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

Section 8.5 Probability 575
(b) The probability that X lies between 4 and 8 is
Ps4 < X < 8d − y 8
f sxd dx − 0.006 y 8
s10x 2 x 2 d dx
4
4
− 0.006f5x 2 2 1 3 x 3 g 4
8
− 0.544
n
Example 2 Phenomena such as waiting times and equipment failure times are commonly
modeled by exponentially decreasing probability density functions. Find the
exact form of such a function.
SOLUTION Think of the random variable as being the time you wait on hold before
an agent of a company you’re telephoning answers your call. So instead of x, let’s use
t to represent time, in minutes. If f is the probability density function and you call at
time t − 0, then, from Definition 1, y 2 f std dt represents the probability that an agent
0
answers within the first two minutes and y 5 f std dt is the probability that your call is
4
answered during the fifth minute.
It’s clear that f std − 0 for t , 0 (the agent can’t answer before you place the call).
For t . 0 we are told to use an exponentially decreasing function, that is, a function of
the form f std − Ae 2ct , where A and c are positive constants. Thus
f std −H 0 Ae 2ct if t , 0
if t > 0
We use Equation 2 to determine the value of A:
1 − y`
2`
f std dt − y 0 f std dt 1 y`
f std dt
2`
0
y
c
f(t)= 0 ce _ct if t<0
if t˘0
− y`
0 Ae2ct dt − lim
x l ` y x
0
− lim
x l `F2 A c e2ctG0
− A c
x
Ae 2ct dt
A
− lim
x l ` c s1 2 e2cx d
Therefore Ayc − 1 and so A − c. Thus every exponential density function has the form
0
FIGURE 2
An exponential density function
t
A typical graph is shown in Figure 2.
f std −H 0 ce 2ct if t , 0
if t > 0
n
Average Values
Suppose you’re waiting for a company to answer your phone call and you wonder how
long, on average, you can expect to wait. Let f std be the corresponding density function,
where t is measured in minutes, and think of a sample of N people who have called this
company. Most likely, none of them had to wait more than an hour, so let’s restrict our
attention to the interval 0 < t < 60. Let’s divide that interval into n intervals of length
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