Chapter 3, Problem 9TP

1.

To determine

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

Answer to Problem 9TP

The value of range is 6, the standard deviation is 2.58 and the sample variance is 6.67.

Explanation of Solution

Given info:

The set of scores is 3,5,7,9.

Calculation:

The formula to calculate 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 9 and the lowest score is 3.

Substitute $9$ for h and $3$ for l in equation (1) to evaluate the value of range.

$\begin{array}{c}r=9-3\\ =6\end{array}$

Thus, the value of range is 6.

The formula to calculate standard deviation is,

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

Where,

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

Mean is given by,

$\begin{array}{c}\overline{X}=\frac{3+5+7+9}{4}\\ =6\end{array}$

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

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
3	6	$-3$	9
5	6	$-1$	1
7	6	1	1
9	6	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 in equation (2) to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{20}{4-1}}\\ =\sqrt{\frac{20}{3}}\\ =\sqrt{6.66}\\ =2.58\end{array}$

Thus, the standard deviation is 2.58.

Variance is the square of standard deviation, so, the variance is,

$\begin{array}{c}{s}^2={\left(2.58\right)}^2\\ =6.67\end{array}$

Thus, the variance is 6.67.

2.

To determine

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

Answer to Problem 9TP

The value of range is 0.6, the standard deviation is 0.26 and the sample variance is 0.07.

Explanation of Solution

Given info:

The set of scores is $0.2,0.4,0.6,0.8$.

Calculation:

The formula to calculate 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 9 and the lowest score is 3.

Substitute $0.8$ for h and $0.2$ for l in equation (1) to evaluate the value of range.

$\begin{array}{c}r=0.8-0.2\\ =0.6\end{array}$

Thus, the value of range is 0.6.

Standard deviation:

The formula to calculate standard deviation is,

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

Where,

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

Mean is given by,

$\begin{array}{c}\overline{X}=\frac{0.2+0.4+0.6+0.8}{4}\\ =0.5\end{array}$

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

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
0.2	0.5	$-0.3$	0.09
0.4	0.5	$-0.1$	0.01
0.6	0.5	0.1	0.01
0.8	0.5	0.3	0.09

The sum of square is,

$\begin{array}{c}{{\displaystyle \sum \left(X-\overline{X}\right)}}^2=0.09+0.01+0.01+0.09\\ =0.2\end{array}$

Substitute 0.2 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 4 for n in equation (2) to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{0.2}{4-1}}\\ =\sqrt{\frac{0.2}{3}}\\ =\sqrt{0.0666}\\ =0.2582\end{array}$

Therefore, for the given set of scores the standard deviation is 0.26.

Variance:

Variance is the square of standard deviation, so, the variance is,

$\begin{array}{c}{s}^2={\left(0.2582\right)}^2\\ =0.067\end{array}$

Thus, the variance is 0.07.

3.

To determine

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

Answer to Problem 9TP

The value of range is 5.8, the standard deviation is 2.22 and the variance is 4.92.

Explanation of Solution

Given info:

The set of scores is $3.5,6.2,9.3,4.1,5.5,7.9$.

Calculation:

The formula to calculate 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 9.3 and the lowest score is 3.5.

Substitute $9.3$ for h and $3.5$ for l in equation (1) to evaluate the value of range.

$\begin{array}{c}r=9.3-3.5\\ =5.8\end{array}$

Thus, the value of range is 5.8.

Standard deviation:

The formula to calculate standard deviation is,

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

Where,

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

Mean is given by,

$\begin{array}{c}\overline{X}=\frac{3.5+6.2+9.3+4.1+5.5+7.9}{6}\\ =6.08\end{array}$

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

X	$\overline{X}$	$X-\overline{X}$	${\left(X-\overline{X}\right)}^2$
3.5	6.08	$-2.58$	6.65
6.2	6.08	$0.12$	0.0144
9.3	6.08	3.22	10.36
4.1	6.08	$-1.98$	3.92
5.5	6.08	$-0.58$	0.336
7.9	6.08	1.82	3.312

The sum of square is,

${{\displaystyle \sum \left(X-\overline{X}\right)}}^2=24.59$

Substitute 24.59 for ${{\displaystyle \sum \left(X-\overline{X}\right)}}^2$ and 6 for n in equation (2) to evaluate the value of standard deviation.

$\begin{array}{c}s=\sqrt{\frac{24.59}{6-1}}\\ =2.22\end{array}$

Thus, the standard deviation is 2.22.

Variance:

Variance is the square of standard deviation, so, the variance is,

$\begin{array}{c}{s}^2={\left(2.22\right)}^2\\ =4.92\end{array}$

Thus, the variance is 4.92.