Econ 250 Fall 2007 s for Assignment 1 1. A local TV station advertises two news-casting positions. If three women (W 1, W 2, W 3 and two men (M 1, M 2 apply what is the sample space of the experiment of hiring two coanchors? Does it matter here that the positions being filled are equivalent? What is the probability that both men get the job? We have to choose two anchors among five individuals. So the experiment effectively entails choosing two elements (without replacement from the following set (W 1, W 2, W 3, M 1, M 2. Thus the sample space is: S {(W 1, W 2, (W 1, W 3, (W 2, W 3, (W 1, M 1, (W 1, M 2, (W 2, M 1, (W 2, M 2, (W 3, M 1, (W 3, M 2, (M 1, M 2 } as there are C 5 2 10 possibilities. The probability that both men get the job is the probability of the outcome (M 1, M 2 which is just 1/10. It does matter that the two positions being filled are equivalent. If the station were seeking to hire, say, a sports announcer and a weather forecaster, the number of possible outcomes would be P 5 2 ( 20 as (W 2, M 1, for example, would represent a different staffing assignment than (M 1, W 2. 2. Consider the linear transformation Y i a bx i Suppose µ x 25 and σ x.05 and you want to have µ y 55 and σ y 1000. (i What values should a and b take? (ii Show what the appropriate transform on Y i is to take it tostandardized form (i.e. µ z 0 and σ z 1 (i µ y a bµ x and σ y b σ x. Thus, b σ y /σ x 1000/0.05 20000 taking the positive root. Then a µ y bµ x 55 (20000(25 445. (ii Note that: Z Y µ y, thus σ y a µ y σ y 55 1000 0.055 b 1 σ y 1 1000 0.001 3. You are given information on the annual returns of thirty stocks. 10%, 8%, 14%, 18%, 1%, 5%, 4%, 4%, 7%, 5%, 17%, 27%, 8%, 0%, 5%, %, 12%, 16%, 1%, 2% %, 5%, 16%, 22%, 15%, 18%, 3%, 5%, 8%, 4% 1
(i Draw a box plot for this sample of returns and interpret the box plot. (ii Calculate the sample mean, sample standard deviation for this sample of bond returns. What is the coefficient of variation and interpret. (i Q 1 0, Q 2 5 and Q 3 20 10 0 10 20 (ii X 4.06667, s.538 and CV 2.83766 4. Consolidated Industries has come under considerable pressure to eliminate its seemingly discriminatory hiring practices. Company officials have agreed that during the next five years, 60% of their new employees will be females and 30% will be minorities. One out of four new employees, though, will be white males. Is this a plausible probability statement. Why or why not? What percentage of their new hires will be minority females? P(male 1 P(female 1 0.6 0.4 (we are assuming that a person can be either male or female but not both. Now, P(male P(male minorityp(male white as males can either be white or minorities. Then using P(male white 0.25, we have that P(male minority 0.4 0.25 0.15. Now, P(minority P(male minorityp(female minority 0.3, since minorities can only be male or female. Therefore P(female minority 0.3 0.15 0.15. 5. Let random variables X, Y have the joint distribution given be the following table: Y X 2 3 4 0 1/48 0 b 1 0 5/48 8/48 2 a 0 /48 3 5/48 12/48 0 (i Find a and b if it is known that P(X Y 1/3. Show all work (ii Find the joint probability table for: P(XY (i P(X Y P(X 2, Y P(X 3, Y 3 a 12/48 1/3. Thus, a 4/48. (ii All the entries in the joint distribution table have to sum to 1. Hence, given the value we calculated above for a, it must be that b is equal to 2/48. 2
6. Six fair dice are rolled at once. What is the probability that each of the six faces appears? The number of ways each of the six faces can appear when six dice are rolled is simply equal to all the re-arrangements of {1, 2, 3, 4, 5, 6}. That is, a total of 6! ways. The size of the sample space is of course 6 6. Therefore, the required probability is 6! 6 6 7. Suppose each of 10 sticks is broken into a long part and a short part. The 20 parts are arranged into 10 pairs and glued back together, so that again there are 10 sticks. What is the probability that each long part will be paired with a short part? (This problem is a model for the effects of radiation on a living cell. Each chromosome, as a result of being struck by ionizing radiation, breaks into two parts, one part containing the centromere. The cell will die unless the fragment containing the centromere recombines with one not containing a centromere. The total number of ways the 20 broken parts can be recombined into pairs is simply ( 20 10. That is, its the same problem as choosing ten parts from a total of 20 without replacement (Why?. Now, the number of ways in which all the broken parts can be recombined and stay matched is 2 10. Hence, the desired probability is ( 210 20 10. 8. Urn I contains five red chips and four white chips; urn II contains four red and five white chips. Two chips are to be transferred from urn I to urn II (i.e. drawn randomly from urn I and placed into urn II. Then a single chip is to be drawn from urn II. What is the probability that the chip drawn from the second urn will be white? Explain the sample space. Let B be the event White chip is drawn from urn II. Let A i, i 0, 1, 2, denote the event i white chips are transferred from urn I to urn II. Then, using the properties of conditional probabilities, P(B P(B A 0 P(A 0 P(B A 1 P(A 1 P(B A 2 P(A 2 Note that P(B A i (5 i/ and that P(A i is given by the hypergeometric distribution formula. Therefore, ( 5 P(B ( 5 53 ( 5 ( 0( 6 ( 10 ( 6 ( 5 ( 1( 1 7 ( 20 ( 7 ( 5 2( 0 ( 6. Suppose that a randomly selected group of k people are brought together. What is the probability that exactly one pair has the same birthday? Start with k 2, then k 3 and so on to see if you can recognize a pattern to write out the general formula 3
in terms of k. If we had 30 people in the room what is the probability that at least one has the same birthday? Picture the k individuals lined up in a row to form an ordered sequence. Omitting leap years, each person might have any one of 5 possible birthdays. Thus, by the multiplication rule, the group as a whole generates a sample space of 5 k birthday sequences. Now, define A to be the event at least two people have the same birthday. If each person is assumed to have the same chance of being born on any given day, the 5 k sequences are equally likely and P(A Number of sequences in A 5 k Counting the number of sequences in the numerator here is prohibitively difficult because of the complexity of the event A. However, counting the number of sequences A C is quite easy. Notice that each birthday sequence in the sample space belongs to exactly one of two categories: (a At least two people have the same birthday (b All k people have different birthdays Therefore, Number of sequences in A 5 k number of sequences where all k people have different birthdays The number of ways to form birthday sequences for k people subject to the restriction that all k birthdays must be different is simply the number of ways to form permutations of length k from a set of 5 distinct objects: Therefore, (5(4...(5 k 1 P(A P(at least two people have the same birthday 5k (5(4...(5 k 1 5 k For k 30, the probability is 0.706316 or nearly 71%! Now, let B be the event exactly two people have the same birthday. That is, we want all the sequences where there is exactly one duplicate. Then, the number of such sequences is ( k (5(4...(5 k 2 Using the same reasoning as before: P(B P(exactly two people have the same birthday ( k 2 (5(4...(5 k 5 k 4
10. Suppose a certain drug test is % accurate, that is, the test will correctly identify a drug user as testing positive % of the time, and will correctly identify a non-user as testing negative % of the time. Let s assume a corporation (say the Atlanta Falcons decides to test its employees for opium use, and 5% of the employees use the drug. Calculate the probability that, given a positive drug test, an employee is actually a drug user. We have, P( positive test drug user 0. P( positive test not a drug user 0.01 P( negative test drug user 0.01 P( negative test not a drug user 0. and P(employee is a drug user 0.05 Now, in order to compute P(employee is a drug user positive test we can use Bayes rule: P(drug user pos. test P(pos. test drug userp(drug user P(pos. test 0. 0.05 P(pos. test drug user P(drug user P(pos. test not drug user P(not drug user 0. 0.05 0. 0.05 0.01 0.5 0.83883 5