Chapter 3, Problem 10TP

1.

To determine

The inclusive range, the sample standard deviation and the sample variance for the given set of scores.

Answer to Problem 10TP

The value of inclusive range is 7, the sample standard deviation is 2.58 and the sample variance is 6.66.

Explanation of Solution

Given info:

The set of scores is 5, 7, 9, and 11.

Calculation:

The formula to calculate inclusive range is,

$r=h-l+1$

Where

• r is the range.
• h is the highest score in data set.
• l is the lowest score in data set.

For the given set, the highest score is 11 and the lowest score is 5.

Substitute $11$ for h and $5$ for l to evaluate the value of inclusive range.

$\begin{array}{c}r=11-5+1\\ =7\end{array}$

Thus, the value of inclusive range is 7.

The formula to calculate standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}{n-1}}$

Where,

•  s is the standard deviation.
•  X is each individual score.
•   $\overline{X}$ is the mean of all score.
•  n is the sample size.

The values of X, $\overline{X}$, $X-\overline{X}$ and ${\left(X-\overline{X}\right)}^2$ is shown in the table below.

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
5	8	$-3$	9
7	8	$-1$	1
9	8	1	1
11	8	3	9

The sum of square is,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=9+1+1+9\\ =20\end{array}$

Substitute 20 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 4 for n to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{20}{4-1}}\\ =\sqrt{\frac{20}{3}}\\ =\sqrt{6.67}\\ =2.58\end{array}$

Thus, the standard deviation is 2.58.

Variance is the square of standard deviation. So, the variance is,

$\begin{array}{c}{s}^2={\sqrt{6.67}}^2\\ =6.67\end{array}$

Thus, the variance is 6.67.

Therefore, for the given set of scores, the value of inclusive range is 6, the sample standard deviation is 2.58 and the sample variance is 6.67.

2.

To determine

The inclusive range, the sample standard deviation and the sample variance for the given set of scores.

Answer to Problem 10TP

The value of inclusive range is 1.6, the sample standard deviation is 0.2497 and the sample variance is 0.0624.

Explanation of Solution

Given information:

The set of scores is 0.3, 0.5, 0.6, 0.9.

Calculation:

The formula to calculate inclusive range is,

$r=h-l+1$

Where

• r is the range.
• h is the highest score in data set.
• l is the lowest score in data set.

For the given set, the highest score is 0.9 and the lowest score is 0.3.

Substitute $0.9$ for h and $0.3$ for l in equation (1) to evaluate the value of inclusive range.

$\begin{array}{c}r=0.9-0.3+1\\ =1.6\end{array}$

Thus, the value of inclusive range is 1.6.

The formula to calculate standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}{n-1}}$

Where,

• s is the standard deviation.
• X is each individual score.
• $\overline{X}$ is the mean of all score.
• n is the sample size.

The values of X, $\overline{X}$, $X-\overline{X}$ and ${\left(X-\overline{X}\right)}^2$ is shown in the table below.

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
0.3	0.575	$-0.275$	0.0756
0.5	0.575	$-0.075$	0.0056
0.6	0.575	0.025	0.0006
0.9	0.575	0.325	0.1056

The sum of square is,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=0.0756+0.0056+0.0006+0.1056\\ =0.1874\end{array}$

Substitute 0.1874 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 4 for n in equation (2) to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{0.1874}{4-1}}\\ =\sqrt{\frac{0.1874}{3}}\\ =\sqrt{0.0624}\\ =0.2497\end{array}$

Thus, the standard deviation is 0.25.

Variance is the square of standard deviation. So, the variance is,

$\begin{array}{c}{s}^2={\left(0.2497\right)}^2\\ =0.0624\end{array}$

Thus, the variance is 0.06.

Therefore, for the given set of scores, the value of inclusive range is 1.6, the sample standard deviation is 0.2497 and the sample variance is 0.0624.

3.

To determine

The inclusive range, the sample standard deviation and the sample variance for the given set of scores.

Answer to Problem 10TP

The value of inclusive range is 4.5, the sample standard deviation is 1.58 and the sample variance is 2.48.

Explanation of Solution

Given info:

The set of scores is 6.1, 7.3, 4.5, 3.8.

Calculation:

The formula to calculate inclusive range is,

$r=h-l+1$

Where

• r is the range.
• h is the highest score in data set.
• l is the lowest score in data set.

For the given set, the highest score is 7.3 and the lowest score is 3.8.

Substitute $7.3$ for h and $3.8$ for l to evaluate the value of inclusive range.

$\begin{array}{c}r=7.3-3.8+1\\ =4.5\end{array}$

Thus, the value of inclusive range is 4.5.

The formula to calculate standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}{n-1}}$

Where,

•  s is the standard deviation.
•  X is each individual score.
•   $\overline{X}$ is the mean of all score.
•  n is the sample size.

The values of X, $\overline{X}$, $X-\overline{X}$ and ${\left(X-\overline{X}\right)}^2$ is shown in the table below.

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
3.8	5.425	$-1.625$	2.64
4.5	5.425	$-0.925$	0.85
6.1	5.425	0.675	0.45
7.3	5.425	1.875	3.51

The sum of square is,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=2.64+0.85+0.45+3.51\\ =7.45\end{array}$

Substitute 7.45 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 4 for n to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{7.45}{4-1}}\\ =\sqrt{\frac{7.45}{3}}\\ =\sqrt{2.48}\\ =1.58\end{array}$

Thus, the standard deviation is 1.58.

Variance is the square of standard deviation. So, the variance is,

$\begin{array}{c}{s}^2={\left(1.58\right)}^2\\ =2.48\end{array}$

Thus, the variance is 2.48.

4.

To determine

The inclusive range, the sample standard deviation and the sample variance for the given set of scores.

Answer to Problem 10TP

The value of inclusive range is 124, the sample standard deviation is 48.23 and the sample variance is 2,326.5.

Explanation of Solution

Given information:

The set of scores is 435, 456, 423, 546, 465.

Calculation:

The formula to calculate inclusive range is,

$r=h-l+1$

Where

• r is the range.
• h is the highest score in data set.
• l is the lowest score in data set.

For the given set, the highest score is 546 and the lowest score is 423.

Substitute $546$ for h and $423$ for l to evaluate the value of inclusive range.

$\begin{array}{c}r=546-423+1\\ =124\end{array}$

Thus, the value of inclusive range is 124.

The formula to calculate standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X-\overline{X}\right)}^2}}{n-1}}$

Where,

•  s is the standard deviation.
•  X is each individual score.
•   $\overline{X}$ is the mean of all score.
•  n is the sample size.

The values of X, $\overline{X}$, $X-\overline{X}$ and ${\left(X-\overline{X}\right)}^2$ is shown in the table below.

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
423	465	$-42$	1,764
435	465	$-30$	900
456	465	$-9$	81
465	465	0	0
546	465	81	6,561

The sum of square is,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=1,764+900+81+0+6,561\\ =9,306\end{array}$

Substitute 9,306 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 5 for n to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{9,306}{5-1}}\\ =\sqrt{\frac{9,306}{4}}\\ =\sqrt{2326.5}\\ =48.23\end{array}$

Thus, the standard deviation is 48.23.

Variance is the square of standard deviation. So, the variance is,

$\begin{array}{c}{s}^2={\left(48.23\right)}^2\\ =2,326.5\end{array}$

Thus, the variance is 2,326.5.