Question

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 1

Answer

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

$\lambda = \dfrac{1}{E(X)}$

$\lambda = \dfrac{1}{10}\ minutes$

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

 $30 \lambda = 30 \frac{1}{10}$

 $30 \lambda = 3$

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

P(Y=y)  = $e^{-3} \dfrac{3^{y}}{y!}$   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

$30 \lambda = 30 \frac{1}{10}$

                           =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) = $\dfrac{e^{-3} 3^{0}}{0!}$

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

$\dfrac{e^{-6x} 6x^{0}}{0!} =0.01$

 $e^{-6x} =0.01$

-6 x = ln (0.01)

$x = \dfrac{-ln(0.01)}{6}$

x =0.7675 hrs converting into minutes

X =46.05 minutes