Chapter 2.4, Problem 18E

The lifetime, in years, of a certain type of pump is a random variable with probability density function

$f\left(x\right)=\{\begin{array}{l}\frac{64}{{\left(x+2\right)}^5}x>0\\ 0x\le 0\end{array}$

1. a. What is the probability that a pump lasts more than two years?
2. b. What is the probability that a pump lasts between two and four years?
3. c. Find the mean lifetime.
4. d. Find the variance of the lifetimes.
5. e. Find the cumulative distribution function of the lifetime.
6. f. Find the median lifetime.
7. g. Find the 60th percentile of the lifetimes.

To determine

Find the probability that a pump lasts more than two years.

Answer to Problem 18E

The probability that a pump lasts more than two years is $\frac{1}{16}$.

Explanation of Solution

Given info:

The lifetime, in years, of a certain type of pump is a random variable with probability density function

$f\left(x\right)=\{\begin{array}{l}\frac{64}{{\left(x+2\right)}^5}x>0\\ \text{0 }x\le 0\end{array}$

Calculation:

Let X be a continuous random variable with probability density function $f\left(x\right)$. Let b be a number and $X>a$ then the probability of $X>a$ can be obtained as,

$P\left(X\ge a\right)=P\left(X>a\right)={\displaystyle {\int }_a^{\infty }f\left(x\right)dx}$

The probability that a pump lasts more than two years can be obtained as,

$\begin{array}{c}P\left(X>2\right)={\displaystyle {\int }_2^{\infty }\frac{64}{{\left(x+2\right)}^5}}dx\\ =-64{\left[\frac{1}{4{\left(x+2\right)}^4}\right]}_2^{\infty }\\ =-16\left[\frac{1}{{\left(\infty +2\right)}^4}-\frac{1}{{\left(2+2\right)}^4}\right]\\ =0+\frac{16}{256}\end{array}$

$=\frac{1}{16}$

Thus, the probability that a pump lasts more than two years is $\frac{1}{16}$.

To determine

Find the probability that a pump lasts between two and four years.

Answer to Problem 18E

The probability that a pump lasts between two and four years is $\frac{65}{1,296}$.

Explanation of Solution

Calculation:

Let X be a continuous random variable with probability density function $f\left(x\right)$. Let a and b be a numbers and $a<X<b$ then the probability of $a<X<b$ can be obtained as,

$P\left(a\le X\le b\right)=P\left(a\le X<b\right)=P\left(a<X\le b\right)=P\left(a<X<b\right)={\displaystyle {\int }_a^{b}f\left(x\right)dx}$

Let X be the lifetime, in years, of a certain type of pump.

The probability that a pump lasts between two and four years can be obtained as,

$\begin{array}{c}P\left(2<X<4\right)={\displaystyle {\int }_2^4\frac{64}{{\left(x+2\right)}^5}}dx\\ =-64{\left[\frac{1}{4{\left(x+2\right)}^4}\right]}_2^4\\ =-16\left[\frac{1}{{\left(4+2\right)}^4}-\frac{1}{{\left(2+2\right)}^4}\right]\\ =-\left(\frac{1}{1,296}-\frac{16}{256}\right)\end{array}$

                       $\begin{array}{c}=-\left(\frac{1}{81}-\frac{1}{16}\right)\\ =\frac{-16+81}{1,296}\\ =\frac{65}{1,296}\end{array}$

Thus, the probability that a pump lasts between two and four years is $\frac{65}{1,296}$.

To determine

Find the mean life time.

Answer to Problem 18E

The mean lifetime is $\frac{2}{3}$.

Explanation of Solution

Calculation:

Mean:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the mean of X is,

${\mu }_X={\displaystyle {\int }_{-\infty }^{\infty }xf\left(x\right)dx}$

The mean lifetime can be obtained as,

${\mu }_X={\displaystyle {\int }_0^{\infty }x\frac{64}{{\left(x+2\right)}^5}}dx$

$\begin{array}{c}\text{Let }x+2=u\\ x=u-2\\ dx=du\end{array}$

Upper limit: if $x=\infty \Rightarrow u-2=\infty \Rightarrow u=\infty$

Lower limit: if $x=0\Rightarrow u-2=0\Rightarrow u=2$

$\begin{array}{c}{\mu }_X={\displaystyle {\int }_2^{\infty }\left(u-2\right)\frac{64}{u^5}}du\\ =64{\displaystyle {\int }_2^{\infty }\left(u-2\right)u^{-5}}du\\ =64{\displaystyle {\int }_2^{\infty }\left(u^{-4}-2u^{-5}\right)}du\end{array}$

      $\begin{array}{c}=64{\left(-\frac{u^{-3}}{3}+\frac{u^{-4}}{2}\right)}_2^{\infty }\\ =64\left(\left(-\frac{{\infty }^{-3}}{3}+\frac{{\infty }^{-4}}{2}\right)-\left(-\frac{2^{-3}}{3}+\frac{{\left(2\right)}^{-4}}{2}\right)\right)\\ =64\left(-0+0+\frac{1}{8}\left(\frac{1}{3}-\frac{1}{4}\right)\right)\\ =\frac{2}{3}\end{array}$

Thus, the mean lifetime is $\frac{2}{3}$.

To determine

Find the variance of the lifetimes.

Answer to Problem 18E

The variance of the lifetimes is $\frac{8}{9}$.

Explanation of Solution

Calculation:

Variance:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the variance of X is,

${\sigma }_{{}_X}^2={\displaystyle {\int }_{-\infty }^{\infty }{x}^{2}f\left(x\right)dx-{\mu }_X^2}$

The variance of the lifetimes can be obtained as,

${\sigma }_{{}_X}^2={\displaystyle {\int }_0^{\infty }{x}^2\frac{64}{{\left(x+2\right)}^5}}dx-{\left(\frac{2}{3}\right)}^2$

$\begin{array}{c}\text{Let }x+2=u\\ x=u-2\\ dx=du\end{array}$

Upper limit: if $x=\infty \Rightarrow u-2=\infty \Rightarrow u=\infty$

Lower limit: if $x=0\Rightarrow u-2=0\Rightarrow u=2$

$\begin{array}{c}{\sigma }_X^2={\displaystyle {\int }_2^{\infty }{\left(u-2\right)}^2\frac{64}{u^5}}du-{\left(\frac{2}{3}\right)}^2\\ =64{\displaystyle {\int }_2^{\infty }\left(u^2-4u+4\right)u^{-5}}du-\frac{4}{9}\\ =64{\displaystyle {\int }_2^{\infty }\left(u^{-3}-4u^{-4}+4u^{-5}\right)}du-\frac{4}{9}\\ =64{\left(-\frac{u^{-2}}{2}+\frac{4u^{-3}}{3}-u^{-4}\right)}_2^{\infty }-\frac{4}{9}\end{array}$

$\begin{array}{c}=64\left(\left(-\frac{{\infty }^{-2}}{2}+\frac{4{\infty }^{-3}}{3}-{\infty }^{-4}\right)-\left(-\frac{2^{-2}}{2}+\frac{4{\left(2\right)}^{-3}}{3}-2^{-4}\right)\right)-\frac{4}{9}\\ =64\left(-0+0-0+\frac{1}{4}\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)\right)-\frac{4}{9}\\ =64\left(\frac{6-8+3}{48}\right)-\frac{4}{9}\\ =\frac{4}{3}-\frac{4}{9}\end{array}$

        $\begin{array}{c}=\frac{12-4}{9}\\ =\frac{8}{9}\end{array}$

Thus, the variance of the lifetimes is $\frac{8}{9}$.

To determine

Find the cumulative distribution function of the lifetimes.

Answer to Problem 18E

The cumulative distribution function of the lifetimes is,

$F\left(x\right)=\{\begin{array}{l}0,\text{ if }x<\text{0}\\ 1-\frac{16}{{\left(x+2\right)}^4}\text{, if }x\ge 0\end{array}$

Explanation of Solution

Calculation:

Cumulative distribution function:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the cumulative distribution function of X is,

$F\left(x\right)=P\left(X\le x\right)={\displaystyle {\int }_{-\infty }^{x}f\left(t\right)dt}$

For $x<0$:

$\begin{array}{c}F\left(x\right)={\displaystyle {\int }_{-\infty }^{x}0}dt\\ =0\end{array}$

For $x\ge 0$:

$\begin{array}{c}F\left(x\right)={\displaystyle {\int }_0^x\frac{64}{{\left(t+2\right)}^5}}dt\\ =64{\left(-\frac{1}{4{\left(t+2\right)}^4}\right)}_0^x\\ =-\left(\frac{16}{{\left(x+2\right)}^4}-\frac{16}{{\left(0+2\right)}^4}\right)\\ =1-\frac{16}{{\left(x+2\right)}^4}\end{array}$

Thus, the cumulative distribution function of the resistance is,

$F\left(x\right)=\{\begin{array}{l}0,\text{ if }x<\text{0}\\ 1-\frac{16}{{\left(x+2\right)}^4}\text{, if }x\ge 0\end{array}$

To determine

Find the median of the lifetimes.

Answer to Problem 18E

The median of the lifetimes is 0.3784.

Explanation of Solution

Calculation:

Median:

The median is the half of the observations. Therefore, $F\left(x_m\right)=0.5$

The median of the lifetimes is,

$\begin{array}{c}1-\frac{16}{{\left(x_m+2\right)}^4}=0.5\\ \frac{16}{{\left(x_m+2\right)}^4}=0.5\\ {\left(x_m+2\right)}^4=32\\ x_m+2=2.3784\\ x_m=0.3784\end{array}$

Thus, the median of the lifetimes is 0.3784.

To determine

Find the 60^th percentile of the lifetimes.

Answer to Problem 18E

The 60^th percentile of the lifetimes is 0.2745.

Explanation of Solution

Calculation:

Percentile:

The n^th percentile can be written as $F\left(x_n\right)=\frac{n}{100}$

The 60^th percentile of the lifetimes is,

$\begin{array}{c}1-\frac{16}{{\left(x_{60}+2\right)}^4}=0.6\\ \frac{16}{{\left(x_{60}+2\right)}^4}=0.6\\ {\left(x_{60}+2\right)}^4=26.6667\\ x_{60}+2=2.2724\\ x_{60}=0.2724\end{array}$

Thus, the 60^th percentile of the lifetimes is 0.2745.