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CHAPTER 8 SECTION 3: CONTINUOUS PROBABILITY DISTRIBUTIONS MULTIPLE CHOICE 116. If the random variable X is exponentially distributed with parameter = 3, then the probability P(X 2) equals: a. 0.3333 b. 0.5000 c. 0.6667 d. 0.0025 ANS: D PTS: 1 REF: SECTION 8.3 117. If the random variable X is exponentially distributed with parameter = 1.5, then the probability P(2 X 4), up to 4 decimal places, is a. 0.6667 b. 0.0473 c. 0.5000 d. 0.2500 ANS: B PTS: 1 REF: SECTION 8.3 118. If the random variable X is exponentially distributed with parameter = 4, then the probability P(X 0.25), up to 4 decimal places, is a. 0.6321 b. 0.3679 c. 0.2500 d. None of these choices. ANS: A PTS: 1 REF: SECTION 8.3 119. The exponential density function f(x): a. is bell-shaped. b. is symmetrical. c. approaches infinity as x approaches zero. d. approaches zero as x approaches infinity. ANS: D PTS: 1 REF: SECTION 8.3 120. If the random variable X is exponentially distributed, then the mean of X will be: a. greater than the median. b. less than the median. c. equal to the median. d. Cannot tell; the answer depends on what is. ANS: A PTS: 1 REF: SECTION 8.3 121. Which of the following is not true for an exponential distribution with parameter ? a. = 1/ b. = 1/ c. The Y-intercept of f(x) is . d. All of these choices are true. ANS: D PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 122. If the mean of an exponential distribution is 2, then the value of the parameter is a. 0 b. 2.0 c. 0.5 d. 1.0 ANS: C PTS: 1 REF: SECTION 8.3 123. If the parameter of an exponential distribution is 1, then which of the following is not true? a. The density function is ex for x 0. b. The mean is equal to 1. c. The standard deviation and variance are both equal to 1. d. All of these choices are true. ANS: D PTS: 1 REF: SECTION 8.3 124. Which of the following can have an exponential distribution? a. Time between phone calls coming in to a technical support desk. b. Time until the first customer arrives at the bank in the morning. c. Lifetime of a new battery. d. All of these choices are true. ANS: D PTS: 1 REF: SECTION 8.3 TRUE/FALSE 125. In the exponential distribution, X takes on an infinite number of possible values in the given range. ANS: T PTS: 1 REF: SECTION 8.3 126. The mean and standard deviation of an exponential random variable are equal to each other. ANS: T PTS: 1 REF: SECTION 8.3 127. If the mean of an exponential distribution is 2, then the value of the parameter is 2.0. ANS: F PTS: 1 REF: SECTION 8.3 128. The mean and the variance of an exponential distribution are equal to each other. ANS: F PTS: 1 REF: SECTION 8.3 129. If the random variable X is exponentially distributed and the parameter of the distribution = 4, then P(X 1) = 0.25. ANS: F PTS: 1 REF: SECTION 8.3 130. The exponential distribution is suitable to model the length of time that elapses before the first telephone call is received by a switchboard. ANS: T PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 131. If the random variable X is exponentially distributed with parameter = 5, then the variance of X, 2 = V(X) = 0.04. ANS: T PTS: 1 REF: SECTION 8.3 132. If the random variable X is exponentially distributed with parameter = 0.05, then the variance of X, 2 = V(X) = 20. ANS: F PTS: 1 REF: SECTION 8.3 133. If the random variable X is exponentially distributed with parameter = 0.05, then the probability P(X > 20) = 0.3679. ANS: T PTS: 1 REF: SECTION 8.3 134. If the random variable X is exponentially distributed with parameter = 0.05, then the probability P(X < 5) = .2865. ANS: F PTS: 1 REF: SECTION 8.3 135. If the random variable X is exponentially distributed with parameter = 2, then the probability that X is between 1 and 2 equals the probability that X is between 2 and 3. ANS: F PTS: 1 REF: SECTION 8.3 COMPLETION 136. A random variable with density function ex for x 0 has an exponential distribution with = ____________________. ANS: one 1 PTS: 1 REF: SECTION 8.3 137. A random variable with density function ex for x 0 has an exponential distribution whose mean is ____________________. ANS: one 1 PTS: 1 REF: SECTION 8.3 138. A random variable with density function 0.01ex/100 for x 0 has an exponential distribution whose mean is ____________________. ANS: 100 PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 139. The shape of the density function for an exponential distribution is ____________________. ANS: skewed positively skewed PTS: 1 REF: SECTION 8.3 140. The mean of an exponential random variable is ____________________ the median. ANS: greater than > PTS: 1 REF: SECTION 8.3 141. If X has an exponential distribution, the possible values of X are from ____________________ to infinity. ANS: zero 0 PTS: 1 REF: SECTION 8.3 142. An exponential random variable is an example of a(n) ____________________ random variable. ANS: continuous PTS: 1 REF: SECTION 8.3 143. If X has an exponential distribution with parameter , then f(0) = ____________________. ANS: PTS: 1 REF: SECTION 8.3 144. If X has an exponential distribution with parameter , then the mean of X is ______________. ANS: 1/ PTS: 1 REF: SECTION 8.3 145. If X has an exponential distribution, then f(x) approaches ____________________ as x approaches infinity. ANS: zero 0 PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 146. The y-intercept of the density function for an exponential distribution with parameter 10 is ____________________. ANS: 10 (0, 10) y = 10 PTS: 1 REF: SECTION 8.3 147. If X has an exponential distribution, its ____________________ is equal to its ____________________. ANS: mean; standard deviation standard deviation; mean PTS: 1 REF: SECTION 8.3 SHORT ANSWER 148. Let X be an exponential random variable with = 1.50. Find the following: a. P(X 2) b. P(X 4) c. P(1 X 3) d. P(X = 1) ANS: a. 0.0498 (note that f(x) = 1.50ex for x 0) b. 0.9975 c. 0.2120 d. 0 PTS: 1 REF: SECTION 8.3 149. Let X be an exponential random variable with = 1.50. Find the following: a. f(x) b. The y-intercept of f(x) ANS: a. f(x) = 1.50ex for x 0 b. (0, 1.50) PTS: 1 REF: SECTION 8.3 150. Suppose X has an exponential distribution with mean 2. Find f(x). ANS: f(x) = 0.50ex for x 0 PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. Car Salesman A used car salesman in a small town states that, on the average, it takes him 5 days to sell a car. Assume that the probability distribution of the length of time between sales is exponentially distributed. 151. {Car Salesman Narrative} What is the probability that he will have to wait at least 8 days before making another sale? ANS: 0.2019 (Note is 1/5 = 0.20 days.) PTS: 1 REF: SECTION 8.3 152. {Car Salesman Narrative} What is the probability that he will have to wait between 6 and 10 days before making another sale? ANS: 0.1659 PTS: 1 REF: SECTION 8.3 Repair Time The time it takes a technician to fix a computer problem is exponentially distributed with a mean of 15 minutes. 153. {Repair Time Narrative} What is the probability density function for the time it takes a technician to fix a computer problem? ANS: f(x) = (1/15)ex/15, x 0 PTS: 1 REF: SECTION 8.3 154. {Repair Time Narrative} What is the probability that it will take a technician less than 10 minutes to fix a computer problem? ANS: 0.4866 PTS: 1 REF: SECTION 8.3 155. {Repair Time Narrative} What is the variance of the time it takes a technician to fix a computer problem? ANS: 225 PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 156. {Repair Time Narrative} What is the probability that it will take a technician between 10 to 15 minutes to fix a computer problem? ANS: 0.1455 PTS: 1 REF: SECTION 8.3 Light Bulb Lifetime The lifetime of a light bulb (in hours) is exponentially distributed with = 0.008. 157. {Light Bulb Lifetime Narrative} What is the mean and standard deviation of the light bulb's lifetime? ANS: = = 1/ = 1/0.008 = 125 hours PTS: 1 REF: SECTION 8.3 158. {Light Bulb Lifetime Narrative} Find the probability that a light bulb will last between 120 and 140 hours. ANS: P(120 X 140) = e0.008(120) e0.008(140) = 0.3829 0.3263 = 0.0566 PTS: 1 REF: SECTION 8.3 159. {Light Bulb Lifetime Narrative} Find the probability that a light bulb will last for: a. more than 125 hours. b. at most 125 hours. c. no more than 125 hours. d. exactly 125 hours. e. less than 125 hours. f. at least 125 hours. g. no less than 125 hours. ANS: a. P(X > 125) = 0.3679 b. P(X 125) = 0.6321 c. P(X 125) = 0.6321 d. P(X = 125) = 0 e. P(X < 125) = 0.6321 f. P(X 125) = 0.3679 g. P(X 125) = 0.3679 PTS: 1 REF: SECTION 8.3 Drive Through Window Suppose that customers arrive at a drive through window at an average rate of three customers per minute and that their arrivals follow the Poisson model. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 160. {Drive Through Window Narrative} Write the probability density function of the distribution of the time that will elapse before the next customer arrives. ANS: Let T = Elapsed time before the next customer arrives. The random variable T follows an exponential distribution where = 3; with mean 1/3 minute between customers. Then the probability density function of T is f(t) = 3et, t 0 minutes. PTS: 1 REF: SECTION 8.3 161. {Drive Through Window Narrative} Use the appropriate exponential distribution to find the probability that the next customer will arrive within 1.5 minutes. ANS: 0.9889 PTS: 1 REF: SECTION 8.3 162. {Drive Through Window Narrative} Use the appropriate exponential distribution to find the probability that the next customer will not arrive within the next 2 minutes. ANS: 0.0025 PTS: 1 REF: SECTION 8.3 Catalog Orders The JC Penney catalog department that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. 163. {Catalog Orders Narrative} What is the value of , the parameter of the exponential distribution in this situation? ANS: Since = 3, then PTS: 1 . REF: SECTION 8.3 164. {Catalog Orders Narrative} What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? ANS: P(X > 1.5) = e0.5 = 0.6065 PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 165. {Catalog Orders Narrative} Find the waiting time at which only 10% of the customers will continue to hold. ANS: P(X > x) = ex ex/3 = .10 x = 6.908 minutes. PTS: 1 REF: SECTION 8.3 166. {Catalog Orders Narrative} Find the time at which 50% of the customers will continue to hold? ANS: P(X > x) = ex ex/3 = .50 x = 2.079 minutes. PTS: 1 REF: SECTION 8.3 167. {Catalog Orders Narrative} What proportion of callers are put on hold longer than 3 minutes? ANS: P(X > 3) = e3/3 = e1 = 0.3679. PTS: 1 REF: SECTION 8.3 168. {Catalog Orders Narrative} What is the probability that a randomly selected caller is placed on hold for fewer than 6 minutes? ANS: P(X < 6) = 1 e6/3 = 1 e2 = 0.8647. PTS: 1 REF: SECTION 8.3 169. {Catalog Orders Narrative} What is the probability that a randomly selected caller is placed on hold for 3 to 6 minutes? ANS: P(3 < X < 6) = e3/3 e6/3 = e1 e2 = 0.2325. PTS: 1 REF: SECTION 8.3 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.