Chapter 11.7, Problem 20PPSE

Solution

To find: a) The mean and the standard deviation of the data for daily energy usage of a small town during ten days in January as follows:

83.8 MWh, 87.1 MWh, 92.5 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

b) It is to be determined how many values in the data set fall within one standard deviation from the mean, within two standard deviations and within three standard deviations.

a) The mean and standard deviation for the given data set are $82.3,4.33$ respectively.

b) The values within one standard deviation of the mean are 83.8 MWh, 80.6 MWh, 82.4 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

The values within two standard deviations of the mean are 83.8 MWh, 87.1 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

The values within three standard deviations of the mean are 83.8 MWh, 87.1 MWh, 92.5 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

Given information:

The given data set is 83.8 MWh, 87.1 MWh, 92.5 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

Calculation:

The mean of the data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{83.8+87.1+92.5+80.6+82.4+77.6+78.9+78.2+81.8+80.1}{10}\\ \overline{x}=\frac{823}{10}\\ \overline{x}=82.3\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
83.8	82.3	1.5	2.25
87.1	82.3	4.8	23.04
92.5	82.3	10.2	104.04
80.6	82.3	-1.7	2.89
82.4	82.3	0.1	0.01
77.6	82.3	-4.7	22.09
78.9	82.3	-3.4	11.56
78.2	82.3	-4.1	16.81
81.8	82.3	-0.5	0.25
80.1	82.3	-2.2	4.84
			187.78

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{187.78}{10}\\ {\sigma }^2\approx 18.78\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{18.78}\\ \sigma =4.33\end{array}$

The minimum and maximum values of the data are 77.6 MWh and 92.5 MWh respectively.

1 standard deviation below the mean: $82.3-4.33=77.97$

1 standard deviation above the mean: $82.3+4.33=86.63$

The values within one standard deviation of the mean are 83.8 MWh, 80.6 MWh, 82.4 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

2 standard deviations below the mean: $82.3-2\left(4.33\right)=73.64$

2 standard deviations above the mean: $82.3+2\left(4.33\right)=90.96$

The values within two standard deviations of the mean are 83.8 MWh, 87.1 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

3 standard deviations below the mean: $82.3-3\left(4.33\right)=69.31$

3 standard deviations above the mean: $82.3+3\left(4.33\right)=95.29$

The values within three standard deviations of the mean are 83.8 MWh, 87.1 MWh, 92.5 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

Conclusion:

The mean and standard deviation for the given data set are $82.3,4.33$ respectively.

The values within one standard deviation of the mean are 83.8 MWh, 80.6 MWh, 82.4 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

The values within two standard deviations of the mean are 83.8 MWh, 87.1 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.

The values within three standard deviations of the mean are 83.8 MWh, 87.1 MWh, 92.5 MWh, 80.6 MWh, 82.4 MWh, 77.6 MWh, 78.9 MWh, 78.2 MWh, 81.8 MWh, 80.1 MWh.