Chapter 4, Problem 12P

The range is completely determined by the two extreme scores in a distribution. The standard deviation, on the other hand, uses every score.

1. a. Compute the range (choose either definition) and the standard deviation for the following sample of n = 5 scores. Note that there are three scores clustered around the mean in the center of the distribution, and two extreme values.

Scores:     0    6    7    8    14

1. a. Now we break up the cluster in the center of the distribution by moving two of the central scores out to the extremes. Once again compute the range and the standard deviation.

New scores:    0    0    7    14    14

1. a. According to the range, how do the two distributions compare in variability? How do they com­pare according to the standard deviation?

To determine

To find: The range, variance and standard deviation for the sample scores.

Answer to Problem 12P

The range for the sample of scores is 14 or 15.

The standard deviation for the sample of scores is 5.

Explanation of Solution

Given info:

The data represents the sample scores. The scores are 0, 6, 7, 8 and 14.

Calculation:

First case of range:

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

The range for the sample of scores is,

From the sample of scores it can be observed that the maximum and minimum values of scores are 14 and 0.

$\begin{array}{c}\text{Range = }{X}_{Max}-X_{Min}\\ =14-0\\ =14\end{array}$

Thus, the range for the sample of scores is 14.

Second case of range:

The second case for range is used, in case of whole numbers.

The general formula for range of whole numbers is,

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

If the scores are whole numbers the range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }+1\\ =11-0+1\\ =15\end{array}$

Thus, the range for the sample of scores is 15.

Standard deviation:

The general formula for standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {(x_i-M)}^2}}{n-1}}$

The mean for the sample scores is,

$\begin{array}{c}M=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{0+6+7+8+14}{5}\\ =\frac{35}{5}\\ =7\end{array}$

Thus the mean for the sample of scores is 7.

The standard deviation for the sample of scores is,

$\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-M\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(0-7\right)}^2+{\left(6-7\right)}^2+{\left(7-7\right)}^2+{\left(8-7\right)}^2+{(14-7)}^2}{5-1}}\\ =\sqrt{25}\\ =5\end{array}$

Thus, the standard deviation for the sample of scores is 5.

To determine

To find: The range and standard deviation for the sample scores of two spreading scores in the middle.

Answer to Problem 12P

The range for the sample scores of two spreading scores in the middle.to each of the sample scores is14 or 15.

The standard deviation for the population scores of two spreading scores in the middle.to each of the sample scores is 7.

Explanation of Solution

Given info:

After spreading the two scores of sample scores. the sample scores are 0, 0, 7, 14, and 14.

Calculation:

First case of range:

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

The range for the sample of scores is,

From the sample of scores after spreading in the middle.to each of the sample score, it can be observed that the maximum and minimum values of scores are 14 and 0.

$\begin{array}{c}\text{Range = }{X}_{Max}-X_{Min}\\ =14-0\\ =14\end{array}$

Thus, the range for the sample of scores is 14.

Second case of range:

The second case for range is used, in case of whole numbers.

The general formula for range of whole numbers is,

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

If the scores are whole numbers the range is,

$\begin{array}{c}\text{Range}=X_{\max }-X_{\min }+1\\ =11-0+1\\ =15\end{array}$

Thus, the range for the sample of scores is 15.

Standard deviation:

The general formula for standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum {(x_i-M)}^2}}{n-1}}$

The mean for the sample scores is,

$\begin{array}{c}M=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{0+0+7+1+14}{5}\\ =\frac{22}{5}\\ =4.4\end{array}$

Thus the mean for the sample of scores is 4.4.

The standard deviation for the sample of scores is,

$\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-M\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(0-4.4\right)}^2+{\left(0-4.4\right)}^2+{\left(14-4.4\right)}^2+{\left(14-4.4\right)}^2}{5-1}}\\ =\sqrt{55.76}\\ =7.46\end{array}$

Thus, the standard deviation for the sample of scores is 7.

Interpretation:

The value of range is not changing even the middle values are clustered out, but the standard deviation for the sample scores are varying.

To determine

To compare: The variability of two distributions by using the range.

To compare: The variability of two distributions by using the standard deviation.

Answer to Problem 12P

The variability of two distributions by using the range indicates no change.

The variability of two distributions by using the standard deviation indicates the variability is multiplied by 2 points.

Explanation of Solution

Justification:

The two distributions comparing variability, the range for part (b) is exactly same as the range of part (a) but in case of standard deviation of two distributions comparing variability is different which means that the standard deviation of part (b) is twice of the standard deviation of part (a).