124 CHAPTER 4 CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
Also, the cumulative distribution function can be used to obtain the same result as follows:
An identical answer is obtained by expressing the mean number of log-ons as 0.417 log-
ons per minute and computing the probability that the time until the next log-on exceeds 6
minutes. Try it.
What is the probability that the time until the next log-on is between 2 and 3 minutes?
Upon converting all units to hours,
An alternative solution is
Determine the interval of time such that the probability that no log-on occurs in the inter-
val is 0.90. The question asks for the length of time xsuch that. Now,
Take the (natural) log of both sides to obtain. Therefore,
x0.00421 hour0.25 minute
25 xln 1 0.90 2 0.1054
P 1 Xx 2 e^25 x0.90
P 1 Xx 2 0.90
P 1 0.033X0.05 2 F 1 0.05 2 F 1 0.033 2 0.152
P 1 0.033X0.05 2
0.05
0.033
25 e^25 x dxe^25 x `
0.05
0.033
0.152
P 1 X0.1 2 1 F 1 0.1 2 e^251 0.1^2
0
0.0
0.4
0.8
1.2
1.6
2.0
2 4 6 8 10 12
x
f(x)
2
0.5
0.1
λ
Figure 4-22 Probability density function of expo-
nential random variables for selected values of .
0.1 x
f(x)
Figure 4-23 Probability for the expo-
nential distribution in Example 4-21.
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