# Solved: The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of

#### ByRaju Chaudhari

Oct 15, 2020

The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes.

a. What is the probability that there are no calls within a 30-minute interval?
b. What is the probability that at least one call arrives within a 10-minute interval?
c. What is the probability that the first call arrives within 5 and 10 minutes after opening?

#### Solution

Let $T$ denote the time until the first call. Then the random variable $T$ is exponentially distributed with $\lambda = 1/15$ calls/minute.

The pdf of $T$ is

 \begin{aligned} f(t) &= \frac{1}{15}e^{-t/15},\; t>0. \end{aligned}

The distribution function of $T$ is

 \begin{aligned} F(t) &= P(T\leq t) = 1- e^{-t/15}. \end{aligned}

(a) The probability that there are no calls within a 30-minute interval is

 \begin{aligned} P(t > 30) &= 1- P(t\leq 30)\\ & = 1- F(30)\\ & = 1- \big[1- e^{-30/15}\big]\\ &= e^{-30/15}\\ & = 0.1353 \end{aligned}

(b) The probability that no call in 10 minutes interval is

 \begin{aligned} P(t > 10) &= 1- P(t < 10)\\ & = 1- F(10)\\ & = 1- \big[1- e^{-10/15}\big]\\ &= e^{-30/15}\\ & = 0.5134 \end{aligned}

Thus, the probability of at least one call in a 10-minute interval equals one minus the probability of zero calls in a 10-minute interval.

The required probability is

 \begin{aligned} P(t < 10) &= 1- P(t > 10)\\ &= 1-0.5134\\ &=0.4866 \end{aligned}

(c) The probability that the first call arrives within 5 and 10 minutes after opening

 \begin{aligned} P(5 < t < 10) &= P(t < 10) - P(t < 5)\\ & = F(10) -F(5)\\ & = \big[1- e^{-10/15}\big] -\big[1- e^{-5/15}\big]\\ &= e^{-5/15} - e^{-10/15}\\ & = 0.7165-0.5134\\ &= 0.2031 \end{aligned}