Homework 6 Solutions

1. Ross 5.5 p. 247

The probability of the gasoline supply being exhausted for a tank capacity c is obtained by integrating from c to 1. Here, we want the probability of the supply exhausted to be 0.01; hence, we must choose c such that

Solving for c gives us , so the tank capacity needs to be about 601.9 gallons in order for the probability of the supply exhausted in a given week to be 0.01.

2. Ross 5.6 p. 248

(changing variables using )

(since the integrand is an odd function)

3. Ross 5.11 p. 248

The statement tells us that X (a random variable representing the location of the point) has a uniform distribution on (0, L).

In order for the ratio of the shorter to the longer segment to be less than 1/4, the shorter segment cannot be more than 1/5 of the total bar length L; in other words, X must either be less than or greater than .

Since X is uniformly distributed, this means the probability that the ratio of the shorter to the longer line segment is less than 1/4 is just .

4. Ross 5.12 p. 248

Let X be the random variable of the distance from the breakdown to the nearest service station.

For the first case, if the bus breaks down between 0 and 25, then it is closest to city A’s station; if it breaks down between 25 and 75, it is closest to the station in the center of the route; and if it breaks down between 75 and 100, it is closest to city B’s station. Thus, the expected distance from a breakdown to the nearest service station in the first case is just

On the other hand, if the 3 service stations are located 25, 50, and 75 miles from city A, then a bus that breaks down between 0 and 37.5 is closest to the 1st station; a bus that breaks down between 37.5 and 62.5 is closest to the 2nd station; and a bus that breaks down between 62.5 and 100 is closest to station 3. Thus, the expected distance from a breakdown to the nearest service station in the second case is

So, the 2nd method yields a smaller expected distance from the breakdown to the nearest service station; hence, the station set-up under the 2nd case is indeed more efficient.

5. Ross 5.20 p. 249

Let X be the number of people in favor of the proposition in a random sample of 100. has a binomial distribution with and . Since is greater than 10, we can use the normal approximation to the binomial distribution. Then X is approximately normal with parameters and .

a) Then the probability that at least 50 favor the proposition is

b) The probability that between 60 and 70 inclusive are in favor of the proposal is

c) The probability that fewer than 75 are in favor is

6. Ross 5.21 p. 249

Let X be a random variable representing the height of a 25-year-old man. Then X has a normal distribution with parameters and .

The probability that a randomly chosen 25-year-old man is over 6’2” tall is just

So, approximately 11.51% of 25-year-old men are over 6’2” tall.

The conditional probability that a 25-year-old man will be over 6’5” tall given that he is over 6’ tall is just

So, about 2.4% of men in the 6’ club are over 6’5” tall.

7. Ross 5.29 p. 250

Let s be the initial price of the stock, and let X denote the number of times (in the 1000 time periods) the stock increases in value; then the price of the stock at the end of the 1000 periods is

For the price to be up at least 30%, we need to have

Substituting in and gives us ; i.e., the stock would have to rise in at least 470 time periods.

Now, X has a binomial distribution with parameters and . Since , we can use the normal approximation to the binomial distribution. So

8. Ross 5.30 p. 250

The probability that a point with a reading of 5 is in the black region is

Thus, we need to find the value of that yields (i.e., the probability of making an error is the same regardless of which region one concludes the point is in).

9. Ross 5.32 p. 250

a) Let X be the time required to repair a machine; then X ~ Exp(1/2). Hence, the probability that a repair time exceeds 2 hours is just

b) The conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours is exactly the probability that a repair takes at least 1 hour (thanks to the memory-less property of exponential random variables):

10. Ross 5.34 p. 250

a) Let X be the total number of miles (in thousands) that a car can be driven before needing to be junked. In this part, X ~ Exp(1/20). The probability that Jones gets at least 20,000 additional miles out of a used car that already has at least 10,000 miles under it is just (thanks again to the memory-less property of exponential random variables) the probability that one gets at least 20,000 miles out of a car before junking it:

b) In this part, instead assume that X ~ Uniform(0, 40). The probability that Jones gets at least 20,000 additional miles out of a used car that already has at least 10,000 miles under it is now