Stat 321 Exam 2 October 26, 2004

1. (12 pts) Recall the investigation dealing with “composite testing.” Continue to suppose that in a group of 10 people, each person has .1 probability of having a certain disease, independently from person to person. Now consider a new plan: We randomly divide the ten people into two groups of five people each. Within each group of five, we pool blood samples from the individuals and do a composite test. Like before, a composite test will be positive if at least one person in the group has the disease, and it will be negative if no person in the group has the disease. Also like before, if the composite test is negative, then we have to go back and do individual tests on every person in the group. Let the random variable T represent the total number of tests to be conducted with this plan.

a) (4 pts) What are the possible values of T?

T can equal 2 (if nobody has the disease), 7 (if one group has nobody with the disease and the other group has somebody with the disease), or 11 (if somebody in both groups has the disease).

b) (8 pts) Determine the probability distribution of T.

P(T=2) = P(nobody has the disease) = (.9)^10 = .349.

P(T=12) = P(both groups have at least one person with disease) = P(first group has at least one person with disease) x P(first group has at least one person with disease) = [1-P(nobody in first group has disease)] x [1-P(nobody in second group has disease)] = [1-(.9)^5][1-(.9)^5] = .168.

P(T=7) = P(one group has nobody with disease and other group has at least one person with disease) = 1-P(T=2)-P(T=7) = 1-.349-.168 = .483.

2. (22 pts) Let the random variable X represent the monetary amount of damage done to a particular car model in a year. Suppose that an insurance company considers a simple model where the probability distribution of X is as follows:

Damage x	$0	$1000	$10,000
Probability p(x)	.80	.15	.05

a) (6 pts) Determine the expected value of the amount of damage in a year.

E(X) = 0(.80) + 1000(.15) + 10000(.05) = 650

b) (4 pts) Write a sentence interpreting what this expected value means from the company’s perspective, assuming that it insures a very large number of these cars.

The long-run average amount of damage will converge to $650 per car, if the company insures a large number of cars.

c) (4 pts) Determine the probability that the damage to a car will be less than its expected value.

The only possible values of X are 0, 1000, and 10000, so P(X<650) = P(X=0) = .80.

d) (8 pts) If the company’s policies carry a $500 deductible (meaning that the insurance company pays only for the damage amount in excess of $500), and if the company wants its expected profit to be $250 per car, how much should it charge for its premium?

Let m represent the premium. The profit will be m with probability .8, (m-500) with probability .15, and (m-9500) with probability .05. The expected value of the profit is therefore .8m + .15(m-500) + .05(m-9500) = m-550. To make the expected profit equal 250, the premium must be m=800.

3. (14 pts) Suppose that you choose a Cal Poly student at random, and let the random variable H represent the student’s studying time (in hours) last week. Suppose that H has a uniform distribution on the interval (25, 35).

a) (4 pts) Determine the probability that a randomly selected student studied for more than 32 hours last week.

P(H>32) = = .3.

b) (6 pts) Now suppose that you randomly select 10 Cal Poly students and that each person’s studying time follows this same probability distribution, independently from person to person. Determine the probability that at least two of the students studied for more than 32 hours last week.

Let X be the number of these 10 students who studied for more than 32 hours last week. Then X has a binomial distribution with n=10 and p=.3. Thus, P(X>=2) = 1 - P(X=0) - P(X=1) = 1 - (.7)^10 – 10(.3)[(.7)^9] = 1 - .028 - .121 = .851.

c) (4 pts) Now suppose that you randomly select Cal Poly students one at a time until you find one who studied more than 32 hours last week. What is the expected number of students that you have to sample before you find one who satisfies this requirement?

Let Y be the number of students that you have to sample before finding one who satisfies the requirement. Then Y has a geometric distribution with p=.3. Thus, E(Y) = (1-p)/p = .7/.3, or about 2.333.

4. (14 pts) On a recent trip I took four flights and encountered eight pilots (two per flight). Seven of these eight pilots were men, and one was a woman. Let the random variable X represent the number of women in a random sample of 8 commercial pilots.

a) (4 pts) Would it be reasonable to model the distribution of X with a binomial distribution? Explain.

Yes. Each pilot is either a man or a woman, we are counting the number of women in a fixed number of trials (n=8), the trials are essentially independent (because the number of commercial pilots is far greater than the sample size of 8), and the probability that a randomly selected commercial pilot is woman is essentially constant (again because the number of commercial pilots is far greater than the sample size of 8).

b) (6 pts) Suppose that half of all commercial pilots are women. Determine the probability that I would encounter one or fewer women in a random sample of eight commercial pilots.

We want P(X<=1), where X has a binomial distribution with n=8 and p=.5. We can calculate P(X<=1) = P(X=0) + P(X=1) = (.5)^8 + 8(.5)(.5)^7 = 9(.5)^8, which is about .035.

c) (4 pts) Is the probability in (b) small enough to convince you that fewer than half of all commercial pilots are women? Explain your reasoning.

This probability is fairly small, so the sample data provide fairly strong evidence that fewer than half of commercial pilots are women. The reasoning is that if half of all commercial pilots were women, there’s only a 3.5% chance of seeing one or fewer women pilots in a random sample of eight. This is quite unlikely, so it seems more plausible that the proportion of females in the population of all commercial pilots is actually less than .5.

5. (22 pts) Suppose that a continuous random variable X has probability density function (pdf) given by: .

a) (6 pts) Determine the value of the constant c that makes this a legitimate pdf. [Hint: You may use geometry or calculus.]

The area of the rectangle between 0 and 1 is c, and the area of the rectangle between 1 and 2 is 2c, so the total are under this pdf is 3c. In order for this area to equal one, c=1/3.

b) (4 pts) Sketch a graph of this pdf.

c) (8 pts) Determine and sketch the cumulative distribution function (cdf) of X.

Let F(x) represent the cdf of X. For x<=0, F(x)=0. For x>=2, F(x)=1. For 0<x<1, F(x)==x/3. For 1<=x<2, F(x)= =(2x-1)/3.

d) (4 pts) Determine the median of X.

The median of X is the value m such that F(m)=.5. From the graph of F(x), it is clear that m>1, so we want (2m-1)/3 = 1/2. Solving gives m=5/4.

6. (16 pts) Suppose that the distance (in hundreds of miles) driven by a trucker in one day is a continuous random variable D whose probability density function (pdf) is given by: .

a) (6 pts) Determine the probability that the trucker travels more than 500 miles in a day.

P(D>5) = =39/64, or about .609.

b) (4 pts) Determine the probability that the trucker travels exactly 500 miles in a day.

P(D=5) = 0 because D is a continuous random variable.

c) (6 pts) Let the random variable T represent the time (in number of days) required for the trucker to make a 3000-mile cross-country trip, so T=30/D. Determine the expected value of T.

E(T) = E(30/D) = = 15/2 = 7.5 days