Chapter 4.3, Problem 33E

The pH of water samples from a specific lake is a random variable Y with probability density function given by

$f\text{(}y\text{)}\text{=}\{\begin{array}{l}(3\text{/}8){(7-y)}^2,5\le y\le 7,\hfill \\ 0,\text{elsewhere}.\hfill \end{array}$

a Find E(Y) and V(Y).

b Find an interval shorter than (5, 7) in which at least three-fourths of the pH measurements must lie.

c Would you expect to see a pH measurement below 5.5 very often? Why?

To determine

Compute the value of $E\left(Y\right)$.

Compute the value of $V\left(Y\right)$.

Answer to Problem 33E

The value of $E\left(Y\right)$ is 5.5 pH.

The value of $V\left(Y\right)$ is 0.15.

Explanation of Solution

The probability density function is given below:

$f\left(y\right)=\{\begin{array}{l}\left(\frac{3}{8}\right){\left(7-y\right)}^2,5\le y\le 7,\\ 0,\text{elsewhere},\end{array}$

The value of $E\left(Y\right)$ is as follows:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}yf\left(y\right)dy}\\ ={\displaystyle \underset{5}{\overset{7}{\int }}y\left(\frac{3}{8}\right){\left(7-y\right)}^{2}dy}\\ ={\displaystyle \underset{5}{\overset{7}{\int }}\left(\frac{3}{8}\right)\left(y^3-14y^2+49y\right)dy}\\ =\left(\frac{3}{8}\right){\left[\frac{1}{4}{y}^4-\left(\frac{14}{3}\right)y^3+\left(\frac{49}{2}\right)y^2\right]}_5^7\end{array}$

$\begin{array}{c}=\left[\frac{y^2\left(3y^2-56y+294\right)}{32}\right]\\ =\frac{11}{2}\\ =5.5\end{array}$

Thus, the value of $E\left(Y\right)$ is 5.5 pH.

The value of $E\left(Y^2\right)$ is,

$\begin{array}{c}E\left(Y^2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^{2}f\left(y\right)dy}\\ ={\displaystyle \underset{5}{\overset{7}{\int }}{y}^2\left(\frac{3}{8}\right){\left(7-y\right)}^{2}dy}\\ ={\displaystyle \underset{5}{\overset{7}{\int }}\left(\frac{3}{8}\right)\left(y^4-14y^5+49y\right)dy}\\ =\left(\frac{3}{8}\right){\left[\frac{1}{4}{y}^4-\left(\frac{14}{3}\right)y^3+\left(\frac{49}{2}\right)y^2\right]}_5^7\end{array}$

$\begin{array}{c}=\frac{3y^5}{40}-\frac{21y^4}{16}+\frac{49y^3}{8}\\ =\frac{y^3\left(6y^2-105y+490\right)}{80}\\ =\frac{152}{5}\\ =30.4\end{array}$

The variance of Y is,

$\begin{array}{c}V\left(Y\right)=E\left(Y^2\right)-{\left[E\left(Y\right)\right]}^2\\ =30.4-{\left(5.5\right)}^2\\ =30.4-\left(30.25\right)\\ =0.15\end{array}$

Thus, the variance of Y is 0.15.

To determine

Compute the interval shorter than (5,7), within which, at least three-fourths of the pH measurements must lie.

Answer to Problem 33E

The interval shorter than (5, 7) at least three-fourths of the pH measurements must lie.

is $5.068\le Y\le 5.932$.

Explanation of Solution

From the given information, let X be the random variable be the spread on either side of $\mu$ From part (a) the results of $E\left(Y\right)$ is 5.5 and $V\left(Y\right)$ is 0.15.

$\begin{array}{c}P\left(\mu -x\le Y\le \mu +x\right)=0.75\\ {\displaystyle {\int }_{\mu -x}^{\mu +x}f\left(t\right)}dy=0.75\\ {\displaystyle {\int }_{\mu -x}^{\mu +x}\left(\frac{3}{8}\right){\left(7-y\right)}^2}dy=0.75\\ \frac{3}{8}{\displaystyle {\int }_{\mu -x}^{\mu +x}\left(49+y^2-14y\right)}dy=0.75\\ \left(\frac{3}{8}\right){\left[\left(\frac{y^3}{3}-14\frac{y^2}{2}+49y\right)\right]}_{\mu -x}^{\mu +x}=0.75\\ \frac{{\left(\mu +x\right)}^3}{3}-7{\left(\mu +x\right)}^2+49\left(\mu +x\right)-\left(\frac{{\left(\mu -x\right)}^3}{3}-7{\left(\mu -x\right)}^2+49\left(\mu -x\right)\right)=2\\ 2x^3+\left(6{\mu }^2-84\mu +294\right)x=6\end{array}$

Suppose $x=0.432$.

Then,

$\begin{array}{c}\mu -0.432\le Y\le \mu +0.432\\ 5.5-0.432\le Y\le 5.5+0.432\\ 5.068\le Y\le 5.932\end{array}$

Thus, the interval shorter than (5, 7) at least three-fourths of the pH measurements must lie.

is $5.068\le Y\le 5.932$.

To determine

Explain whether one can expect to see a pH measurement below 5.5.

Answer to Problem 33E

One may expect to see a pH measurement below 5.5 with probability 0.5781.

Explanation of Solution

From the given information, let X be the random variable that is the spread on either side of $\mu$. From part (a) the results are, that the value of $E\left(Y\right)$ is 5.5 pH and $V\left(Y\right)$ is 0.15.

The expect to see a pH measurement below 5.5 very often is obtained below:

$\begin{array}{c}P\left(Y<5.5\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}\\ ={\displaystyle \underset{5}{\overset{5.5}{\int }}\left(\frac{3}{8}\right){\left(7-y\right)}^{2}dy}\\ ={\displaystyle \underset{5}{\overset{5.5}{\int }}\left(\frac{3{\left(7-y\right)}^2}{8}\right)dy}\\ ={\left(\frac{3{\left(7-y\right)}^2}{8}\right)}_5^{5.5}\end{array}$

         $\begin{array}{c}=\frac{y\left(y^2-21y+147\right)}{8}\\ =\frac{37}{64}\\ =0.5781\end{array}$

Thus, one may expect to see a pH measurement below 5.5 with probability 0.5781.