CHAPTER 2. Descriptive Statistics I: Elementary Data Presentation and Description ASSIGNMENT CLASSIFICATION TABLE (BY LEARNING OBJECTIVE)

CHAPTER Descriptive Statistics I: Elementary Data Presentation and Description ASSIGNMENT CLASSIFICATION TABLE (BY LEARNING OBJECTIVE) Level of Difficulty (Moderate to Challenging) Learning Objectives 1 3 1. Compute and interpret the three main measures of central tendency. 1,,3,4 5,6,7,11,1,13,14. Compute and interpret the four main measures of dispersion. 8,9,1,5,53 3. Summarize data in a frequency distribution. 15,16,17 4. Summarize data in a relative frequency distribution. 7,8,9,3, 33,64, 5. Build and interpret a cumulative frequency distribution. 11,1,13,14,51,5, 54,55,56, 18,19,,1,,3, 4,5,6,59,6,61 3,31,34,35,36,65 37,38,66 39,4,41,67,68,69 7 57,58 6,63 6. Analyze grouped data and show that data in a histogram. 4,43,44,45 46,47,48,49,71,7, 73,74,75,76 77,78

- Chapter CHAPTER 6 3 5 9 1 5 1. Mean ( ) = = 3.5. This is the center of the data 1 set insofar as it represents the balance point for the data. The ordered list is, 1,,,, 3, 5, 5, 6, 9. 1 1 The median is the middle value; in this case, it s.5, the value in position = 5.5 or halfway between the 5 th and 6 th values in the ordered list.) At least half the values in the data set are at or above.5; at least half of the values are at or below.5. Mode =. This is the most frequently occurring value. c) mode = median =.5 5 1 3 5 6 9 mean = 3.5 The high value, 9, exerts a great deal of leverage in setting the balance point (that is, the mean) because it sits so far to the right.. 471 3... 67 Mean ( ) = = $533.5 million. This is the center of the data set 1 insofar as it represents the balance point for the data. The ordered list is 1 3 4 5 6 7 8 9 1 11 1 $67, $3, $333, $384, $471, $479, $495, $59, $614, $757, $787, $9 1 1 The median is the middle value. It s the value in position = 6.5 in the ordered list, halfway between the 6 th and the 7 th value: 479 and 495. We can calculate

-3 Chapter 479 495 its value as = $487. Interpretation: At least half the movies had revenues of $487 million or more; at least half had revenues of $487 million or less. There s no mode, since no value appears more than once. 3. a) Mean ( ) =15.9 This is the center of the data set insofar as it represents the balance point for the data. The ordered list is 11, 11, 1, 1, 14, 15, 17, 17, 17,,. 11 1 The median is the middle value; in this case, 15 (the value in position = 6 in the ordered list) At least half the months had 15 or more bankruptcies; at least half had 15 or less. The mode (or modal value) is 17, the most frequently occurring value. c) median =15 mode = 17 17 11 1 17 11 1 14 15 17 mean = 15.9 4. Mean ( x ) =14. This is the center of the data set insofar as it represents the balance point for the data. The ordered list is 4, 99, 115, 13, 13, 19, 9. 7 1 The median is the middle value here, 13 (the value in position = 4 in the ordered list) At least half the scores were 13 or more; at least half the scores were 13 or less. The mode is 13 the most frequent score.

-4 Chapter b) A typical or representative score is about 13. There s a chance for an occasional great game and an occasional dismal performance. (For purposes of comparison, a professional bowler will average above 3.) 5. Mean ( ) =1.411. This is the center of the data set insofar as it represents the balance point for the data. The ordered list is:.1.3.6 1.43 1.64.4.5.69 3.6 1 1 The median is the middle value here, 1.535 (the value in position = 5.5 in the ordered list, or halfway between the 5 th and 6 th 1.43 1.64 value: = 1.535. ) At least half the values were 1.535% or more; at least half the scores were 1.535% or less. There is no mode here. 6. Mean ( ) =$6,1. This is the center of the data set insofar as it represents the balance point for the data. The ordered list is: 55,43 57,118 6,55 6,451 6,946 63,11 63,57 7 1 The median is the middle value here, 6,451 (the value in position = 4 in the ordered list, 6,451. At least half the values were $6,451 or more; at least half the scores were $6,451 or less. There is no mode. 7. This suggests that there are a relatively few very high volume texters, pulling the average well to the right of (that is, above) the median. The mean is generally more sensitive to extreme values than is the median. 8. Range: =

-5 Chapter Interpretation: The difference between the largest value and the smallest value is. MAD: Given that the mean of the data is 9, MAD = - 9 + 1-9 + 6-9 + 4-9 8-9 + 1-9 = 3 5.33 6 6 Interpretation: The average difference between the individual values and the overall average for the data is 5.33. Variance: = = ( - 9) + (1-9) + (6-9) +... + (1-9) 6 43.667 6 6 Standard Deviation: = 43.667 6.61 Interpretation: Roughly speaking, the individual values are, on average, about 6.61 units away from the overall mean for the data. As is typically the case, the standard deviation here, 6.61, is greater than the MAD, 5.33. 9. Range: 116 111 = 5 yen Interpretation: The difference between the highest exchange rate and the lowest exchange rate was 5 yen. MAD: Given that the mean of the data is 114, MAD = 11-114 + 115-114 + 111-114 +... + 7 11-114 = 14 = yen 7 Interpretation: The average difference between the daily exchange rate and the overall average exchange rate for the week was yen. Variance: = = (11-114) + (115-114) + (111-114) +... + (11-114) 3 = 4.86 7 7

-6 Chapter Interpretation: The average squared difference between the daily exchange rate and the overall average exchange rate for the week was 4.86 (yen ). Standard Deviation: = 4.86 =.7 Interpretation: Roughly speaking, the daily exchange rate is, on average, about.7 yen away from the overall mean exchange rate for the week (114 yen/dollar). As is typically the case, the standard deviation here,.7, is slightly greater than the MAD,.. 1. Range: 146 1 = $4 Interpretation: The difference between the highest price and the lowest price is $4. MAD: Given that the mean of the data is $18, MAD = 1-18 + 14-18 + 17-18 +... + 134-18 6 = = $5 1 1 Interpretation: The average difference between the individual competitor prices and the overall average competitor price is $5. Variance: = (1-18) + (14-18) + (17-18) +... + (134-18) 53 = 44.33 1 Interpretation: The average squared difference between the individual competitor prices and the overall average competitor price is 44.33 (squared dollars). Standard Deviation: = 44.33 $6.66 Interpretation: Roughly speaking, the price charged by individual competitors is, on average, about $6.66 away from the overall average price charged by the group. 1 11. YEAR 3 4 5 6 7 UNITS 7.56 7.48 7.66 7.76 7.56 YEAR 8 9 1 11 1 UNITS 6.77 5.4 5.64 6.9 7.4

-7 Chapter Mean: 6.916 Range: 7.76 5.4 =.36 million cars sold. Interpretation: The difference between the highest sales year and the lowest sales year was.36 million cars. MAD: Given that the mean is 6.916 million cars, MAD = 7.56-6.916 + 7.48-6.916 + 7.66-6.916 +... + 7.4-6.916 1 7.58 = =.758 million cars (= 75,8 cars) 1 Interpretation: The average difference between individual year sales and the overall average sales level for the 1-year period is approximately.753 million cars. Variance: = (7.56-6.916) + (7.48-6.916) + (7.66-6.916) +... + (7.4-6.916) 1 7.148 = =. 715 1 Interpretation: The average squared difference between individual year sales and the overall average sales level is.715 million cars. Standard Deviation: =.7148.845 million cars. (845, cars) Interpretation: : Roughly speaking, individual year sales are, on average, about 845, cars away from overall average car sales for the 1-year period. 1. a) Range: 9 4 = 5 pins Interpretation: The difference between the highest score and the lowest score rate is 5. MAD: Given that the mean score is 14 pins, MAD = 13-14 + 99-14 + 19-14 +... + 7 115-14 = 39 = 56 pins. 7

-8 Chapter Interpretation: The average difference between the individual scores and the overall average score for the sample is 56 pins. Variance: s = (13-14) + (99-14) + (19-14) +... + (115-14) 7-1 37478 = = 646.33 6 Interpretation: If we looked at the population being represented here, the average squared difference between the individual scores and the overall average score in that population would be about 646 (pins ). Standard Deviation: s = 646.33 79. pins. Interpretation: Roughly speaking, the scores in the sample suggest that if we looked at the population being represented here, the individual scores in that population would be, on average, about 79 pins away from the overall population average score. b) Your bowling looks to be pretty erratic. 13. a) Range: 397 = 397 megawatts Interpretation: The difference between the highest capacity and the lowest is 397 megawatts. MAD: Given that the mean is 134, MAD = 1-134 + 138-134 + 397-134 +... + 573-134 = 16 = 8 8 137.75 megawatts. Interpretation: The average difference between the individual capacities and the overall average capacity is 137.75 megawatts. Variance: = = (1-134) + (138-134) + (397-134) +... + (573-134) 16584496 = 736 8 8 Interpretation: The average squared difference between the individual values and the overall average value is about,73,6 (megawatts ).

-9 Chapter Standard Deviation: = 736 1439.8 megawatts. Interpretation: Roughly speaking, the individual state capacities are, on average, about 144 megawatts away from the overall average capacity of 134 megawatts. 14. a) Range: 4.5.6 = 39.9 $billion Interpretation: The difference between the highest assets and the lowest is 39.9 $billion. MAD: Given that the mean is 9.7 $billion, MAD = $billion. 4.5-9.7 + 5.1-9.7 + 9.9-9.7 +... +.6-9.7 1 = 96.58 1 = 9.658 Interpretation: The average difference between the individual assets and the overall average asset value is 9.658 $billion. Variance: = = (4.5-9.7) + (4.5-9.7) + (4.5-9.7) +... + (4.5-9.7) 1588.5 158.85 1 1 Interpretation: The average squared difference between the individual asset values and the overall average asset value is 158.85 $billion. Standard Deviation: = 158.85 1.64 $billion. Interpretation: Roughly speaking, the individual asset values are, on average, about 1.64 $billion away from the overall average asset value of 9.7 $billion.

Frequency -1 Chapter 15. Rating x Frequency f(x) 3 8 4 16 5 1 Stock Rating 15 1 5 3 4 5 More rating 16. a) errors x frequency f(x) 1 1 1 1 3 4 5 3 6 6 7 4 b)

Frequency 1 3 4 5 6 7 More Frequency -11 Chapter Audit Results 7 6 5 4 3 1 errors c) The distribution is negatively skewed and unimodal. 17. a) score x frequency f(x) 5 3 6 6 7 1 8 4 9 6 b) Test Results 8 6 4 5 6 7 8 9 More score c) The distribution is bi-modal.

9 1 11 1 13 14 15 16 17 18 19 More No. of Players No. of bars -1 Chapter 18. calories x frequency f(x) 19 5 1 5 3 3 1 4 5 1 6 7 1 8 6 5 4 3 1 19 Energy Bar Calories 1 3 4 calories 5 6 7 8 More 19. No. of touchdowns Players x f(x) 9 8 1 3 11 6 1 1 13 14 1 15 16 1 17 18 1 19 1 9 8 7 6 5 4 3 1 Touchdowns

No. of States -13 Chapter. No. of In-State Cities w/pop > 5, x No. of States f(x) 1 13 1 3 4 4 1 5 6 7 8 9 1 1 11 1 13 1 Total = 5 5 15 1 5 1 3 4 5 6 7 8 9 1 11 1 13 No. of In-State Cities w/pop > 5,

-14 Chapter 1. a) 1.9 defectives b) median = value halfway between the 1 th and 13 th value in the ordered list = 1 c) = 1.71; = 1. 71 = 1.31 defectives The detailed calculations are shown below: x f(x) xf(x) x (x ) (x ) f(x) 9 1.9 1.66 14.97 1 6 6.9.8.5 4 8.71.5. 3 3 9 1.71.9 8.77 4 8.71 7.3441 14.69 totals 4 31 4.96 =31/4 =1.9 = 1.71. a) =.56 household members b) median = approximately the 5 million th ((14.6 million +1)/) value in the ordered list = c) = 1.99; = 1. 99 1.41 household members The detailed calculations are shown below: x f(x) (millions) xf(x) x (x ) (x ) f(x) 1 6.7 6.7 1.56.43 64.98 34.7 69.4.56.31 1.88 3 17. 51.6.44.19 3.33 4 15.3 61. 1.44.7 31.73 5 6.9 34.5.44 5.95 41.8 6.4 14.4 3.44 11.83 8.4 7 1.4 9.8 4.44 19.71 7.6 totals 14.6 67.6 8 67.6/14.6 =.56 = 8/14.6 = 1.99

-15 Chapter 3. a) x = 6. calls b) median = the value halfway between the 7 th and 8 th value ((14+1)/) in the ordered list = 6 c) s = 1.41; s = 1. 41 1.19 calls The detailed calculations are shown below: x f(x) xf(x) x x (x x ) (x x ) f(x) 4 1 4. 4.84 4.84 5 3 15 1. 1.44 4.3 6 4 4..4.16 7 4 8.8.64.56 8 16 1.8 3.4 6.48 totals 14 87 18.36 x = 87/14 = 6. s = 18.36/(14-1) =1.41 4. a) =.8 ships b) median = 183 rd value ((365+1)/ = 183) in the ordered list = 3. c) = 1.55; = 1. 55 1.4 ships The detailed calculations are shown below: x f(x) xf(x) x (x ) (x ) f(x) 14.8 7.84 19.76 1 47 47 1.8 3.4 15.8 71 14.8.64 45.44 3 13 369..4 4.9 4 8 38 1. 1.44 118.8 5 8 14. 4.84 135.5 totals 365 16 566. = 16/365 =.8 = 566/365 = 1.55

-16 Chapter 5. a) b) = 1.75 years c) median = 11.5th value ((+1)/ = 1.5) in the ordered list, halfway between the 1 th and the 11 st value = 1 years. d) =.7; =. 7 1.65 years The detailed calculations are shown below: x (age) f(x) (no. of aircraft) xf(x) x (x ) (x ) f(x) 8 7 16.75 7.565 4.188 9 37 333 1.75 3.65 113.313 1 14 14.75.565 7.875 11 1 31.5.65 1.315 1 84 18 1.5 1.565 131.5 13 17 1.5 5.65 86.65 totals 149 544. = 147/ = 544/ =.7 = 1.75

-17 Chapter 6. b) = $13. c) median = 5.5th value ((1+1)/ = 5.5) in the ordered list, halfway between the 5 th and the 6 th value = $1. d) = 1.96; = 1. 96 $3.31 x (price) f(x) no. of IPOs xf(x) x (x ) (x ) f(x) 9 1 9 4. 17.64 17.64 1 3. 1.4.48 11 1 11. 4.84 4.84 1 4 1. 1.44.88 16 3.8 7.84 15.68 17 1 17 3.8 14.44 14.44 19 1 19 5.8 33.64 33.64 totals 1 13 19.6 = 13/1 = 19.6/1 = 13. = 1.96 7. a) Index value x proportion of months p(x) 86 / =.1 87 3/ =.15 88 4/ =. 89 4/ =. 9 4/ =. 91 3/ =.15

proportion of days Proportion of months -18 Chapter b).5 Confidence Index..15.1.5 86 87 88 89 9 91 Index Value 8. a) complaints x proportion of days p(x).333 1.38.38 3.143 8.48 b) 1-day Complaint Record.35.3.5..15.1.5 1 3 4 5 6 7 8 complaints

proportion of cities -19 Chapter 9. a) rating x proportion of cities p(x) 1.1. 3. 4.5 5.5 b) Livability Ratings for Cities.3.5..15.1.5 1 3 4 5 rating 3. Offers x No. of Students f(x) Proportion of Students p(x) 8.8 1 88.88 367.367 3 1.1 4 81.81 5 54.54 6 8.8 1 1.

Proportion of students P(x) Proportion of students P(x) - Chapter.4.3..1 1 3 4 5 6 Offers (x) 31. Interviews x No. of Students x Proportion of Students p(x) 3 43.43 4 95.95 5 136.136 6 19.19 7 35.35 8 177.177 9 1.1 1 1..3..1 3 4 5 6 7 8 9 Interviews(x) 3. median = 7. Summing down the relative frequency (proportion) column, we reach a sum of.5 part way through the 7s (. + part of.36). The median, then, is 7).

-1 Chapter x p(x) 6. 7.36 8. 9.1 1.1 1. 33. a) =.5 days (See detailed calculations below.) b) median = days (Summing down the relative frequency (proportion) column, we reach a sum of.5 part way through the s (.39 + part of.7). The median, then, is. c) = 1.98; = 1. 98 1.38 days 34. a) x p(x) xp(x) (x ) (x ) (x ) p(x) 1.39.39 1.5 1.563.69.7.54.5.63.17 3.17.51.75.563.96 4.8.3 1.75 3.63.45 5.5.5.75 7.563.378 6.4.4 3.75 14.63.563 =.5 =1.98 Games Won Proportion of Award Winning Pitchers 17 4/5=.8 18 7/5=.14 19 9/5=.18 11/5=. 1 8/5=.16 5/5=.1 3 4/5=.8 4 /5=.4

- Chapter b) =.6 games x p(x) xp(x) 17.8 1.36 18.14.5 19.18 3.4. 4.4 1.16 3.36.1. 3.8 1.84 4.4.96 =.6 c) Median = games (Summing down the relative frequency (proportion) column, we reach a sum of.5 part way through the s (.8 +.14 + +.18 + part of.). The median, then, is.) 35. a) =.3 offers b) Median = offers (Summing down the relative frequency (proportion) column, we reach a sum of.5 part way through the s (.8 +.88 + part of.367). The median, then, is.) c) = 1.667. = 1. 667 1.91 offers x p(x) xp(x) x (x (x p(x).8.3 4.19.3967 1.88.88 1.3 1.69.35539.367.734.3.9.33 3.1.366.97.949.11479 4.81.34 1.97 3.889.314353 5.54.7.97 8.89.47639 6.8.48 3.97 15.769.1687 =.3 1.6671 mean variance

-3 Chapter 36. a) = 6.5 interviews b) Median = 7 interviews (Summing down the relative frequency (proportion) column, we reach a sum of.5 part way through the 7s (.43 +.95 +.136 +.19 + part of.35). The median, then, is 7.) c) =.694. =. 694 1.641 interviews x p(x) xp(x) x (x (x p(x) 3.43.19 3.5 1.5.5675 4.95.38.5 6.5.59375 5.136.68 1.5.5.36 6.19 1.15.5.5.48 7.35 1.645.5.5.5875 8.177 1.416 1.5.5.3985 9.1 1.98.5 6.5.765 = 6.5 =.694 mean variance 37. No. of customers who made x or x fewer trips/ week. f(trips < x) 6 1 9 4 3 49 4 5

cum no. of customers -4 Chapter Shopping Survey-Cumulative Distribution 6 5 4 3 1 1 3 4 trips 38. Just divide cumulative frequencies by 5, the total number of customers in the survey. Proportion of customers who made x or fewer trips/ week. x f(trips < x) 6/5=.1 1 9/5 =.58 4/5=.8 3 49/5=.98 4 5/5 = 1. 39. a) x No. of checks where line length was less than or equal to x. Cum. f(line length < x) 1 13 3 3 46 4 53 5 57 6 6

checks -5 Chapter Number of checks w here line length w as less than or equal to x. 7 6 5 4 3 1 1 3 4 5 6 x (line length) b) To produce relative frequencies here, divide cumulative frequencies by 6. x Proportion of checks where line length was less than or equal to x. Cum. p(line length < x).33 1.17.533 3.767 4.883 5.95 6 1.

no. of units prop of checks -6 Chapter Proportion of checks where line length was less than or equal to x. 1. 1..8.6.4.. 1 3 4 5 6 x(length) 4. a) Number of Defects x No of Units f(x) No of units that had x or more defects f(defects >x) 66 1 1 1 34 94 3 5 16 4 45 74 5 3 9 6 6 6 1 1 1 8 6 4 No. of units with x or more defects 1 3 4 5 6 x (no. of defects) b)

Proportion of units -7 Chapter Number of Defects x No of Units f(x) No of units that had x or more defects f(defects >x) Proportion of units that had x or more defects p(defects >x) 66 1 1. 1 1 34.34 94. 3 5 16.16 4 45 74.74 5 3 9.9 6 6 6.6 1 1.8.6.4. Proportion of units with x or more defects 1 3 4 5 6 No. of defects (x) 41. a) Number of Complaints (x) No. of Days f(x) No. of days that had x or more complaints f(complaints >x) 9 1 18 68 14 5 3 6 36 4 3 5 7 3 6 6 3 7 9 17 8 5 8 9 3 3 Total = 9

No. of Days -8 Chapter 1 9 8 7 6 5 4 3 1 No. of days where complaints were > x 1 3 4 5 6 7 8 9 No. of Complaints (x) b) Number of Complaints (x) No. of Days f(x) No. of days that had x or more complaints f(complaints >x) Proportion. of days that had x or more complaints p(complaints >x) 9 1. 1 18 68.756 14 5.556 3 6 36.4 4 3.333 5 7 3.333 6 6 3.56 7 9 17.189 8 5 8.89 9 3 3.33 Total = 9

No. of Days -9 Chapter Proportion of days where complaints were > x 1..8.6.4.. 1 3 4 5 6 7 8 9 No. of complaints (x) 4. a) Class Midpoint Frequency 1 to under 14 13 7 14 to under 16 15 9 16 to under 18 17 7 18 to under 19 6 to under 1 6 b) Assembly Times 8 4 1 14 16 18 time in minutes

-3 Chapter c) Estimated Mean = 16.7 Estimated Variance = 7.58 Estimated Standard Deviation = Variance.75 m (midpoint) f(x) mf(x) m (m ) (m ) f(x) 13 7 91 3.7 13.69 95.83 15 9 135 1.7.89 6.1 17 7 119.3.9.63 19 6 114.3 5.9 31.74 1 6 16 4.3 18.49 11.94 totals 35 585 65.15 = 585/35 = 65.15/35 = 16.7 = 7.58 43. a) rating midpoint frequency relative freq 3 to under 4 35 4.1 4 to under 5 45 5.15 5 to under 6 55 5.15 6 to under 7 65 7.175 7 to under 8 75 6.15 8 to under 9 85 8. 9 to 1 95 5.15 4 1. Product Ratings..1 3 4 5 6 7 8 9 1 rating minutrat

-31 Chapter 44. a) Air Quality days 1 6 5 1 15 index value b) Estimated Mean = 8.9 Estimated Variance = 53.6 Estimated Standard Deviation = Variance = 45.3 m(midpoint) f(x) mf(x) m (m ) (m ) f(x) 5 74 185 55.9 314.81 3135.9 75 117 8775 5.9 34.81 47.77 15 58 75 44.1 1944.81 11799 175 4 4 94.1 8854.81 1515.4 totals 73 75 5663.1 = 8.9 =53.6

proportion of stocks -3 Chapter 45. a).6 Share Price Changes.5.4.3..1 $ -.6 $. $.6 $ 1. $1.8 price change x b) Estimated Mean = $.57 Estimated Variance =.55 Estimated Standard Deviation = Variance $.55 46. a) m (midpoint) p(x) mp(x) m (m ) (m ) p(x).3.1.3.87.757.76.3.48.144.7.73.35.9.9.61.33.19.3 1.5.13.195.93.865.11 mean =.57 variance =.55

prop or students -33 Chapter Weekly TV Viewing.6.5.4.3..1 1 3 4 viewing hours x b) Estimated Mean = 13.5 hours Estimated Variance = 94.75 Estimated Standard Deviation = Variance 9.73 hours m(midpoint) p(x) mp(x) m (m ) (m ) p(x) 5.48.4 8.5 7.5 34.68 15.7 4.5 1.5.5.68 5.17 4.5 11.5 13.5.483 35.8.8 1.5 46.5 36.98 totals 1. 13.5 94.75 = 13.5 = 94.75 = 9.73 47. a) To produce relative frequencies, divide the frequency column by the total projected population of 357.4 (million). Age Group mid f(x) p(x) xp(x) to under 1 94.3.64.64 to under 4 3 9.9.6 7.8 4 to under 6 5 85.38 11.89 6 to under 8 7 7.3.197 13.77 8 to under 1 9 14.8.41 3.73 1 to 1 11.1..3

no. of movies -34 Chapter Total = 357.4 mean = 39.85 b) Estimated Mean age = 39.85 Estimated Variance = 569.15 Estimated Standard Deviation = Variance = 3.86 48. a) m f(x) p(x) mp(x) (m (m p(x)(m ) 1 94.3.64.64 9.85 891. 35.1 3 9.9.6 7.8 9.85 97. 5. 5 85.38 11.89 1.15 13. 4.5 7 7.3.197 13.77 3.15 99. 178.8 9 14.8.41 3.73 5.15 515. 14.15 11.1..3 7.15 491. 1.38 Total = 357.4 = 39.85 Var = 569.15 8 7 6 5 4 3 1 Top Movies of All Time 6 8 1 1 14 16 18 Gross Revenue ($millions) b)

No. of States -35 Chapter Interval $millions midpoint (m) frequency f(x) mf(x) m (m ) (m ) f(x) 6-8 7 6 4 6 676 456 8-1, 9 7 63 6 36 5 1,-1, 11 5 55 14 196 98 1,-1,4 13 34 1156 1,4-1,6 15 1 15 54 916 916 1,6-1,8 17 1 17 74 5476 5476 19 1368 = 19/ = 96 = 1368/ = 684 =61.5 Estimated Mean = $96 million Estimated Variance = 684 Estimated Standard Deviation = 684 = $61.5 million 49. Interval Frequency (% change) (No. of States) 6 to 3 4 3+ to + to 3 1 3+ to 6 14 6+ to 9 15 9+ to 1 5 1+ to 15 1 16 14 1 1 8 6 4 % change in food stamp participation (Jan 11-Jan 1) -6-3 3 6 9 1 15 % change

-36 Chapter 5. a) = 1 strokes.1 here represents the central tendency of your golf scores insofar as it serves as a balance point for the scores b) The ordered list is 6, 8, 9, 1, 1, 11, 1, 14. Median = 1 strokes, halfway between the 4 th and 5 th values (here, these values are both 1). It s the middle value insofar as at least half the scores are at or above this value and at least half the scores are at or below this value. Mode=1 strokes. The most frequent score. c) range = 14 6 = 8 strokes. The difference between the highest and lowest score. MAD = 17.5 strokes. The average difference between the individual scores and the overall average score. = 55. The average squared difference between your individual scores and the overall average score. = 55 =.9 strokes. Roughly the average difference between the individual scores and the overall average score. x x lx l (x ) 1 9 1 1 1 11 1 1 1 8 4 1 4 14 4 4 16 1 6 4 4 16 8 14 4 = 8/8 =1 MAD=14/8 = 17.5 =4/8 =55 51. a) = $17.51 billion (See detailed calculation below.) b) The ordered list is 13.7, 15, 16.3, 18.6, 19, 19.,.8 Median = $18.6 billion, the 4 th value No Mode. c) range =.8 13.7 = $7.1 billion; MAD = $.16 billion = 5.6; = 5. 6 = $.37 billion

-37 Chapter x Ix I (x ) 13.7 3.81 14.5 15.51 6.3 16.3 1.1 1.46 18.6 1.9 1.19 19. 1.69.86 19 1.49..8 3.9 1.8 totals 1.6 15.9 39.37 17.51.16 5.6 mean MAD Variance 5. a) = 7 hours (See detailed calculation below.) b) The ordered list is 3, 5, 5, 6, 7, 1, 13 Median = 6 hours, the 4 th value ((7+1)/ = 4) Mode =5 hours. c) range = 13 3 = 1 hours; MAD =.57 hours = 1; = 1 = 3.16 hours x x lx l (x ) 5 4 6 1 1 1 3 4 4 16 5 4 7 1 3 3 9 13 6 6 36 49 18 7 = 49/7 = 7 MAD = 18/7 =.57 = 7/7 = 1 53. a) x = 16. (See the detailed calculations below.) b) The ordered list is 89, 93, 95, 97, 99, 1, 1, 1, 16, 181 Median = 99.5, a value halfway between the 5 th and 6 th values in the ordered list. Mode = 1. c) Range = 181 89 = 9; MAD = 15 Variance= 715.8 ; Standard deviation = 715. 8 = 6.75.

-38 Chapter d) The median or mode would better represent the typical time since these are measures less influenced by the one extreme of 181. x x x lx x l (x x ) 16 1 6 6 36 1 6 6 36 97 9 9 81 89 17 17 89 95 11 11 11 93 13 13 169 181 75 75 565 99 7 7 49 1 6 6 36 16 15 644 x =16/1 = 16 MAD = 15/1 = 15 s = 644/9 = 715.8 54. a) = $4,358. (See detailed calculations below.) b) The ordered list is: 3647, 36919, 41661, 4578, 445. Median = $41,661, the 3 th value in the ordered list. No mode. c) range = 445 3647 = $7,778; MAD = $,948 = 9,741,; = 9741= $3,11 x Ix I (x ) 445 3847 1479949 4578 4984 41661 133 169789 36919 3439 118671 3647 3931 1545761 Totals 179 1474 48751 4358 948 9741 mean MAD Variance

-39 Chapter 55. In the chapter, the mean was described as a balance point for a set of data, equalizing distances (deviations) of values to the left of the mean with distances of values to the right. Show that this is the case for the data in a) Exercise 5: The mean, is 1, so summing the x distances gives ( x ) = + ( 1) + 1 + ( ) + + 4 + + ( 4) = b) Exercise 53: The mean, x, here is 16, so summing the x x distances gives ( x x) = +( 6)+( 6)+( 9)+( 17)+( 11)+( 13)+75+( 7)+( 6) = 56. The central tendency of team performance is the same for both teams: the mean and the median in each case are 75. The teams obviously differ in terms of consistency: The range for Team A is 1, vs. 5 for Team B. The standard deviation for Team A is 3.16, while the standard deviation for Team B is 18.4. Most would choose the more consistent team, Team A. However, an argument could be made that even though assigning the job to Team B is more risky, Team B also offers the chance for superior performance, as indicated by the high score of 9. Team A appears the safer bet, but has a lower ceiling. Team B has a much higher upside, but a more severe downside. 57. a) The median must be halfway between 13 and 5: 19 b) If the MAD is 9, then the distance of each value from the mean must be 9 units. The sum of the squared distances, then, must be 9 + 9 = 16. The variance, therefore, is 16/ = 81, making the standard deviation 81= 9. c) The standard deviation is 4. For a data set of two values, the standard deviation is precisely equal to the average distance of the two values from the mean. (See part b).)the values must be 4 = 16 and + 4 = 4. 58. a) The MAD must be 4. b) The range must be 1. c) If the mode is, there must be two values of, making the third value 45 in order to have the mean of the three numbers be 15. The sum of the squared distances from the mean, therefore, must be ( 15) + ( 15) + 3 = 135. The standard deviation, then, is 135 = 1.. 3

no. of nurses -4 Chapter 59. a) Nurse's Shifts for the Past Month 5 45 4 35 3 5 15 1 5 1 3 4 no. of shifts b) Mean = 1.4 shifts; Variance = 1.4; standard deviation = 1. 4 = 1.19 shifts. (See the details below.) c) Median = 1 shifts, the value halfway between the 65 th and 66 th values ((13+1)/ = 65.5) in the list. Mode= 1. d) This is a unimodal, positively skewed distribution x f(x) xf(x) x (x ) (x ) f(x) 31 6 1.4 1.96 6.76 1 46 966.4.16 7.36 8 616.6.36 1.8 3 15 345 1.6.56 38.4 4 1 4.6 6.76 67.6 totals 13 787 184. = 787/13 = 1.4 = 184/13 = 1.4

no. of shifts -41 Chapter 6. a) Staffing Chart 14 1 1 8 6 4 4 41 4 43 44 45 no. of officers b) Mean = 4.5 officers; Variance = 3.93; standard deviation = 3. 93 = 1.98 officers. (See the details below.) c) Median = 4.5, the value halfway between the 5 th and 6 th values ((5+1)/ = 5.5) in the list. Modes are 4 and 45. d) This is a symmetric, bi modal distribution. x f(x) xf(x) x (x ) (x ) f(x) 4 13 5.5 6.5 81.5 41 7 87 1.5.5 15.75 4 5 1.5.5 1.5 43 5 15.5.5 1.5 44 7 38 1.5.5 15.75 45 13 585.5 6.5 81.5 totals 5 15 196.5 = 15/5 = 4.5 = 196.5/5 = 3.93

no. of players -4 Chapter 61. a) NHL Lost Teeth Chart 5 15 1 5 1 3 4 5 6 no. of lost teeth b) Mean = 3.88 teeth; Variance =.; standard deviation =. = 1.41 teeth. (See the details below.). c) Median = 4, the 38 th value ((75+1)/ = 38) in the list. Mode = 4. x f(x) xf(x) x (x ) (x ) f(x) 1 3.88 15.5 15.5 1 4 4.88 8.9 33.18 8 16 1.88 3.53 8.8 3 13 39.88.77 1.7 4 88.1.1.3 5 18 9 1.1 1.5.58 6 9 54.1 4.49 4.45 totals 75 91 15 =91/75 = 3.88 = 15/75 =. 6. a) The chart shows two distinct groups of applicants, one that performed rather poorly on the test and another that performed fairly well. It might be useful for the company to try to identify what particular factors caused this sort of result: Was it a difference in education levels for the two groups? A difference in prior experience? A difference in the conditions under which the tests were administered? The particular administrators who conducted the tests? Etc.

proportion of retailers -43 Chapter b) The mean, as the balance point for the data, appears to be around 6. The median the 5-5 marker for the data appears to be 8. At least half the values look to be at or below 8; at least half look to be at or above 8. c) The MAD for the distribution appears to be approximately 3. It s the average distance of the values in the data set from the mean. Among the other possible answers here, 1 appears to be too small, and 7 and 1 appear too large to fit this definition. d) The standard deviation for the distribution appears to be approximately 4. It should be roughly equal to the MAD, but will almost always be larger. 63. a) Clearly the company isn t meeting the 3-day standard for a significant number of its deliveries. In fact, in a number of instances, delivery time was at least double the 3-day standard. b) The mean, as the balance point for the data, appears to be about halfway between 3 and 4 days. The median the 5-5 marker for the data appears to be 3 days. At least half the times look to be at or below 3; at least half look to be at or above 3. c) The MAD for the distribution looks to be approximately 1.5 days. It s the average distance of the values in the data set from the mean. The other possible answers here, 5.5, 8.5 and 15, all appear too large to fit this definition. d) The standard deviation for the distribution looks to be approximately days. It should be roughly equal to the MAD, but will almost always be larger. 64. a) Retailer Prices.6.5.4.3..1 $99.95 19.95 119.95 19.95 price b) Mean = $113.95; Variance = 64; standard deviation = 64 = $8. (See the details below.).

proportion of days -44 Chapter 65. a) x p(x) xp(x) x (x ) (x ) p(x) 99.95.1 9.995 14 196 19.6 19.95.5 54.975 4 16 8 119.95.3 35.985 6 36 1.8 19.95.1 1.995 16 56 5.6 = 113.95 = 64 -day Stoppage Record.4.35.3.5..15.1.5 1 3 4 5 no. of stoppages b) Mean = 1.85 stoppages per day; Variance = 1.548; standard deviation = 1.4 stoppages. (See the details below.) x p(x) xp(x) x (x ) (x ) p(x).1 1.85 3.43.34 1.38.38.85.73.75.3.46.15.3.5 3.19.57 1.15 1.33.51 4.6.4.15 4.63.77 5.4. 3.15 9.93.397 = 1.85 = 1.548

cum. prop. of traders -45 Chapter 66. a) Prediction ( No. of Months) x No. of Traders whose predictions were less than or equal to x months Cum. f(predict < x) 3 1 11 3 6 4 8 5 9 6 3 Stock Market Predictions 35 3 5 15 1 5 1 3 4 5 6 months of increase 67. a) b) Flaws x No. of panels with no more than x flaws cum. f(flaws<x) 6 1 15 18 3 3 4 4 5 5

cum no. of panels -46 Chapter Panel Flaws Record 3 5 15 1 5 1 3 4 5 no. of flaw s 68. Just divide the table values from Exercise 67 by the total number of panels, 5. Flaws x Proportion of panels with no more than x flaws Cum. p(flaws<x).4 1.6.7 3.9 4.96 5 1. 69. a) No. of Commercials x Proportion of Nights where no more than x commercials appeared Cum. p(commercials<x) 4/3 =.133 1 16/3 =.533 3/3 =.767 3 8/3 =.933 4 3/3 = 1.

-47 Chapter b) No. of Commercials x Proportion of Nights where at least x commercials appeared cum p(commercials>x) 3/3 = 1. 1 6/3 =.867 14/3 =.467 3 7/3 =.33 4 /3 =.67 7. a) Number of Firearms x Proportion of Households Owning the Indicated Number of Firearms p(x).57 1.79.57 =..93.79 =.14 4.97.93 =.4 5.99.97 =. 6 1..99 =.1 b) x p(x) xp(x) x (x ) (x ) p(x).57.8.67.383 1...18.3.7.14.8 1.18 1.39.195 4.4.16 3.18 1.11.44 5..1 4.18 17.47.349 6.1.6 5.18 6.83.68 1. =.8 = 1.68 71. a) Class midpoint Frequency $ to under $ 1. 16 $ to under $4 3. 7 $4 to under $6 5. 1 $6 to under $8 7. 13 $8 to under $1 9. 7

-48 Chapter b) freq Equivalent Wages 1 4 6 8 1 wages ($/hour) value c) Estimated mean = 4.63 (The actual mean of the raw data is 4.5) Estimated variance = 6.86 (The actual variance of the raw data is 7.6) Estimated standard deviation = Variance =.6 (Actual is.66) m (midpoint) f(x) mf(x) m (m ) (m ) f(x) 1 16 16 3.63 13.177 1.83 3 7 1 1.63.657 18.598 5 1 15.37.137.875 7 13 91.37 5.617 73. 9 7 63 4.37 19.97 133.678 totals 96 439.1 = 96/64 = 439/64 = 4.63 = 6.86

-49 Chapter 7. class midpoint frequency to under 3 1.5 3 to under 6 4.5 4 6 to under 9 7.5 17 9 to under 1 1.5 3 a) Equivalent Wages 1 3 6 9 1 wages ($/hour) value b) Estimated mean = 4.64 Estimated variance = 6.87 Estimated standard deviation = Variance =.6 m(midpoint) f(x) mf(x) m (m ) (m ) f(x) 1.5 3 3.14 9.86 197.19 4.5 4 18.14..47 7.5 17 17.5.86 8.18 139.5 1.5 3 31.5 5.86 34.34 13. totals 64 97 439.73 = 97/64 439.73/64 = 4.64 = 6.87

-5 Chapter 73. a) class midpoint frequency to under 5.5 34 5 to under 1 7.5 3 Equivalent Wages 3 15 5 1 wages ($/hour) value b) Estimated mean = 4.84 Estimated variance = 6.3 Estimated standard deviation = Variance =.5 m(midpoint) f(x) mf(x) m (m ) (m ) f(x).5 34 85.34 5.48 186.17 7.5 3 5.66 7.8 1.7 totals 64 31 398.44 = 31/64 =398.44/64 = 4.84 = 6.3

-51 Chapter 74. a). Number of Employees 4 1 3 4 5 6 7 employees b) Estimated Mean = 5.7 employees. Estimated Variance = 5.89 Estimated Standard Deviation = Variance = 15.9 employees. m (midpoint) f(x) mf(x) m (m ) (m fx) 5 1..7 48.49 8569.8 15 5 75. 1.7 114.49 574.5 5 3 75..7.49 14.7 35 7. 9.3 86.49 179.8 45 15 675. 19.3 37.49 5587.35 55 1 55. 9.3 858.49 8584.9 65 5 35. 39.3 1544.49 77.45 totals 15 385. 37933.5 = 385/15 = 37933.5/15 = 5.7 = 5.89

-5 Chapter 75. a) Age of Tabloid Buyers..1 5 35 45 55 65 8 age * Note: The area of each bar in a histogram should be proportional to the fraction of the observations that fall in the bar s class. Above we re showing a graph in which heights are used to represent relative frequency, even though we have classes of unequal width. Although technically incorrect, this is fairly common practice. The graph below shows the more technically correct picture, with areas representing frequency. To give the proper proportional area, the first bar height has been raised in response to the bar s relatively narrow width; the height of the last bar has been lowered because of its relatively greater width. These adjustments give a picture in which areas and not heights are proportional to the frequency of class membership.. Proper bar heights can be set by dividing class frequencies by class widths. (The vertical scale on the histogram would be a somewhat cumbersome frequency per unit width. ) Age of Tabloid Buyers 5 35 45 55 65 8 age

-53 Chapter b) Estimated Mean = 43.4 years of age. Estimated Variance = 56.16 Estimated Standard Deviation = 56. 16 = 16. m (midpoint) p(x) (m)p(x) m (m ) (m ) p(x) 1.5.147 3.16 1.9 479.61 7.5 3.194 5.8 13.4 179.56 34.83 4.56 1.4 3.4 11.56.96 5.17 8.5 6.6 43.56 7.41 6.11 6.6 16.6 75.56 7.83 7.5.133 9.64 9.1 846.81 11.63 = 43.4 = 56.16 76. a) Household Income.1.1.6. 1 3 4 5 6 75 1 15 15 $1s * As noted in the solution to Exercise 57, the area of each bar in a histogram should be proportional to the fraction of the observations that fall in its class. Above we re showing a graph in which heights are used to represent relative frequency, even though we have classes of unequal width. Although technically incorrect, this is fairly common practice. The graph below shows the more technically correct picture, with areas representing frequency. Notice that the bar heights have been lowered as the classes get wider so that area is kept proportional to frequency. Proper bar heights can be set by dividing class frequencies by class widths. (The vertical scale on the histogram would be a somewhat cumbersome relative frequency per unit width. )

-54 Chapter Household Income 1 3 4 5 6 75 1 15 15 $1s b) Estimated Mean = 54.77 (times $1). Estimated Variance = 311.4 (times $1 ). Estimated Standard Deviation = Variance = 48.8 (times $1). m (midpoint) p(x) (m)p(x) m (m ) (m ) p(x) 5.95.48 49.77 477.5 35.3 15.16 1.89 39.77 1581.65 199.9 5.13 3.5 9.77 886.5 115.1 35.13 4.31 19.77 39.85 48.7 45.17 4.8 9.77 95.45 1.1 55.9 4.95.3.5. 67.5.14 7. 1.73 16.5 16.85 87.5.1 8.93 3.73 171.5 19.7 11.5.5 5.85 57.73 333.75 173.3 137.5.5 3.44 8.73 6844.5 171.11 175. 3.85 1.3 14455.5 318. 5.4 6. 195.3 38114.75 914.75 = 54.77 = 311.4

-55 Chapter 77. a) There is a significant cluster of customers in the 1 to 18 age group and another significant cluster of customers in their late s and early 3s. There is a substantial gap in the late teens/early s age group. b) There is a clear indication here that the store attracts high income shoppers. The number of customers with family incomes of $6, or less is relatively small. c) The store appears to attract customers who are relatively infrequent visitors to the mall. More than 5% of the shoppers made no more than 1 visits a rate of less than once a month. (Of course it s possible that there just aren t that many people in general who visit the mall more than 1 time or so in a year.) Based on what you see in these charts, what recommendations might you make to the owner of Kari H Junior Fashions? Kari H might find a line that appeals to shoppers in their late teens and early s. Clearly this is an age group that is not well represented among Kari H shoppers. Furthermore, the store can either continue to cultivate its appeal to higher end shoppers or try to find ways to reach low to moderate income shoppers as well. Finally, the store doesn t appear to be attracting the more frequent mall visitors, which may mean more frequent changes in displays and/or merchandise may be appropriate. 78. a) 3 looks to be the approximate balance point for the data. b) 7 appears to be the best answer among the possibilities. More so than any of the other choices, it looks like the approximate 5-5 marker. At least half the values are at or above; at least half are at or below. c) The standard deviation roughly the average distance of the values from the mean looks to be around 6. Among the other answers, 3 looks much too small to be the average distance; 1 and 18 look too large. 79. a) 1 looks to be the approximate balance point for the data. b) 7 appears to be the best answer among the possibilities. More so than any of the other choices, it looks like the approximate 5-5 marker: at least half the values appear to be at or above that point; at least half at or below. c) The standard deviation roughly the average distance of the values from the mean looks to be closest to 13. The other answers all appear to be much too small to be measuring, even roughly, this sort of average distance.