80. a. These cases are a population because that these cases account for each of the 25 tables that are filled on the night in question. b. Mean = (28 + 39 + 23 + 67 +37 + 28 +56 + 40 +28+50+ 51 + 45 + 44 + 65 + 61 + 27 + 24 + 61 + 34 + 44 + 64 + 25+ 24 + 27 + 29) / 25 = 40.84 Median = The median value is the same as the middle value, and this is the same as the 13th highest number = 39 c. The biggest value = 67, lowest value = 23; so, the range is 67 – 23 = 44 Standard deviation = sqrt([(28 - 40.84)^2 + (39 - 40.84)^2 + (...same pattern) + (29 - 40.84)^2] / 25 ) =14.55 82. a. Mean = (3 x 90 + 8 x 110 + 12 x 130 + 16 x 150 + 7 x 170 + 4 x 190) / 50 = 141.2 b. Standard deviation =sqrt((3(90 - 141.2)^2 + 8(110 - 141.2)^2+ (...same pattern) +4(190 141.2)^2) / 50) =26 c. Within two standard deviations  95% 141.2 – 2 x 26.2, 141.2 + 2 x 26.2 Limits between 88.8 and 193.6 87. I. a. Mean = Sum of all selling prices / 105 = 221.1 Median = 213.6 (middle value after sorting from lowest to highest) Standard deviation = sqrt(((263.1 - 221.1)^2 + (182.4 - 221.1)^2+ (...same pattern) + (188.3 221.1)^2) / 105) = 47 b. The mean price is about $221,100; this means that, the median selling price is lower $213,600. There is standard deviation of about $47,100. II. a. Mean = sum of the entire home area / 105 = 2223.8

Median = 2220 (see attached spreadsheet) Standard deviation = 248.65 (see attached spreadsheet) b. The mean footage is about 2224 sq. ft.; but, the median sq. ft. is smaller 2200 sq. ft. This standard deviation is about 249 sq. ft. 34. P(a3|b1) = [P(a3)P(b1|a3)] / [P(a1)P(b1|a1) + P(a2)P(b1|a2) + P(a3)P(b1|a3)] = 0.1x 0.4 / (0.25 x 0.2 + 0.05 x 0.4 + 0.1 x 0.4) = 0.3636 36. Per Bayes' Theorem P(A|C) = P(C|A)P(A) / [P(C|A)P(A) + P(C|B)P(B)] = 0.9 x 0.8/ [0.9 x 0.8 + 0.6 x 0.2] = 0.857143 38. (1/4) x 0.05 / [(1/4) x 0.05 + (3/4) x 0.01] = 0.625 45. a. p(x=5) = C(10,5)