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Unformatted text preview: Problem 1. (5 pts) Multiple choice. A random sample of size n = 200 was drawn from a population. A histogram of the data is shown in the figure. The sample was most likely drawn from which distribution? A. T with 3 degrees of freedom B. Poisson with λ = 3 C. Skewed normal with μ = 3 D. Chi-squared with 3 degrees of freedom E. Exponential with E [ X ] = 1/3 Problem 2. (5 pts) Multiple choice. Let X 1 , X 2 , X 3 be a random sample from a uniform distribution with unknown mean μ and unknown variance σ 2 . Knowing full well that a valid confidence interval based on the t statistic is based on the assumption that the sample is drawn from a normally distributed population, a probabilist constructs a “100(1– α )% confidence interval” for μ as 3 / 2 , 2 / S t x α ± . Then, ) 3 / 3 / ( 2 , 2 / 2 , 2 / S t X S t X P α α μ + ≤ ≤- would be: A. 1 – α B. Less than 1 – α C. Greater than 1 – α Problem 3. (5 pts) A company uses a urine test to screen job applicants for marijuana use. The test correctly gives positive results for 93% of marijuana users and correctly gives negative results for 98% of non-users. Assume that 33% of college students in the U.S. use marijuana, what is the probability that a randomly selected student who tests positive is actually a marijuana user? Let T be the event that the student tests positive. Let U be the event that the student uses marijuana. Given: 93 . ) | ( = U T P , 98 . ) | ( = ′ ′ U T P , and 33 . ) ( = U P . Find: % 8 . 95 ) 67 (. 02 . ) 33 (. 93 . ) 33 (. 93 . ) ( ) | ( ) ( ) | ( ) ( ) | ( ) | ( = + = ′ ′ + = U P U T P U P U T P U P U T P T U P 1 10 5 30 20 10 X Frequency Problem 4. (5 pts) A teller at JFK airport’s foreign exchange window knows that if a customer wishes to convert British pounds ( X ) into American dollars( Y ), the formula for the conversion is Y = 1.64 X – 2. (Each pound is worth $1.64, and there is a $2 service charge.) The average amount presented for exchange is 140 pounds, with standard deviation 48 pounds. Find the standard deviation of the amount of dollars given out. 72 . 78 $ 48 64 . 1 ) ( 64 . 1 ) 2 64 . 1 ( ) ( 2 = ⋅ = =- = X V X V Y V Problem 5. (15 points total) a. (5 pts) Suppose the number of fish a fisherman catches in an hour is a Poisson random variable with an average rate of 2 fish per hour. If he fishes for one hour tomorrow morning, what is the probability he will catch fewer than 2 fish? Let X be a r.v. for the number of fish caught in an hour. Then, assume X ~ Poisson(2). The probability that he will catch fewer than two fish in one hour would then be: 406 . ! 1 2 ! 2 ) 1 ( ) ( ) 2 ( 1 2 2 = + = = + = = <-- e e X P X P X P b. (5 pts) What conditions would guarantee that the distribution would be Poisson?...
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This note was uploaded on 05/14/2008 for the course ENGRD 2700 taught by Professor Staff during the Fall '05 term at Cornell.
- Fall '05