James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 8.5 Probability 577
To evaluate this integral we use integration by parts, with u − t and dv − ce 2ct dt, so
du − dt and v − 2e 2ct :
y`
0
tce 2ct dt − lim
x l ` y x
tce 2ct dt − lim
0
x l `S2te 2ct g x 1 0 y x
e dtD
2ct 0
The limit of the first term is 0 by
l’Hospital’s Rule.
− lim
x l `S2xe 2cx 1 1 c 2 e2cx
c
D − 1 c
The mean is − 1yc, so we can rewrite the probability density function as
f std −H 0 21 e 2ty if t , 0
if t > 0
n
Example 4 Suppose the average waiting time for a customer’s call to be answered
by a company representative is five minutes.
(a) Find the probability that a call is answered during the first minute, assuming that an
exponential distribution is appropriate.
(b) Find the probability that a customer waits more than five minutes to be answered.
SOLUTION
(a) We are given that the mean of the exponential distribution is − 5 min and so,
from the result of Example 3, we know that the probability density function is
f std −H 0 0.2e 2ty5 if t , 0
if t > 0
where t is measured in minutes. Thus the probability that a call is answered during the
first minute is
Ps0 < T < 1d − y 1
f std dt
0
− y 1
0.2e 2ty5 dt − 0.2s25de 2ty5 1
g 0
0
− 1 2 e 21y5 < 0.1813
So about 18% of customers’ calls are answered during the first minute.
(b) The probability that a customer waits more than five minutes is
PsT . 5d − y`
f std dt − y`
0.2e 2ty5 dt
5
5
− lim y x
0.2e 2ty5 dt − lim
x l ` 5
x l ` se21 2 e 2xy5 d
− 1 e 2 0 < 0.368
About 37% of customers wait more than five minutes before their calls are answered.
n
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