ECON1003: Analysis of Economic Data Fall 2003 Answers to Quiz #2 11:40a.m. 12:25p.m. (45 minutes) Tuesday, October 28, 2003


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1 ECON1003: Analysis of Economic Data Fall 2003 Answers to Quiz #2 11:40a.m. 12:25p.m. (45 minutes) Tuesday, October 28, (4 points) The number of claims for missing baggage for a wellknown airline in a small city averages six per day. What is the probability that, on a given day, there will be: a. (2 points) Fewer than three claims made? First, we know that the number of claims for missing baggage in a given day (x) follows Poisson distribution. Using Poisson distribution, we can compute P(x=0), P(x=1) and P(x=2). The probability that, on a given day there will be fewer than three claims made = P(x<3) = P(x=0) + P(x=1) + P(x=2). b. (2 points) Three or more claims made? The probability that, on a given day there will be three or more claims made = P(x>=3) = P(x=3) + P(x=4) + P(x=5) + = 1 P(x<3). 2. (7 points) An insurance company will insure a $50,000 diamond for its full value against theft at a premium of $400 per year. Suppose that the probability that the diamond will be stolen is 0.005, and let x denote the insurance company s profit. a. (2 points) Set up the probability distribution of the random variable x. Let Z be the premium charged. Event P(event) Profit x Stolen Z Not stolen Z b. (2 points) Calculate the insurance company s expected profit. Expected profit = P(stolen) * Profit(stolen) + P(Not stolen) * Profit(Not stolen) = * (Z 50000) *Z = Z * c. (3 points) Find the premium that the insurance company should charge if it wants its expected profit to be $1,000. From (b), expected profit = Z * To get an expected profit of 1000, we must charge Z = * Page 1 of 5
2 3. (6 points) In the past several years, credit card companies have made an aggressive effort to solicit new accounts from college students. Suppose that a sample of 200 students at your college indicated that following information as to whether the student possessed a bank card and / or a travel and entertainment credit card: Travel and entertainment credit card Bank credit card Yes No Yes No a. (2 points) Assume you know that the student has a bank credit card. What, then, is the probability that he or she has a travel and entertainment card? Consider a general case: Travel and entertainment credit card Bank credit card Yes No Yes A B No C D Conditional on that the student has a bank credit card, the probability that he or she has a travel and entertainment card = A/(A+B) b. (2 points) Assume you know that the student does not have a travel and entertainment card. What, then, is the probability that he or she has a bank credit card? Conditional on that the student does not have a travel and entertainment card, the probability that he or she has a bank credit card = A/(A+C) c. (2 points) Are the two events, having a bank credit card and having a travel and entertainment card, statistically independent? Explain. The two events are independent if P(having a bank card & having a travel card) = P(having a bank card) * P(having a travel card). P(having a bank card & having a travel card) = A/(A+B+C+D) P(having a bank card) = (A+B)/* (A+B+C+D) P(having a travel card) = (A+C)/* (A+B+C+D) To check independence, we only need to check whether the above definition is satisfied numerically. Page 2 of 5
3 4. (9 points) A recent poll about the economic performance and whether the governor should step down results in a set of probability estimates. Estimates of the probabilities that a randomly selected citizen would give the economic performance a ratings 5 (good), 4, 3, 2 and 1 (bad) are P(5) = 0.18, P(4) = 0.27, P(3) = 0.28, P(2) = 0.08, and P(1) = The estimates of the conditional probabilities are P(D 5) = 0, P(D 4) =0.04, P(D 3) = 0.25, P(D 2) = 0.25, and P(D 1) = a. (2 points) In a random sample of 10 citizens who give the economic performance a rating of 5, what is the probability that exactly five of them want the governor to step down? We know that the number of citizens who want the governor to step down follows a Binomial distribution. Using the Binomial formula, we can compute P(x=5 rating = 5). b. (2 points) In a random sample of 10 citizens who give the economic performance a rating of 1, what is the probability that nine or less of them want the governor to step down? We know that the number of citizens who want the governor to step down follows a Binomial distribution. Using the Binomial formula, we can compute P(x=10 rating =1). The probability that nine or less of the citizens want the governor to step down = P(x<=9 rating =1) = 1 P(x=10 rating =1). c. (5 points) Find the estimate of the probability that a randomly selected citizen gives 1 in his/her rating of the economic performance given that the selected citizen wants the governor to step down. That is, find the estimate of P(1 D). P(1 D) = P(1&D) / P(D). P(1&D) = P(D 1) * P(1) Similarly P(2&D) = P(D 2) * P(2), P(3&D) = P(D 3) * P(3), P(4&D) = P(D 4) * P(4), P(5&D) = P(D 5) * P(5), P(D) = P(1&D) + P(2&D) + P(3&D) + P(4&D) + P(5&D) Page 3 of 5
4 5. (7 points) In a survey, the respondents were asked a sensitive question about whether they have ever used illegal drugs. In order to get the respondent to tell the truth, the survey question was formulated as: Roll a die. If the result is {1, 2, 3, or 4}, answer the question marked A. If the result is {5 or 6}, answer the question marked B. A. Have you ever used illegal drugs? B. Roll a die. Did you get an odd number? Your answer: (0=No, 1=yes) With this formulation, the respondents should have no incentive to lie because no one knows whether he/she is answering A or B. Let p be the proportion of respondents who have ever used illegal drugs. a. (3 points) Draw a tree diagram describing probability of these events. Let p be the proportion of respondents who have ever used illegal drugs; q be the probability that the respondents will have to answer question A; r be the probability that the respondents will choose Y given that they have to answer question B. p Y q A 1p N r Y 1q B 1r N b. (4 points) Suppose 40% respondents answer YES to this survey question. Estimate p. Equating the survey result and the probability from the tree diagram, we have 0.4 = q*p + (1q)*r. Given r and q, we can estimate p. If it is difficult for us to think in probability terms, we can imagine we have 1000 respondents. 1000*q will answer question A and 1000*(1q) will answer question B. [1000*q*p (1q)*r ] will answer Y. Hence the proportion who answer Y is [1000*q*p (1q)*r ]/1000. Equating the survey result and the probability from the tree diagram, we have 0.4 = q*p + (1q)*r. Given r and q, we can estimate p. Page 4 of 5
5 6. (7 points) Consider two identical independent random variables X 1 and X 2 with mean 1 and variance 2/3. That is, E(X 1) = E(X 2)= 1, V(X 1) = V(X 2) = 2/3. a. (2 points) Compute the mean and variance of a new random variable Y = 3X 1. E(Y) = E(3X 1) = 3+E(X 1) V(Y) = V(3+X 1 ) = V(X 1 ) b. (2 points) Compute the mean and variance of a new random variable Y = 2X 1. E(Y) = E(2X 1 ) = 2E(X 1 ) V(Y) = V(2X 1 ) = V(2X 1 ) = 4V(X 1 ) c. (3 points) Compute the mean and variance of a new random variable Y = 2X 1 + X 2. E(Y) = E(2X 1 + X 2 ) = E(2X 1 ) + E(X 2 ) = 2E(X 1 ) + E(X 2 ) V(Y) = V(2X 1 + X 2) = V(2X 1) + V(X 2)= 4V(X 1) + V(X 2) [Hint: suppose the random variable (both X 1 and X 2) takes values 0, 1 and 2 with equal probability.] Statistical formula for reference: 1. Binomial probability distribution: P(x) = n C x π x (1π) nx where C denotes a combination; n is the number of trials; x is the number of successes; π is the probability of a success on each trial. 2. Hypergeometric distribution: P(x) = ( S C x )( NS C nx ) / N C n where N is the size of the population; S is the number of successes in the population; x is the number of successes in the sample (0,1,2,3, ); n is the size of the sample or the number of trials; C is the symbol for a combination. 3. Poisson distribution: P(x) = µ x e µ / x! where µ is the mean number of occurrence (successes) in a particular interval; e is the constant ; x is the number of occurrences (successes). 4. The combination operator nc x is defined as nc x =n!/[x! (nx)!], where n! = [n*(n1)* *2*1] and 0! = 1.  END  Page 5 of 5
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