Chapter 7.3, Problem 20E

(a)

To determine

To calculate the population, mean and population standard deviation.

Answer to Problem 20E

The population mean is $27.2$.

The value of population standard deviation is $3.97$.

Explanation of Solution

Given information:

Year: Winner	Age
$\begin{array}{l}2013:\text{ Fun}\\ 2012:\text{ Bon Iver}\\ 2011:\text{ Esperanza Spalding}\\ 2010:\text{ Zac Brown Band}\\ 2009:\text{ Adele }\end{array}$	$\begin{array}{l}29\\ 30\\ 26\\ 31\\ 20\end{array}$

Formula used:

Obtain the population mean by using the formula is given below:

$\mu =\frac{{\displaystyle \sum x}}{N}$

The formula for population standard deviation is given below.

$\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{N}}$

Calculation:

Obtain the population mean by using the formula is given below:

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x}}{N}\\ =\frac{29+30+26+31+20}{5}\\ =\frac{136}{5}\\ =27.2\end{array}$

Thus, the population mean is $27.2$.

Compute tile population standard deviation

Tile formula for population standard deviation is given below:

$\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{N}}$

$X$	$\left(x-\mu \right)$	${\left(x-\mu \right)}^2$
$29$	$1.8$	$3.24$
$30$	$2.8$	$7.84$
$26$	$-1.2$	$1.44$
$31$	$3.8$	$14.44$
$20$	$-7.2$	$51.84$
		${\displaystyle \sum {\left(x-\mu \right)}^2=78.8}$

Substitute $78.8$ for ${\displaystyle \sum {\left(x-\mu \right)}^2}\text{ and 5 for }N$ in the above standard deviation formula.

$\begin{array}{c}\sigma =\sqrt{\frac{78.8}{5}}\\ =\sqrt{15.76}\\ =3.97\end{array}$

Thus, the value of population standard deviation is $3.97$.

(b)

To determine

To calculate the sample space.

Answer to Problem 20E

The simple random sample of size $2$ with replacement is listed below.

$\left\{\begin{array}{l}\left(29,29\right),\left(29,30\right),\left(29,26\right),\left(29,31\right),\left(29,20\right)\\ \left(30,29\right),\left(30,30\right),\left(30,26\right),\left(30,31\right),\left(30,20\right)\\ \left(26,29\right),\left(26,30\right),\left(26,26\right),\left(26,31\right),\left(26,20\right)\\ \left(31,29\right),\left(31,30\right),\left(31,26\right),\left(31,31\right),\left(31,20\right)\\ \left(20,29\right),\left(20,30\right),\left(20,26\right),\left(20,31\right),\left(20,20\right)\end{array}\right\}$

Explanation of Solution

Formula used:

$\text{Age}:29,30,26,31,20$

Calculation:

The simple random sample of size $2$ with replacement is listed below.

$\left\{\begin{array}{l}\left(29,29\right),\left(29,30\right),\left(29,26\right),\left(29,31\right),\left(29,20\right)\\ \left(30,29\right),\left(30,30\right),\left(30,26\right),\left(30,31\right),\left(30,20\right)\\ \left(26,29\right),\left(26,30\right),\left(26,26\right),\left(26,31\right),\left(26,20\right)\\ \left(31,29\right),\left(31,30\right),\left(31,26\right),\left(31,31\right),\left(31,20\right)\\ \left(20,29\right),\left(20,30\right),\left(20,26\right),\left(20,31\right),\left(20,20\right)\end{array}\right\}$

(c)

To determine

To calculate the mean and standard deviation.

Answer to Problem 20E

${\sigma }_x=2.81$

Explanation of Solution

Formula used:

$\overline{x}=\frac{sample1+sample2}{2}$

Calculation:

Compute the sample mean $\overline{x}$ for each of the $25$ samples of size $2$.

Sample $1$	Sample $2$	$\overline{x}$	Sample $1$	Sample $2$	$\overline{x}$
$29$	$29$	$29$	$26$	$31$	$28.5$
$29$	$30$	$29.5$	$26$	$20$	$23$
$29$	$26$	$27.5$	$31$	$29$	$30$
$29$	$31$	$30$	$31$	$30$	$30.5$
$29$	$20$	$24.5$	$31$	$26$	$28.5$
$30$	$29$	$29.5$	$31$	$31$	$31$
$30$	$30$	$30$	$31$	$20$	$25.5$
$30$	$26$	$28$	$20$	$29$	$24.5$
$30$	$31$	$30.5$	$20$	$30$	$25$
$30$	$20$	$25$	$20$	$26$	$23$
$26$	$29$	$27.5$	$20$	$31$	$25.5$
$26$	$30$	$28$	$20$	$20$	$20$
$26$	$26$	$26$

Compute the mean ${\mu }_{\overline{x}}$

Sample $1$	Sample $2$	$\overline{x}$	$\left(x-\mu \right)$	${\left(x-\mu \right)}^2$
$29$	$29$	$29$	$1.8$	$3.24$
$29$	$30$	$29.5$	$2.3$	$5.29$
$29$	$26$	$27.5$	$0.3$	$0.09$
$29$	$31$	$30$	$2.8$	$7.84$
$29$	$20$	$24.5$	$-2.7$	$7.29$
$30$	$29$	$29.5$	$2.3$	$5.29$
$30$	$30$	$30$	$2.8$	$7.84$
$30$	$26$	$28$	$0.8$	$0.64$
$30$	$20$	$25$	$-2.2$	$4.84$
$26$	$29$	$27.5$	$0.3$	$0.09$
$26$	$30$	$28$	$0.8$	$0.64$
$26$	$26$	$26$	$-1.2$	$1.44$
$26$	$31$	$28.5$	$1.3$	$1.69$
$26$	$20$	$23$	$-4.2$	$17.64$
$31$	$29$	$30$	$2.8$	$7.84$
$31$	$30$	$30.5$	$3.3$	$10.89$
$31$	$26$	$28.5$	$1.3$	$1.69$
$31$	$31$	$31$	$3.8$	$14.44$
$31$	$20$	$25.5$	$-1.7$	$2.89$
$20$	$29$	$24.5$	$-2.7$	$7.29$
$20$	$30$	$25$	$-2.2$	$4.84$
$20$	$26$	$23$	$-4.2$	$17.64$
$20$	$31$	$25.5$	$-1.7$	$2.89$
$20$	$20$	$20$	$-7.2$	$51.84$
		Mean $=27.2$		${\displaystyle \sum {\left(x-\mu \right)}^2}=197$

Compute the standard deviation ${\sigma }_{\overline{x}}$ of the sample means.

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{197}{25}}\\ =\sqrt{7.88}\\ =2.81\end{array}$

(d)

To determine

To calculate the sample, mean for all 25 samples for size 2 and also calculate the ${\mu }_{\overline{x}}$ and ${\sigma }_{\overline{x}}$.

Answer to Problem 20E

$\begin{array}{l}{\mu }_{\overline{x}}=\mu \text{ and }{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{2}}\\ {\mu }_{\overline{x}}=27.2\end{array}$

Explanation of Solution

Formula used:

The formula for population standard deviation is given below.

$\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{N}}$

Calculation:

Verify that ${\mu }_{\overline{x}}=\mu \text{ and }{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{2}}$

${\mu }_{\overline{x}}=\mu$, which is independent of sample size.

Hence ${\mu }_{\overline{x}}=27.2$.

Find the ${\sigma }_{\overline{x}}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{3.97}{\sqrt{2}}\\ =\frac{3.97}{1.414}\\ =2.81\end{array}$

Thus, the standard deviation of $\overline{x}\text{ is }2.81$. Therefore, ${\sigma }_{\overline{x}}\left(=2.81\right)=\frac{\sigma }{\sqrt{2}}\left(=2.81\right)$.

Hence, ${\mu }_{\overline{x}}=\mu \text{ and }{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{2}}$