Chapter 6, Problem 39RE

a.

To determine

To find: the probability that a customer is served in the first 3 minutes.

Answer to Problem 39RE

  $P(0\le x\le 3)\approx 0.3127$ .

Explanation of Solution

Given information: The length of time spent waiting in line at a certain bank is

modeled by an exponential density function with mean 8 minutes.

Calculation:

The probability density function is:

  $\begin{array}{l}f(t)=0,ift<0\\ f(t)=\frac{1}{8}{e}^{-\frac{t}{8}},ift\ge 0\end{array}$

  $\begin{array}{l}P(0\le x\le 3)={\displaystyle \underset{0}{\overset{3}{\int }}\frac{1}{8}}{e}^{\frac{-t}{8}}dt=[-e\frac{-t}{8}]\begin{array}{c}3\\ 0\end{array}=-e^{\frac{-3}{8}}-(-1)\approx 0.3127\\ P(0\le x\le 3)\approx 0.3127\end{array}$

b.

To determine

To find: the probability that a customer has to wait more than 10 minutes.

Answer to Problem 39RE

  $P(X>10)=0.287$

Explanation of Solution

Given information:The length of time spent waiting in line at a certain bank is

modeled by an exponential density function with mean 8 minutes.

Calculation:

The probability density function is:

  $\begin{array}{l}f(t)=0,ift<0\\ f(t)=\frac{1}{8}{e}^{-\frac{t}{8}},ift\ge 0\end{array}$

  $\begin{array}{c}P(X>10)=1-P(x\le 10)\\ =1-{\displaystyle \underset{0}{\overset{10}{\int }}f(x)dt}\\ =1-{\displaystyle \underset{0}{\overset{10}{\int }}\frac{1}{8}}{e}^{-\frac{t}{8}}dt\\ =1+{\displaystyle \underset{0}{\overset{10}{\int }}\frac{1}{8}}{e}^{-\frac{t}{8}}(-\frac{1}{8})dt\\ \text{Let}u=-\frac{1}{8}t\text{then}du=\frac{1}{8}dt\\ P(X>10)=1+{\displaystyle \underset{0}{\overset{10}{\int }}{e}^u}du\\ =1+e^u|\begin{array}{c}10\\ 0\end{array}\\ =1+e^{-\frac{1}{8}t}|\begin{array}{c}10\\ 0\end{array}\\ =1+[e^{-\frac{1}{8}10}-e^{-\frac{1}{8}0}]\\ =1+e^{-\frac{5}{4}}-1\\ =e^{-\frac{5}{4}}\\ =0.287\\ P(X>10)=\mathrm{0.287.}\end{array}$

c.

To determine

To find: the median waiting time.

Answer to Problem 39RE

The median waiting time is 5.55.

Explanation of Solution

Given information:The length of time spent waiting in line at a certain bank is

modeled by an exponential density function with mean 8 minutes.

Calculation:

The probability density function is:

  $\begin{array}{l}f(t)=0,ift<0\\ f(t)=\frac{1}{8}{e}^{-\frac{t}{8}},ift\ge 0\end{array}$

The median of a probability function is a number m such that:

  ${\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=\frac{1}{2}}$ .

  $\begin{array}{c}{\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=-\underset{t\to \infty }{\lim }}[{\displaystyle \underset{m}{\overset{t}{\int }}-e^{-\frac{1}{8}t}}(-\frac{1}{8})]dt\\ {\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=-\underset{t\to \infty }{\lim }}[-e^{-\frac{1}{8}t}]|\begin{array}{c}t\\ m\end{array}\\ {\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=-\underset{t\to \infty }{\lim }}[-e^{-\frac{1}{8}t}+e^{-\frac{1}{8}m}]\\ {\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=0+}{e}^{-\frac{1}{8}m}\\ {\displaystyle \underset{m}{\overset{\infty }{\int }}f(x)dx=}{e}^{-\frac{1}{8}m}\end{array}$

  $\begin{array}{l}\frac{1}{2}=e^{-\frac{1}{8}m}\\ 2^{-1}=e^{-\frac{1}{8}m}\\ \text{Take}\ln \text{of}\text{both}\text{side}\\ \ln 2^{-1}=\ln e^{-\frac{1}{8}m}\\ -\ln 2=-\frac{1}{8}m\\ m=8\ln 2\\ m=8(0.6932)\\ m=5.55\end{array}$