Chapter 4, Problem 21P

For the following population of N = 6 scores: 2, 9, 6, 8, 9, 8

1. a. Calculate the range and the standard deviation. (Use either definition for the range.)
2. b. Add 2 points to each score and compute the range and standard deviation again. Describe how adding a constant to each score influences measures of variability.

To determine

To find: The range and standard deviation for the population of scores.

Answer to Problem 21P

The range and standard deviation for the population of scores is 7 or 8 and 2.45.

Explanation of Solution

Given info:

The set of scores are 2, 9, 6, 8, 9, and 8.

Calculation:

The formula for range set of scores is,

$\text{Range}=X_{\max }-X_{\min }$

If the scores are whole numbers, the formula for range set for scores is,

$\text{Range}=X_{\max }-X_{\min }+1$

The formula of standard deviation is,

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

From the set of scores, the maximum and minimum values are 9 and 2.

The range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }\\ =\text{9}-2\\ =7\end{array}$

If the scores are whole numbers, the range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }+1\\ =\text{9}-2+1\\ =8\end{array}$

The mean is,

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x_i}}{N}\\ =\frac{2+9+6+8+9+8}{6}\\ =\frac{42}{6}\\ =7\end{array}$

The standard deviation is,

$\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(x_i-\mu )}^2}}{N}}\\ =\sqrt{\frac{{(2-7)}^2+{(9-7\sout{)}}^2+{(6-7)}^2+{(8-7)}^2+{(9-7)}^2+{(8-7)}^2}{6}}\\ =\sqrt{6}\\ =2.45\end{array}$

Conclusion:

The obtained range and standard deviation for the scores is 7 or 8 and 2.45.

To determine

To find: The range and standard deviation for the population of scores.

Answer to Problem 21P

After adding 2 points for each score, the range and standard deviation for the population of scores is 7 or 8 and 2.45.

Explanation of Solution

Given info:

The set of scores are 2, 9, 6, 8, 9, and 8.

Calculation:

The formula for range set of scores is,

$\text{Range}=X_{\max }-X_{\min }$

If the scores are whole numbers, the formula for range set for scores is,

$\text{Range}=X_{\max }-X_{\min }+1$

The formula of standard deviation is,

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

First add 2 points for each score, the new data set of scores is, 4, 11, 8, 10, 11, and 10

The maximum and minimum values for set of scores are 4 and 11.

The range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }\\ =11-4\\ =7\end{array}$

If the scores are whole numbers, the range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }+1\\ =11-4+1\\ =8\end{array}$

The mean is,

$\begin{array}{c}\mu =\frac{{\displaystyle \sum x_i}}{N}\\ =\frac{4+11+8+10+11+10}{6}\\ =\frac{54}{6}\\ =9\end{array}$

The standard deviation is,

$\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {(x_i-\mu )}^2}}{N}}\\ =\sqrt{\frac{{(4-9)}^2+{(11-9)}^2+{(8-9)}^2+{(10-9)}^2+{(11-9)}^2+{(10-9)}^2}{6}}\\ =\sqrt{6}\\ =2.45\end{array}$

Conclusion:

The required new range and standard deviation for the scores is 7 or 8 and 2.45.

Based on the results, it can be concluded that the range and standard deviation does not change when adding the two points for each score.