Let X ~ Geom(p), let n be a non-negative integer, and let Y ~ Bin(n, p). Show that P(X = n) = (1/n)P(Y = 1).
What will be an ideal response?
(a) The proportion of strengths that are less than 65 is 0.10. Therefore 65 is the 10th percentile of strength. The z-score of the 10th percentile is approximately z= ?1.28. Let ?be the required standard deviation. The value of ? must be chosen so that the z-score for 65 is ?1.28. Therefore ?1.28 = (65 ? 75)/?. Solving for ? yields ?= 7.8125 N/.
(b) Let ? be the required standard deviation. The value of ?must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 = (65?75)/?. Solving for ?yields ?= 4.292 N/.
(c) Let ? be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z= ?2.33. Therefore ?2.33 = (65 ? ?)/5. Solving or ? yields ?= 76.65 N/.
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