Chapter 7.1, Problem 23E

(a)

To determine

To find:

The standard deviation of the scores of golfer A.

Answer to Problem 23E

  $4.63$

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

The mean is

  $\begin{array}{l}\overline{x}=\frac{900}{10}\\ \overline{x}=90\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{49+36+\mathrm{...........}+25}{10}\\ =21.4\end{array}$

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$83$	$90$	$-7$	$49$
$84$	$90$	$-6$	$36$
$91$	$90$	$1$	$1$
$90$	$90$	$0$	$0$
$98$	$90$	$8$	$64$
$88$	$90$	$-2$	$4$
$95$	$90$	$5$	$25$
$89$	$90$	$-1$	$1$
$87$	$90$	$-3$	$9$
$95$	$90$	$5$	$25$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{49+36+\mathrm{...........}+25}{10}}\\ \sigma =\sqrt{\frac{214}{10}}\\ \sigma \approx 4.63\end{array}$

The score of golfer A differs from the mean by about $4.63$ points

(b)

To determine

To find:

The standard deviation of the scores of golfer B.

Answer to Problem 23E

  $2.53$

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

The mean is

  $\begin{array}{l}\overline{x}=\frac{910}{10}\\ \overline{x}=91\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{4+4+\mathrm{...........}+1}{10}\\ =6.4\end{array}$

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$89$	$91$	$-2$	$4$
$93$	$91$	$2$	$4$
$92$	$91$	$1$	$1$
$88$	$91$	$-3$	$9$
$89$	$91$	$-2$	$4$
$87$	$91$	$-4$	$16$
$95$	$91$	$4$	$16$
$94$	$91$	$3$	$9$
$91$	$91$	$0$	$0$
$92$	$91$	$1$	$1$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{4+4+\mathrm{...........}+1}{10}}\\ \sigma =\sqrt{\frac{64}{10}}\\ \sigma \approx 2.53\end{array}$

The score of golfer B differs from the mean by about $2.53$ points

(c)

To determine

To explain:

The standard deviation for golfer A and golferB.

Answer to Problem 23E

The scores of golfer A are more spread out than the scores of golfer B

Explanation of Solution

Given:

The table

  $x,\overline{x},x-\overline{x},{\left(x-\overline{x}\right)}^2$

Concept used:

The mean of the squared deviation or variance

  $=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}$

The square root of the variance

  $\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}$

Calculation:

The mean is

  $\begin{array}{l}\overline{x}=\frac{900}{10}\\ \overline{x}=90\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{49+36+\mathrm{...........}+25}{10}\\ =21.4\end{array}$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{49+36+\mathrm{...........}+25}{10}}\\ \sigma =\sqrt{\frac{214}{10}}\\ \sigma \approx 4.63\end{array}$

The mean is

  $\begin{array}{l}\overline{x}=\frac{910}{10}\\ \overline{x}=91\end{array}$

The mean of the squared deviation or variance

  $\begin{array}{l}=\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}\\ =\frac{4+4+\mathrm{...........}+1}{10}\\ =6.4\end{array}$

The square root of the variance

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{.............}+{\left(x_n-\overline{x}\right)}^2}{n}}\\ \sigma =\sqrt{\frac{4+4+\mathrm{...........}+1}{10}}\\ \sigma =\sqrt{\frac{64}{10}}\\ \sigma \approx 2.53\end{array}$

The scores of golfer A are more spread out than the scores of golfer B