Download CHAPTER 8 SECTION 3: CONTINUOUS PROBABILITY

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
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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 ex 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
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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 ex for x  0 has an exponential distribution with  =
____________________.
ANS:
one
1
PTS: 1
REF: SECTION 8.3
137. A random variable with density function ex 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 0.01ex/100 for x  0 has an exponential distribution whose
mean is ____________________.
ANS: 100
PTS: 1
REF: SECTION 8.3
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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
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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) = 1.50ex 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) = 1.50ex 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) = 0.50ex for x  0
PTS: 1
REF: SECTION 8.3
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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)ex/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
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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) = e0.008(120)  e0.008(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.
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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) = 3et, 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) = e0.5 = 0.6065
PTS: 1
REF: SECTION 8.3
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165. {Catalog Orders Narrative} Find the waiting time at which only 10% of the customers will continue to
hold.
ANS:
P(X > x) = ex  ex/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) = ex  ex/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) = e3/3 = e1 = 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  e6/3 = 1  e2 = 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) = e3/3  e6/3 = e1  e2 = 0.2325.
PTS: 1
REF: SECTION 8.3
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copied, or distributed without the prior consent of the publisher.