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Chapter 3 Data Description Section 3-2 Measures of Central Tendency Chapter 3 Data Description Section 3-2 Exercise #3 61, 11, 1, 3, 2, 30, 18, 3, 7 The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities. Find (a) the mean. Find (b) the median. Find (c) the mode. Find (d) the midrange. 61, 11, 1, 3, 2, 30, 18, 3, 7 The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities. Find (a) the mean. x X= n = 136 9 = 15.1 61, 11, 1, 3, 2, 30, 18, 3, 7 The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities. Find (b) the median. MD: 1, 2, 3, 3, 7, 11, 18, 30, 61 MD: 7 61, 11, 1, 3, 2, 30, 18, 3, 7 The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities. Find (c) the mode. Mode = 3 61, 11, 1, 3, 2, 30, 18, 3, 7 The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities. Find (d) the midrange. MR = 1 + 61 2 = 31 61, 11, 1, 3, 2, 30, 18, 3, 7 Which measure of average might be the best in this case? Explain your answer. x X= n = 136 = 15.1 9 MD: 1, 2, 3, 3, 7, 11, 18, 30, 61 MD = 7 Mode = 3 MR = 31 61, 11, 1, 3, 2, 30, 18, 3, 7 Which measure of average might be the best in this case? Explain your answer. The median is probably the best measure of average because 61 is an extremely large data value and makes the mean artificially high. Chapter 3 Data Description Section 3-2 Exercise #5 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 Find (a) the mean, (b) the median, (c) the mode, and (d) the midrange. A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 (a) 229 239 663 465 372 136 1,202 88 189 75 X X = 4349, n = 12 X= n X= 4349 12 - 362 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 (b) Arrange data in increasing order: 75 88 117 136 189 229 239 372 465 574 663 1202 median A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 (b) Arrange data in increasing order: MD = 229 + 239 2 = 234 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 (c) No data are repeated. No mode. 1,202 88 189 75 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 (d) Arrange data in increasing order. 75 88 117 136 189 229 239 372 465 574 663 1202 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 (d) Arrange data in increasing order. MR = 75 + 1202 2 = 638.5 A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve were randomly selected, and the number of victims of identity theft in each city is shown. 574 117 229 239 663 465 372 136 1,202 88 189 75 Can you conclude that the researcher was correct? X = 362, MD = 234, no mode, MR = 638.5 It seems that the average number of identity thefts is higher than 300. Chapter 3 Data Description Section 3-2 Exercise #17 Eighty randomly selected light bulbs were tested to determine their lifetimes (in hours). This frequency distribution was obtained. Find the (a) mean and (b) modal class. Class Boundaries Frequency 52.5-63.5 6 63.5-74.5 12 74.5-85.5 25 85.5-96.5 18 96.5-107.5 14 107.5-118.5 5 Boundaries Xm f 52.5 – 63.5 58 6 348 63.5 – 74.5 69 12 828 74.5 – 85.5 80 25 2000 85.5 – 96.5 91 18 1638 96.5 – 107.5 102 14 1428 107.5 – 118.5 113 5 565 80 6807 f • Xm Eighty randomly selected light bulbs were tested to determine their lifetimes (in hours). This frequency distribution was obtained. Find the (a) mean and (b) modal class. f • X m 6807 = 80 = 85.1 X = n modal class: 74.5 – 85.5 Chapter 3 Data Description Section 3-2 Exercise #19 The cost per load (in cents) of 35 laundry detergents tested by a consumer organization is shown. Find the (a) mean and (b) modal class. Class Limits 13-19 20-26 27-33 34-40 41-47 48-54 55-61 62-68 Frequency 2 7 12 5 6 1 0 2 Class limits 13 – 19 Boundarie s 12.5 – 19.5 Xm f 16 2 32 20 – 26 19.5 – 26.5 23 7 161 27 – 33 26.5 – 33.5 30 12 360 34 – 40 33.5 – 40.5 37 5 185 41 – 47 40.5 – 47.5 44 6 264 48 – 54 47.5 – 54.5 51 1 51 55 – 61 54.5 – 61.5 58 0 0 62 – 68 61.5 – 68.5 65 2 130 35 1183 f • Xm The cost per load (in cents) of 35 laundry detergents tested by a consumer organization is shown. Find the (a) mean and (b) modal class. f •Xm 1183 = 35 = 33.8 X = n modal class: 26.5 – 33.5 Chapter 3 Data Description Section 3-3 Exercise #7 The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data represent samples. Find the range. range = 48 – 0 = 48 The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data represent samples. Use the shortcut formula for the unbiased estimator to compute the variance and standard deviation. s2 = X2 – X = 133 X n–1 2 / n 2 = 4061 n = 10 X The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data represent samples. Use the shortcut formula for the unbiased estimator to compute the variance and standard deviation. 2 4061 – 133 / 10 2 s = - 254.7 10 – 1 s- 254.7 - 16 The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Is the data consistent or does it vary? Explain your answer. range = 48 s2 = 254.7 s = 16 By any of these measures, it can be said that the data can vary. Chapter 3 Data Description Section 3-3 Exercise #21 The data shows the number of murders in 25 selected cities. Find the variance and Number f standard deviation. 27-90 13 91-154 155-218 219-282 283-346 347-410 411-474 475-539 539-602 2 0 5 0 2 0 1 2 Class Xm f • Xm f • X 2 m 13 760.5 44,489.25 f 27-90 58.5 91-154 122.5 2 245 30,012.5 155-218 186.5 0 0 0 219-282 250.5 5 1252.5 313,751.25 283-346 314.5 0 0 0 347-410 378.5 2 757 286,524.5 411-474 442.5 0 0 0 475-538 506.5 1 506.5 256,542.25 539-602 570.5 2 1141 650,940.5 The data shows the number of murders in 25 selected cities. Find the variance and standard deviation. f • X m = 4662.5 s2 = 2– f • X 2 = 1,582,260.25 f • X m f • X n–1 n 2 The data shows the number of murders in 25 selected cities. Find the variance and standard deviation. f • X m = 4662.5 s2 = 1, 582, 260.25 – 2 = 1,582,260.25 f • X m 4662.5 24 = 29,696 s = 29,696 - 172.3 25 2 Chapter 3 Data Description Section 3-3 Exercise #33 The mean of a distribution is 20 and the standard deviation is 2. Answer each. Use Chebyshev’s theorem. a. At least what percentage of the values will fall between 10 and 30? b. At least what percentage of the values will fall between 12 and 28? a. Subtract the mean from the larger value: 30 – 20 = 10 10 Divide by the standard deviation to get k: =5 2 1 1– = 0.96 or 96% 52 b. Subtract the mean from the larger value: 28 – 20 = 8. Divide by the standard 8 deviation to get k: = 4 2 1– 1 42 = 0.9375 or 93.75% Chapter 3 Data Description Section 3-3 Exercise #41 The average U.S. yearly per capita consumption of citrus fruits is 26.8 pounds. Suppose that the distribution of fruit amounts consumed is bell-shaped with a standard deviation equal to 4.2 pounds. What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year? By the Empirical Rule, 68% of consumption is within 1 standard deviation of the mean. Then 1/2 of 32%, or 16%, of consumption would be more than 31 pounds of citrus fruit per year. = 4.2 = 26.8 34% 34% 16% 22.6 26.8 16% 31 Chapter 3 Data Description Section 3-4 Exercise #13 Which of the following exam scores has a better relative position? a. A score of 42 on an exam with X = 39 and s = 4 z= 42 – 39 4 3 = 4 b. A score of 76 on an exam with X = 71 and s = 3 z= 76 – 71 3 = 5 3 Chapter 3 Data Description Section 3-4 Exercise #22 Find the percentile ranks of each weight in the data set. The weights are in pounds. Data: 78, 82, 86, 88, 92, 97 Percentile = number of values below + 0.5 total number of values Data: 78, 82, 86, 88, 92, 97 0 + 0.5 100% = 8th percentile For 78, 6 th For 82, 1 + 0.5 = 25 percentile 100% 6 nd For 86, 2 + 0.5 = 100% 42 percentile 6 100% Percentile = number of values below + 0.5 total number of values Data: 78, 82, 86, 88, 92, 97 For 88, 3 + 0.5 100% = 58th percentile 6 th For 92, 4 + 0.5 = 75 percentile 100% 6 nd For 97, 5 + 0.5 92 percentile = 100% 6 100% Chapter 3 Data Description Section 3-4 Exercise #23 What value corresponds to the 30th percentile? Find the percentile ranks of each weight in the data set. The weights are in pounds. 78, 82, 86, 88, 92, 97 6(30) c= 100 = 1.8 or 2 Therefore, the answer is the 2nd in the series, or 82. Chapter 3 Data Description Section 3-5 Exercise #1 Identify the five number summary and find the interquartile range. 8, 12, 32, 6, 27, 19, 54 Data arranged in order: 6, 8, 12, 19, 27, 32, 54 Minimum: 6 Median: 19 Maximum: 54 Q1: 8 Q3: 32 Interquartile Range: 32 – 8 = 24 Chapter 3 Data Description Section 3-5 Exercise #9 Use the boxplot to identify the maximum value, minimum value, median, first quartile, third quartile, and interquartile range. 50 55 60 65 70 75 80 85 90 95 100 Minimum: 55 Maximum: 95 Median: 70 Interquartile Range: Q1: 65 Q3: 90 90 – 65 = 25 Chapter 3 Data Description Section 3-5 Exercise #15 9.8 15.9 8.0 3.2 13.9 11.7 4.4 24.8 3.9 34.1 21.7 17.6 These data are the number of inches of snow reported in randomly selected cities for September 1 through January 10. Construct a boxplot and comment on the skewness of the data. 0 5 10 15 20 25 30 35 Data arranged in order : 9.8 15.9 8.0 3.2 13.9 11.7 Minimum: 3.2 4.4 24.8 3.9 34.1 Maximum: 34.1 MD: 11.7 + 13.9 = 12.8 2 Q1: 4.4 + 8.0 = 6.2 2 + 21.7 17.6 Q3: = 19.65 2 21.7 17.6 Comment on the skewness of the data. 3.2 6.2 12.8 0 5 10 34.1 19.65 15 The distribution is positively skewed. 20 25 30 35 Chapter 3 Data Description Section 3-5 Exercise #16 These data represent the volumes in cubic yards of the largest dams in the United States and in South America. Construct a boxplot of the data for each region and compare the distributions. United States 125,628 92,000 78,008 77,700 66,500 62,850 52,435 50,000 South America 311,539 274,026 105,944 102,014 56,242 46,563 For USA: Min = 50,000 Max = 125,628 MD = 72,100 Q1 = 57,642.5 Q3 = 85,004 72,100 125,628 50 100 150 United States 125,628 92,000 78,008 77,700 66,500 62,850 52,435 50,000 200 250 300 350 For South Min = 46,563 Max = 311,539 America: MD = 103,979 Q = 56,242 1 Q3 = 274,026 South America 311,539 274,026 105,944 102,014 56,242 46,563 103,979 50 100 311,539 150 200 250 300 Compare the distributions: 72,100 125,628 USA 103,979 50 100 311,539 150 200 South America 250 300

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