A firm has monitored the duration of longdistance telephone

A firm has monitored the duration of long-distance telephone calls placed by its employees to help it decide which lonng-distance package to purchase. The duration of calls was found to be exponentially distribution, with a mean call length of 5 minutes.

(a) What is the probability a call lasts more than 2 minutes?

(b) What is the probability a call lasts between 5 and 8 minutes?

(c) If a call has lasted 3 minutes, what is the probability it lasts an additional 4 minutes?

Solution

Given X~exponentially distribution with mean=5

F(x)=1-exp(-x/5) for x>0

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(a) What is the probability a call lasts more than 2 minutes?

P(X>2)=1-P(X<=2)

=1-F(2)

=1-(1-exp(-2/5))

=0.67032

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(b) What is the probability a call lasts between 5 and 8 minutes?

P(5<=X<=8) = F(8)- F(5)

=(1-exp(-8/5)) - (1-exp(-5/5))

=0.1659829

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(c) If a call has lasted 3 minutes, what is the probability it lasts an additional 4 minutes?

P(X=4| X=3) = P(X=4)

=1-exp(-4/5)

=0.55067