Pr(+|notD)Pr(notD) = .95*.005+.03*.995=0.0346. Using properties of binomial, E(
Y) = np = 100*0.0346=3.46. b. If three randomly selected people were tested, ...
Stat 515 final exam: Tues, Dec 14th. 1:30-‐3:30PM. LGRT 103. Format: Open book and open note. Calculators are fine, but they are not necessary. Coverage: • Chapter 2 (omit 2.9 and 2.12) • Chapter 3 (omit moment generating functions and 3.10) • Chapter 4 (omit moment generating functions and 4.11) • Chapter 5 (omit 5.9, 5.10, and 5.11) • Chapter 7 (omit 7.4) The exam will emphasize chapters 4, 5, and 7, but there will be one problem from chapters 2 and/or 3. There are six problems on the exam, and about 80-‐85% of the points are similar to the questions below or a question from one of the previous exams. Note, if you recognize a probability distribution function (e.g. uniform, normal, etc), you may use facts about that distribution (e.g. mean, cdf, etc) without proof or derivation. Suggested practice problems: 1. A medical test has the following properties. Given that someone has the disease, it says that she does (a “positive” test) with 95% probability. Given that someone does not have the disease, it says that she does not (a “negative” test) with 97% probability. Suppose 0.5% of the population has the disease. Note, Pr(+|D) = .95, Pr(-|notD) = .97, Pr(D) = 0.005. a. If 100 randomly selected people were tested, how many would be expected to receive positive tests? Let Y be number tested. Y is binomial(100,p) where p is probability a person gets a positive test. p=Pr(+) = Pr(+,D)+P(+,noD) = Pr(+|D)Pr(D) + Pr(+|notD)Pr(notD) = .95*.005+.03*.995=0.0346. Using properties of binomial, E(Y) = np = 100*0.0346=3.46. b. If three randomly selected people were tested, what is the probability that 2 were positive and 1 was negative? Now, Y is Bin(3,0.0346). Pr(Y=2) = (3 choose 2)0.03462(1-0.0346)= 0.003467215. c. Given that a randomly selected person receives a positive test, what is the probability that she has the disease? Want Pr(D|+) = Pr(D,+)/Pr(+) = Pr(+|D)Pr(D)/Pr(+) = .95*.005/0.0346 = 0.1372832. d. Given that a randomly selected person receives a negative test, what is the probability that she does not have the disease? Want Pr(notD|-) = Pr(notD,-)/Pr(-) = Pr(-|notD)Pr(notD)/Pr(-) = .97*(1-.005)/(1-0.0346) = 0.999741. e. What would need to happen to make both answers to c and d one? Pr(+|D), and Pr(-|notD) could become close to one. Also, Pr(D) increasing to 0.5 would “help.”
2. A warehouse has 10 diesel generators in it. Two of them do not work. Three different generators are randomly chosen. Let Y be the number of generators that work in a sample of size 3, chosen without replacement. Y is hypergeometric with population size N=10, M=8 of the type in which we’re interested, and sample size, n=3. a. What is the probability that they all work? Pr(Y=2) = (8 choose 3)*(2 choose 0)/(10 choose 3) = 0.47. b. What is the probability that at least 2 work? Pr(Y=0 or Y=1) = 0.47+(8 choose 2)*(2 choose 1)/(10 choose 3) = 0.47+0.47. c. What is the expected number of generators in the sample of three that work? E(Y) = 3*8/10 = 2.4. 3. A random variable Y has can take on three values: 1, 2, and 3 with Pr(Y=1) = 0.5, Pr(Y=2) = 0.3. a. What is Pr(Y=3)? Pr(Y=1)+Pr(Y=2)+Pr(Y=3) = 1, so Pr(Y=3) = 0.2. b. What is Pr(Y>2)? Pr(Y>2) = Pr(Y=2)+Pr(Y=3) = 0.5 c. What is E(Y)? E(Y) = 1*Pr(Y=1)+ 2*Pr(Y=2)+3*Pr(Y=3)=1.7 d. What is Var(Y)? Var(Y) = E((Y-EY)2)=(1-1.7)2*Pr(Y=1)+ (2-1.7)2*Pr(Y=2)+ (3- 1.7)2*Pr(Y=3)=0.61. e. What is Pr(Y2
