6.Mean () =$60,010. This is the center of the data set insofar as it represents the balance for the data. The ordered list is: 55,430 57,118 60,055 60,451 60,946 63,011 63,057 The median is the middle value—here, 60,451 (the value in position = 4 in the ordered list, 60,451. At least half the values were $60,451 or more; at least half the scores were $60,451 or less. There is no mode. 7. This suggests that 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. 22 – 2 = 20 Interpretation: The difference between the largest value and the smallest value is 20. MAD: Given that the mean of the data is 9, MAD = = ≈ 5.33 Interpretation: The average difference between the individual values and the overall average for the data is 5.33. Variance: 2 = = ≈43.667 Standard Deviation:  = ≈ 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.
b)  = $13.20 c) median = 5.5th value ((10+1)/2 = 5.5) in the ordered list, halfway between the 5th and the 6th value = $12. d) 2 = 10.96;  = ≈ $3.31 x (price) f(x) no. of IPOs xf(x) x− (x−)2 (x−)2 f(x) 9 1 9 −4.2 17.64 17.64 10 2 20 −3.2 10.24 20.48 11 1 11 −2.2 4.84 4.84 12 2 24 −1.2 1.44 2.88 16 2 32 2.8 7.84 15.68 17 1 17 3.8 14.44 14.44 19 1 19 5.8 33.64 33.64 totals 10 132 109.6  = 132/10 2 = 109.6/10 = 10.96 = 13.2 27. a) Index value x proportion of months p(x) 86 2/20 = 0.10 87 3/20 = 0.15 88 4/20 = 0.2 89 4/20 = 0.2 90 4/20 = 0.2 91 3/20 = 0.15 b) 28. a) complaints x proportion of days p(x) 0 0.333 1 0.238 2 0.238 3 0.143
8 0.048 b) 29. a) rating x proportion of cities p(x) 1 0.1 2 0.2 3 0.2 4 0.25 5 0.25 b) 30. Offers x No. of Students f(x) Proportion of Students p(x) 0 80 0.08 1 288 0.288 2 367 0.367 3 122 0.122 4 81 0.081 5 54 0.054 6 8 0.008 1000 1.000
b) Number of Defects x No of Units f(x) No of that had x or more defects f(defects >x) Proportion of units that had x or more defects p(defects >x) 0 660 1000 1.00 1 120 340 0.34 2 94 220 0.22 3 52 126 0.126 4 45 74 0.074 5 23 29 0.029 6 6 6 0.006 1000 41. a) Number of Complaints (x) No. of Days f(x) No. of days that had x or more complaints f(complaints >x) 0 22 90 1 18 68 2 14 50 3 6 36
Interval $millions midpoint (m) frequency f(x) mf(x) m − (m – )2 (m – )2f(x) 600-800 700 6 4200 −260 67600 405600 800-1,000 900 7 6300 −60 3600 25200 1,000-1,200 1100 5 5500 140 19600 1,200-1,400 1300 0 0 340 115600 0 1,400-1,600 1500 1 1500 540 291600 291600 1,600-1,800 1700 1 1700 740 547600 547600 19200 1368000  = 19200/ 20 =960 2= 1368000/20 =261.5 Estimated Mean = $960 million Estimated Variance = 68400 Estimated Standard Deviation = = $261.5 million Interval (% change) Frequency (No. of States) −6 to −3 4 −3+ to 0 0 0+ to 3 12 3+ to 6 14 6+ to 9 15 9+ to 12 5 12+ to 15 1 49. 50. a)  = 100 strokes.100 here represents the central tendency of your golf scores
25 30 750.00 −0.7 0.49 14.70 35 20 700.00 9.3 86.49 1729.80 45 15 675.00 19.3 372.49 5587.35 55 10 550.00 29.3 858.49 8584.90 65 5 325.00 39.3 1544.49 7722.45 totals 150 3850.00 37933.50  = 3850/150 2 = 37933.5/150 = 25.7 = 252.89 75. a) * 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’reshowing a graph in which heightsare 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 narrowwidth; the height of the last bar has been lowered because of itsrelatively 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.”)
b) Estimated Mean = 43.42 years of age. Estimated Variance = 256.16 Estimated Standard Deviation = = 16. m (midpoint) p(x) (m)p(x) m− (m−)2 (m−)2p(x) 21.5 0.147 3.16 −21.9 479.61 70.50 30 0.194 5.82 −13.4 179.56 34.83 40 0.256 10.24 −3.4 11.56 2.96 50 0.17 8.50 6.6 43.56 7.41 60 0.101 6.06 16.6 275.56 27.83 72.5 0.133 9.64 29.1 846.81 112.63  = 43.42 2 = 256.16 76. a)