number of canvassers. The total collected is $3,768 so the mean = $188.4. The median is is $3,768 20 the average of the tenth and eleventh highest amounts collected so the $163 median is $140 + = $151.5. The 2 sample standard deviation is the square root of the sum of the squared deviations from the mean divided by 19. The sum of squared deviations is 362,689. Dividing by 19 we get about 19,088.89 and taking the square root we get about $138.16. b. Q1 = 95 and Q3 = 231. The IQR = 136. Any outliers will be less than 95 – 1.5(136) = -109 which is impossible, or greater than 231 + 1.5(136) = 435 so 440, 445, and 500 are outliers. c. Using the data points minus the three outliers and the same process we get the mean to be $140.18, the median to be $140, and the standard deviation to be $77.83. 28. Answers vary. Sample: The numbers representing age group categories, or the numbers representing gender categories, or the numbers representing education categories. 29. year; voting-age population (millions), percent of voting-age population 25 Functions, Statistics and Trigonometry Solution Manual
reporting they voted, age, gender, education 30. It means 28.4% of 18 to 20 year olds reported that they voted. 31. We find that there were 111.9 million women and that 60.1% of them reported that they voted in 2004. 60.1% of 111.9 million is about 67.25 million women. 32. We find that there were 65.3 million people that had graduated high school or gotten their GED and 57.5% of them reported that they voted in 1992. 57.5% of 65.3 million people is about 37.55 million people. 33. In each reported year between 1992 and 2004 the percentage of people aged 65 and over reporting that they voted was the largest. 34. The standard deviation of the normal temperatures in Minneapolis-St. Paul is about 21.2, whereas it is 11.0 for Juneau. This indicates a much greater variability in temperatures in Minneapolis St. Paul than in Juneau. 35. Answers vary. Sample: The mean, median, IQR, and standard deviations differ very little between the males and females. These statistics, as well as the box plots and histograms indicate little to no difference in the two-mile running times of males and females. 36.
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c. 15 students scored more than 80 35.71%. points, so 15 42
37.
38. a. The largest bar for both men and women is in the age group 30-34, so that age group has the largest population. Adding them we find a population in this age group of about 22 million. b. The seventh largest bar for both men and women in India is in the age group 30-34, so that age group has the seventh largest population. Adding them we find a population in this age group of about 90 million. c. Answers vary. Sample: India has about 6 times the population of the U.S. in the age interval 0-9. In addition, age groups 0-4 and 5-9 are the largest age groups in India, while age groups 25-29 and 30-34 are the largest groups in the U.S. d. Answers vary. Sample: Due to rapid increases in the general standard of living and health care, India has a rapidly expanding population. The population of the U.S. is stable or declining with a greater percent of the population in the oldest age groups. This may indicate that the standard of living and healthcare are at a generally high level so the rate of improvement is not as dramatic as India’s. 39. a. Add the heights of the intervals in the histogram to get 42. b. False. The median score will be between the 21st and 22nd scores. Looking at the histogram one can see that these scores are in the interval of 70-80. 26
Functions, Statistics and Trigonometry Solution Manual
40. a. $83 b. $91 c. 50%; This is the interquartile range, which holds the middle 50% of the population. d. This is the group of the 25% of people who spent the least at the baseball game. They spent between $49 and $62. e. The second quartile, between Q1 and Q2 , has the greatest spread because it is the longest quartile on the box plot. 41.
42.
43. a. Month 1 2 3 4 5 6
Cumulative Wages $3,250 $7,375 $10,125 $15,500 $20,000 $23,800
b. Total sales are $23,800 so half of total sales is $11,900. Looking at the frequency table we see that the first month with more than this much in total sales was month 4. c.
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44. a. Score 1 2 3 4 5 6 7 8 9 10
Frequency 1 0 2 1 3 3 5 4 0 1
Relative Frequency 0.05 0 0.1 0.05 0.15 0.15 0.25 0.2 0 0.05
b.
27
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Chapter 1, Chapter Review
