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.

AnswerLet X be the time between calls to a corporate office. X has exponential distribution with mean 10 minutes [tex]\lambda = \dfrac{1}{E(X)}[/tex] [tex]\lambda = \dfrac{1}{10}\ minutes[/tex] Let Y be the number of calls arrive in one half hour. So Y follows Poisson distribution with parameter  [tex]30 \lambda = 30 \frac{1}{10} [/tex]  [tex]30 \lambda = 3 [/tex]Y follows Poisson distribution with parameter 3. The probability distribution function of Y is ;P(Y=y)  = [tex]e^{-3} \dfrac{3^{y}}{y!}[/tex]   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[tex]30 \lambda = 30 \frac{1}{10}[/tex]                            =3P(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- 0.6472P(Y>3) = 0.3528b) Probability that there are no calls within one-half hourP(Y =0) = [tex]\dfrac{e^{-3} 3^{0}}{0!}[/tex] P(Y =0) = 0.0498c) Let x be the number for which probability that there will be no call within x hours is 0.01[tex]\dfrac{e^{-6x} 6x^{0}}{0!} =0.01[/tex]  [tex]e^{-6x} =0.01[/tex]-6 x = ln (0.01)[tex]x = \dfrac{-ln(0.01)}{6}[/tex]x =0.7675 hrs converting into minutesX =46.05 minutes