Questions 1 Through 4 Are Based on the Following Information

EF 507 PS-Chapter 6 FALL 2008

QUESTIONS 1 THROUGH 4 ARE BASED ON THE FOLLOWING INFORMATION: The amount of time you have to wait at a particular stoplight is uniformly distributed between zero and two minutes.

1. What is the probability that you have to wait more than 30 seconds for the light? A) 0.25 B) 0.50 C) 0.75 D) 1.01 ANSWER: C

2. What is the probability that you have to wait between 15 and 45 seconds for the light? A) 0.15 B) 0.25 C) 0.35 D) 0.45 ANSWER: B

3. Eighty percent of the time, the light will change before you have to wait how long?

A) 90 seconds B) 24 seconds C) 30 seconds D) 96 seconds ANSWER: D

4. Sixty percent of the time, the light will change before you have to wait how long?

A) 72 seconds B) 60 seconds C) 48 seconds D) 36 seconds ANSWER: A

QUESTIONS 5 THROUGH 7 ARE BASED ON THE FOLLOWING INFORMATION: Let the random variable Z follow a standard normal distribution. 5. What is P(Z > 1.2)? A) 0.1112 B) 0.8849 C) 0.1151 D) 0.6112 ANSWER: C

1 6. What is P(Z > -0.21)? A) 0.4207 B) 0.4168 C) 0.5793 D) 0.5832 ANSWER: D

7. What is P(0.33 < Z < 0.45)? A) 0.5443 B) 0.0443 C) 0.4557 D) 0.1515 ANSWER: B

QUESTIONS 8 AND 10 ARE BASED ON THE FOLLOWING INFORMATION: Let the random variable X follow a normal distribution with a mean of 61.7 and a standard deviation of 5.2. 8. What is the value of k such that P(X > k) = 0.63? A) 59.984 B) 66.446 C) 62.830 D) 67.576 ANSWER: A

9. What is the value of k such that P(59 < X < k) = 0.54? A) 65.8 B) 64.6 C) 63.7 D) 66.9 ANSWER: D

10. Let the random variable Z follow a standard normal distribution. Find P(0 < Z < 0.57). A) 0.2843 B) 0.7843 C) 0.2157 D) 0.7157 ANSWER: C

11. Let the random variable Z follow a standard normal distribution. Find P(-2.21 < Z < 0). A) 0.9864 B) 0.4864 C) 0.0136 D) 0.5136 ANSWER: B

2 QUESTIONS 12 AND 13 ARE BASED ON THE FOLLOWING INFORMATION: During a professor’s office hours, students arrive, on average, every ten minutes. Assume that the distribution of the time between arrivals follows an exponential distribution. Suppose that a student has just left. 12. What is the probability that the professor has more than 20 minutes before the next student shows up? A) 0.8187 B) 0.8647 C) 0.1353 D) 0.1813 ANSWER: C 13. Suppose that a student has just left. Only 25% of the time will the professor have to wait approximately how long or longer before the next student shows up? A) 13.9 minutes B) 14.6 minutes C) 2.9 minutes D) 15.3 minutes ANSWER: A

14. Let the random variable Z follow a standard normal distribution, and let z1 be a possible value of Z that is representing the 10th percentile of the standard

normal distribution. Find the value of z1 . A) 0.255 B) -0.255 C) 1.28 D) -1.28 ANSWER: D

15. A very large logging operation has serious problems keeping their skidders operating properly. The equipment fails at the rate of 3 breakdowns every 48 hours. Assume that x is time between breakdowns and is exponentially distributed. The probability of two or less breakdowns in the next 48-hour period is A) 0.9672. B) 0.4230. C) 0.2231. D) 0.7769. ANSWER: B

16. If the random variable x is exponentially distributed with parameter l = 1.5, then the probability P(2 X 4) is equal to A) 0.6667. B) 0.5000. C) 0.0473. D) 0.2500. ANSWER: C

3 QUESTIONS 17 THROUGH 19 ARE BASED ON THE FOLLOWING INFORMATION: The amount of time you have to wait at a dentist office before you called in is uniformly distributed between zero and twenty minutes. 17. What is the probability that you have to wait more than 8 minutes? ANSWER: P(X > 8) = (20-8) / (20-0) = 12 / 20 = 0.60 18. What is the probability that you have to wait between 10 and 15 minutes? ANSWER: P(10 < X < 15) = (15-10) / (20-0) = 5 / 20 = 0.25 19. Seventy percent of the time, you will be called in before you have to wait how long? ANSWER: The range is a total of 20 minutes. Seventy percent of the time, you should be called in before you wait 14 minutes

20. Suppose that 24% of all sales in a grocery store are for amounts greater than $100. In a random sample of 50 invoices, what is the probability that more than ten of the invoices are for over $100? Use the normal approximation to the binomial distribution, both with and without the continuity correction. ANSWER: P = 0.23, n = 50, m = E(X) = nP = 12, s 2 = Var(X) = nP(1-P) = 9.12, s = 3.02 Without continuity correction P(X>10) = P(Z>-0.66) = 0.7454. 21. It has been found that 62.1% of all unsolicited third class mail delivered to households goes unread. If, over the course of a month, a household receives 150 pieces of unsolicited mail, what is the probability that the household discards more than 80 pieces of the mail without reading it? Use the normal approximation to the binomial distribution without the continuity correction. ANSWER: P = 0.621, n = 150, m =E(X) = 93.15, s 2 =Var(X) = nP(1-P) = 35.3039, s X = 5.942 P(X > 80) = P(Z > -2.21) = 0.9864. 22. You are the Webmaster for your firm’s Website. From your records, you know that the probability that a visitor will buy something from your firm is 0.23. If the number of visitors in one day is 952, what is the probability that less than 200 of them will buy something from your firm? Use the normal approximation to the binomial without the continuity correction. ANSWER: n = 952, P = 0.23, m =E(X) = 218.96, s 2 =Var(X) = nP(1-P) = 168.5992, s = 12.985 P(X<200) =P(Z<-1.46) = 0.0721 QUESTIONS 23 AND 24 ARE BASED ON THE FOLLOWING INFORMATION: During hours, students arrive, on average, every ten minutes. Assume that the distribution of the time between arrivals follows an exponential distribution. Suppose that a customer has just left the gas station.

4 23. What is the probability that the cashier at the gas station has more than 15 minutes before the next customer arrive? ANSWER: Exponential with mean 1/ l =10, then, l = 0.10, and P(T>15) = e-15l = e - 1.5 = 0.223 24. The probability is 0.30 that the cashier has to wait at least how long or longer before the next customer arrives? ANSWER: P(T > t) = elt= e-0.10 t = 0.30 -0.10t = -1.204 t =12.04 minutes 25. The total cost for a production process is equal to $1,200 plus 2.5 times the number of units produced. The mean and variance for the number of units produced are 520 and 840, respectively. Find the mean and variance of the total cost.

ANSWER: Let X = number of units produced, and T = total cost for the production process. 2 Since T = 1200 + 2.5X, mX = 520 and s X = 840, then 2 2 2 mT=1200 + 2.5 m X = 1200 + (2.5)(520) = 2500, and sT=(2.5) s X = (6.25)(840) = 5250.

26. A random variable X is normally distributed with mean of 50 and variance of 50, and a random variable Y is normally distributed with mean of 100 and variance of 200. Given the random variables X and Y have a correlation coefficient equal to 0.50, find the mean and variance of the random variable W = 4X+3Y.

ANSWER: 2 2 mX=50, s X = 50, m Y = 100, s Y = 200, Corr(X , Y ) = 0.50

mW=4 m X + 3 m Y = (4)(50) + (3)(100) = 500 2 2 2 2 2 sW=(4) s X + (3) s Y + 2(4)(3)Corr(X , Y ) s X s Y = (16)(50) +(9)(200) + (24)(0.50)(7.071)(14.142) = 3,799.977

27. A normal random variable x has an unknown mean  and standard deviation   2.5 If the probability that x exceeds 7.5 is 0.8289, find .

ANSWER: It is given that x is normally distributed with  = 2.5 but with unknown mean  , and that P (x > 7.5) = 0.8289. In terms of the standard normal random variable z, we can write P( X> 7.5) = P [ Z > (7.5 -m ) / 2.5] = 0.8289 Since the area to the right of (7.5  ) / 2.5 is greater than 0.5, then (7.5  ) / 2.5 must be negative, and that P[ (7.5  ) / 2.5 < z < 0] = 0.3289. Hence, (7.5  ) / 2.5 = -.095. This implies  = 9.875.