exam 2 solutions

exam 2 solutions - Stat 4105 Second In-Class Exam Solutions...

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Unformatted text preview: Stat 4105 Second In-Class Exam Solutions 1. At the annual spring dance, a bored stat student with leg in a cast estimates the following probabilities for various numbers of dancers to be on the floor at random times when waltzes are being played. Let the probability mass function [p(x) = P(X = x | waltz)] be x p(x) F(x) 2 .12 .12 4 .07 .19 6 .19 .38 8 .25 .63 10 .21 .84 12 .16 1.00 x) a) (5 pts.) Fill in the blanks for the cumulative distribution function F(x) = P(X b) (5 pts.) What is the probability of more than 6 dancers on the floor? P(X>6) = 1 F(6) = 1 .38 = .62 c) (5 pts.) Find the expected value E(X) of the number of dancers on the floor. E(X) = 2x.2 + 4x.07 + 6x.19 + 8x.25 + 10x.21 + 12x.16 = 7.68 2. A random sample of Tech students are interviewed and asked 2 questions: First, do you often listen to rap music? Second, are you an engineering major? The results were that there were 3 engineering majors who listen to rap, 15 other majors who listen to rap, 19 other majors who do not listen to rap, and 10 engineering majors who do not listen to rap. (10 pts.) You may be surprised that so few engineering majors listen to rap. Use the Fisher exact test to calculate the probability of so few, if there is in fact no connection between music habits and major. What do you conclude? Putting the data in a table is handy: Taste Rap Major Engi Other total 3 15 18 No 10 19 29 total 13 34 47 If the two issues are unrelated, we have a hypergeometric model in which (for example) W = 18, B = 29, and n = 13. Then the probability of observing so few Engineering rap fans is P(X 3) = p(0) + p(1) + p(2) + p(3) 18 29 18 29 18 29 18 29 0 13 1 12 2 11 3 10 = + + + 47 47 47 47 13 13 13 13 =.0005 + .0066 + .0376 + .1162 = .1609 This is usually thought of as not too small a probability, so we have little evidence that taste and major are related 3. Many years ago, I became enamored of the Loafer Gumshoe series of 22 mystery novels. I remember a particularly entertaining minor character, Polly Ester Pantsuit, who appeared in 5 of those novels. Unfortunately, I cannot remember which novels she appeared in. I decide to reread the novels, in no particular order, until I find one in which she appears. a) (5 pts.) What is the probability that I will read exactly 6 novels in which Polly does not appear before I find one in which she does? The number of non-Polly novels read follows a "discrete geometric" N(W = 17, B = 6 5, b = 1) probability law. Thus p 6 = 22 = .0518 . 7 17 5 b) (5 pts.) What is the probability that I will have to read at least five novels in which Polly does not appear before I find one in which she does? P X 5 = P first 5 are non-Polly = 17 22 5 5 = .235 c) d) (5 pts.) On average, how many novels total will I have read, by the time I finish the first one I read in which Polly appears? , so E(Read) = 2.83 + 1(Polly) = 3.83 E non-Polly = E X = 171 = 2.83 5+1 4. We are going to invent a new family of random variables that I will call exhaustive sampling E(n, m) variables. n will stand for the number of individuals in a population that you wish to locate, and m for the number of individuals you have no interest in locating. You choose individuals at random, without replacement, until you have found all n individuals of interest. Then the random variable 0 X m is the number of individuals you had no interest in locating that you accidentally chose along the way. a) (5 pts.) Derive the probability mass function p(x) = P[X = x | E(n, m)] for this family. x+n-1 x Since the x+nth is interesting, and all the rest are not, we find p x = n + m m b) (5 pts.) Derive the cumulative distribution function F(x) = P[X anywhere in it. Easy way is to work backwards through the complete list of n + m: the last m x individuals on your list are all individuals of no interest. So F(x) = P(m x in a row of no interest) = m m-x . n + m m-x x | E(n, m)]. Your formula must be in closed form, which means you do not used summation symbols or ... c) (5 pts.) The E(n, m) family is actually a special case of a family we have studied that has more parameters. Show me how this works, and which parameters correspond to which others. Use this fact to give a formula for E(X). This is N(W=m, B=n, b=n). Thus E(X) = Wb/(B+1) = mn/(n+1) ...
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