Chapter 11.7, Problem 15PPSE

Solution

To calculate: The mean and standard deviation of given data set and also find the number of data values between one standard deviation from the mean, within two standard deviations from the mean, and within three standard deviations from the mean and explain how a graph help to describe how many standard deviations a data value is from the mean.

The mean and standard deviation of the given data is 53.8 and 3.60 respectively and the points $(50.20,57.40)(46.60,61.00)(43.00,64.60)$ are the points within 1 standard deviation. 2 standard deviations and 3 standard deviations from the mean. There are 7 data values within one standard deviation from the mean, there are 10 data values within two and three standard deviations from the mean. The graph is used to determine Z score which is the number of standard deviations a data value is from the mean.

Given:

The data for daily energy usage of a small town in 10 days in June is shown below

51.8 MWh	53.6 MWh	54.7 MWh	51.9 MWh	49.3 MWh
52.0 MWh	53.5 MWh	51.2 MWh	60.7 MWh	59.3 MWh

Formula used:

The formula used to calculate the mean and standard deviation are shown below?

  $\begin{array}{l}Mean=\frac{{\displaystyle \sum X_i}}{n}\\ \text{Standard deviation}=\sqrt{\frac{{{\displaystyle \sum \left(X_i-\stackrel{\_\_}{X}\right)}}^2}{n-1}}\end{array}$

Calculation:

The mean of the given data is calculated as shown below

  $\begin{array}{c}Mean=\frac{{\displaystyle \sum X_i}}{n}\\ =\frac{51.8+53.6+54.7+51.9+49.3+52+53.5+51.2+60.7+59.3}{10}\\ =53.8\end{array}$

The standard deviation is the average of the deviations of all data points from the mean and is calculated using the following table

X	$\left(X-\stackrel{\_\_}{X}\right)$	${\left(X-\stackrel{\_\_}{X}\right)}^2$
51.8	$$$51.8-53.8=-2$	4
53.6	$53.6-53.8=-0.2$	0.04
54.7	$54.7-53.8=0.9$	0.81
51.9	$51.9-53.8=-1.9$	3.61
49.3	$49.3-53.8=-4.5$	20.25
52	$52-53.8=-1.8$	3.24
53.5	$53.5-53.8=-0.3$	0.09
51.2	$51.2-53.8=-2.6$	6.76
60.7	$60.7-53.8=6.9$	47.61
59.3	$59.3-53.8=5.5$	30.25

The summation of squared deviations from the mean is calculated as shown below

  ${{\displaystyle \sum \left(X-\stackrel{\_\_}{X}\right)}}^2=4+0.04+0.81+\mathrm{...}+30.25=116.66$

The standard deviation is calculated as shown below

  $\begin{array}{l}\text{Standard deviation}=\sqrt{\frac{{{\displaystyle \sum \left(X_i-\stackrel{\_\_}{X}\right)}}^2}{n-1}}\\ =\sqrt{\frac{116.66}{10-1}}\\ =3.60\end{array}$

The one, two and three standard deviations from the mean data points are calculated as shown below

  $\begin{array}{c}\mu \pm \sigma =53.8\pm 3.60\\ =\left(53.8-3.60,53.8+3.60\right)\\ =(50.2,57.4)\end{array}$

  $\begin{array}{l}\mu \pm 2\sigma =53.8\pm 2\times 3.60\\ =\left(53.8-7.20,53.8+7.20\right)\\ =(46.6,61)\end{array}$

  $\begin{array}{l}\mu \pm 3\sigma =53.8\pm 3\times 3.60\\ =\left(53.8-10.80,53.8+10.80\right)\\ =(43,64.6)\end{array}$

Thus with reference to given data points, within one standard deviation of mean there are 7 data points and within two and three standard deviation of mean there are all 10 data points.

The graph of the standard normal curve has Z scores and the probabilities, Z score is the number of standard deviations of a data point from mean. The standard normal curve is shown below

  

With reference to Z scores and their probabilities, the number of data points within given standard deviations of mean can be determined using the graph.