Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes.

a. Define the random variable. $X=$

b. Is X continuous or discrete?

c. $X~$

d. $\mu =$

e. $\sigma =$

f. Draw a graph of the probability distribution. Label the axes.

g. Find the probability that a phone call lasts less than nine minutes.

h. Find the probability that a phone call lasts more than nine minutes.

i. Find the probability that a phone call lasts between seven and nine minutes.

j. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why?

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The random variable, $\mathrm{X}$can be defined as,

$\mathrm{X}=$The duration of long-distance calls.

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

Because the random variable X has an exponentially distributed distribution, X will be continuous.

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

According to the information provided, the random variable X has an exponential distribution.

$$\begin{gathered} \mathrm{X}~Exp\left(\frac{1}{8}\right) \\ \mathrm{X}~Exp(0.125) \end{gathered}$$

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The formula for calculating the mean is as follows:

$\mu =8.$

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The formula for calculating standard deviation is:

$$\begin{gathered} \sigma =\mu \\ \sigma =8 \end{gathered}$$

Therefore standard deviation is $\sigma =8$.

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

We know that,

$X~Exp(0.125),\mathrm{m}=0.125$

Probability density function of exponential distribution is given by,

$$\begin{gathered} f\left(x\right)=m{e}^{-mx} \\ f\left(x\right)=0.125e^{-0.125x} \end{gathered}$$

From above equation which represents the probability density function, make the graph by using the above obtained probability density function.

The value of maximum f(x) will lie on y axis and at x=0 is given by,

$$\begin{gathered} \mathrm{f}\left(0\right)=m \\ \mathrm{f}\left(0\right)=0.125{\mathrm{e}}^{-0.125\times 0} \\ \mathrm{f}\left(0\right)=0.125{\mathrm{e}}^0 \\ \mathrm{f}\left(0\right)=0.125 \end{gathered}$$

Graph can be written as below,

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

Calculate the required probability using,

$$\begin{gathered} P(X<x)=1-m{e}^{-mx} \\ P(X<9)=1-e^{-0.125\times 9} \\ P(X<9)=1-0.3247 \\ P(X<9)=0.6753 \end{gathered}$$

Therefore , The chances of a phone call lasting less than nine minutes are slim. $P(X<9)=0.6753$.

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The required probability can be calculated using the following formula:

$$\begin{gathered} P(X>x)=1-P(x<9) \\ P(X>9)=1-0.6753 \\ P(X>9)=0.3247 \\ P(X<9)=0.6753 \end{gathered}$$

Therefore the probability that phone call lasts more than nine minutes $P(X>9)=0.3247$.

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The required probability can be calculated using the following formula:

$$\begin{gathered} \mathrm{P}(7<\mathrm{x}<9)=\mathrm{P}(\mathrm{x}<9)-\mathrm{P}(\mathrm{x}<7) \\ \mathrm{P}(7<\mathrm{x}<9)=1-{\mathrm{e}}^{-0.125\times 9}-\left(1-{\mathrm{e}}^{-0.125\times 7}\right) \\ \mathrm{P}(7<\mathrm{x}<9)={\mathrm{e}}^{-0..875}-{\mathrm{e}}^{-1.125} \\ \mathrm{P}(7<\mathrm{x}<9)=0.4169-0.3247 \\ \mathrm{P}(7<\mathrm{x}<9)=0.0922 \end{gathered}$$

As a result, the likelihood of a phone call lasting between seven and nine minutes is high. $\mathrm{P}(7<\mathrm{x}<9)=0.0922$

The average time of a long distance phone call is eight minutes, according to an exponential distribution of call lengths measured in minutes.

The average length of a phone call is 8 minutes, thus if 25 calls are made, the total call time will be.

$25\times 8$

$=200minutes$