Student enrolled in a university ends up graduating from college, it’s reasonable to expect that there is percentage students dropping out every year. Also, freshmen on average will be younger than sophomores, sophomores younger than juniors, and so on. Thus, this distribution if plotted will approximate a rightskewed distribution. As with other right skewed distributions, the mean should be larger than the median. #1 page 82 On a certain exam, the average of the scores was 50 and the SD was 10 x = 50 & σ = 10 . I will be assuming a normal distribution. (a) Convert each of the following scores to standard units: 60, 45, 75. (number mean) / SD = Standard unit (S.U.) For 60 → (60 50) / 10 = +1 S.U. For 45 → (45 50) / 10 = 0.5 S.U. For 75 → (75 50) / 10 = +2.5 S.U. (b) Find the scores which in standard units are: 0, +1.5, 2.8. score = mean + (S.U. * SD) For 0 → 50 + (0 * 10) = 50 For 1.5 → 50 + (1.5 * 10) = 65 For 75 → 50 + (2.8 * 10) = 22 #1a,b,c page 84 Find the area under the normal curve (a) to the right of 1.25 100% (50% + ((area under 1.25) / 2)) 100% (50% + (78.87% / 2)) = 100% 89.44% = 10.5% (b) to the left of 0.40 100% (50% + (area under 0.40) / 2)) 100% (50% + 31.08% / 2) = 100% 65.54% = 34.46% (c) to the left of 0.80 50% + ((area under 0.80) / 2) 50% + (57.63% / 2) = 78.81% #1 page 93 review exercises (in class we did 1.33 SD) The following list of test scores has an average of 50 and and SD of 10: 39 41 47 58 65 37 37 49 56 59 62 36 48 52 64 29 44 47 49 52 53 54 72 50 50 mean = 50 SD = 10 (a) Use the normal approximation to estimate the number of scores within 1.25 SDs of the average 1.25 = 79% n = 25 numbers 1.25 SDs = * 0.79 = 19.75 (b) How many scores really were within 1.25 SD of the average? 1.25 * SD = 1.25 * 10 = 12.5 50 12.5 = 37.5 50 + 12.5 = 62.5 37.5 < Numbers < 62.5 Numbers = 18 numbers {39, 41, 47, 58, 49, 56, 59, 62, 48, 52, 44, → 47, 49, 52, 53, 54, 50, 50} #11 page 95 One term, about 700 Statistics 2 students at the University of California, Berkeley, were asked how many college mathematics courses they had taken, other than Statistics 2. The average number of courses was about 1.1; the SD was about 1.5. Would the histogram for the data look like (i), (ii), or (iii)? Why? It would look like graph (i). We can deduce this from the nature of the problem: having an average of 1.1 and SD of 1.5 we infer that the data spreads with certain sparsity away from the 1.1 point. But, as it disperses on both sides of the mean to the left of the mean the lowest possible value it can attain is 0, there can be no negative value as the number of courses taken. Yet, the SD is larger than the distance from the mean (1.1) to 0. We can infer then conclude that the sparsity extendeds then to the only other direction possible, to the far right. This would mean we would end up with a right
