The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes.
(a) what is the probability that there are more than three calls in one-half hour?
(b) what is the probability that there are no calls within one half hour?
(c) determine x such that the probability that there are no calls within x hours is 0.01.

Answer

Let X be the time between calls to a corporate office. X has exponential distribution with mean 10 minutes

Let Y be the number of calls arrive in one half hour. So Y follows Poisson distribution with parameter

Y follows Poisson distribution with parameter 3. The probability distribution function of Y is ;

P(Y=y)  =   for y=0, 1, 2, 3, ....

a) Probability that there are more than three calls in one-half hour

  The number of calls arrive in one hour Y₁ is

                           =3

P(Y > 3) = 1  -  P(Y ≤ 3)

= 1 - [P(Y = 0) +P(Y = 1) +P( Y = 2) +P( Y = 3)]

=1-[0.0497+ 0.1494+0.2240+0.2240]

= 1- 0.6472

P(Y>3) = 0.3528

b) Probability that there are no calls within one-half hour

P(Y =0) =

P(Y =0) = 0.0498

c) Let x be the number for which probability that there will be no call within x hours is 0.01

-6 x = ln (0.01)

x =0.7675 hrs converting into minutes

X =46.05 minutes

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step-by-step explanation:

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