Question

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?

Short answer

a. $X=Lengthofphonecalls$

b. $X$is continuous

c. $X~0.125$

d. $\mu =8$

e.$\sigma =8$

f. Graph the condition : $f\left(x\right)=0.125e^{(-0.125x)}$.

g. The probability that a phone call lasts less than nine minutes is $0.675$.

h. The probability that a phone call lasts more than nine minutes is $0.325$

i. The probability that a phone call lasts between seven and nine minutes is $0.092$.

j. If $25$phone calls are made one after another, on average, the total will be$200$minutes.

Step-by-step solution

Defining Exponential distribution

In probability theory and statistics, theExponential distribution is the probability distribution of the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate and is given by :

$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}\lambda e^{-\lambda x},x\ge 0\\ 0,\mathrm{elsewhere}\end{array}\right.$

Let $X$has Exponential distribution with parameter $\lambda$

So that$E\left(X\right)=\frac{1}{\lambda }\mathrm{and}Var\left(X\right)=\frac{1}{{\lambda }^2}$

Defining the random variable  X

(a) According to the given question, we can define the random variable $X$ as the "Length of long distance phone calls" .

Determine if X continuous or discrete

(b) Since $X$is known to have an exponential distribution, which is a continuous distribution, we conclude that$X$is a continuous variable.

Finding X ~

(c) To find the parameter of the said distribution, follow these steps :

$$\begin{gathered} 8=E\left(X\right) \\ =\frac{1}{\lambda } \\ \Rightarrow \lambda =\frac{1}{8} \\ =0.125 \end{gathered}$$

So,$X~Exp(0.125)$

Finding μ

(d) According to the question, we have the average time as$8$minutes.

Finding σ

(e) We can find the Standard deviation by :

$$\begin{gathered} Var\left(X\right)=\frac{1}{{\lambda }^2}\mathrm{and}\mathrm{\sigma }=\sqrt{\mathrm{Var}\left(\mathrm{X}\right)} \\ \Rightarrow \mathrm{\sigma }=\sqrt{\frac{1}{{\mathrm{\lambda }}^2}} \\ =\frac{1}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{\sigma }=8 \end{gathered}$$

Graphing the probability distribution.

(f) the probability Density function is given by :

$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}0.125e^{(-0.125x)},x\ge 0\\ 0,\mathrm{elsewhere}\end{array}\right.$

Graphing this function :

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

(g) The required probability is :

$$\begin{gathered} P(X<9)={\int }_{-\infty }^{9}f\left(X\right).dx \\ ={\int }_{-\infty }^{0}0.dx+{\int }_0^{9}0.125e^{-0.125x}.dx \\ \boxed{t=0.125x,dt=0.125dx} \\ =0+{\int }_0^{1.125}{e}^{-t}.dt \\ =-e_0^{1.125} \\ =0.675 \end{gathered}$$

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

(h) The required probability is :

$$\begin{gathered} P(X>9)={\int }_9^{\infty }f\left(X\right).dx \\ ={\int }_9^{\infty }0.125e^{-0.125x}.dx \\ \boxed{t=0.125x,dt=0.125dx} \\ =0+{\int }_{1.125}^{\infty }{e}^{-t}.dt \\ =-e_{1.125}^{\infty } \\ =0.325 \end{gathered}$$

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

(i) The required probability is :

$$\begin{gathered} P(7<X<9)={\int }_7^{9}f\left(X\right).dx \\ ={\int }_7^{9}0.125e^{-0.125x}.dx \\ \boxed{t=0.125x,dt=0.125dx} \\ =0+{\int }_{0.875}^{1.125}{e}^{-t}.dt \\ =-e_{0.875}^{1.125} \\ =0.092 \end{gathered}$$

Finding if 25 phone calls are made one after another, on average, what would the total be

(j) Since one phone-call lasts average of$8\mathrm{minutes}$, so, $25$phone-calls would last an average of$25\times 8=200$minutes