Chapter 11.7, Problem 11PPSE

Solution

To find: The mean and the standard deviation for the data set of the Dow Jones Industrial average for the first 12 weeks of 2008 as follows:

  

The mean and standard deviation for the given data set are $12287.58,289.49$ respectively.

Given information:

The given data set is $\begin{array}{l}12800.18,12606.3,12099.3,12207.17,12743.19,12182.13,12348.21,12381.02,12266.39,11893.69,\\ 11951.09,11972.25\end{array}$ .

Calculation:

The mean of the data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}12800.18+12606.3+12099.3+12207.17+12743.19+12182.13\\ +12348.21+12381.02+12266.39+11893.69+11951.09+11972.25\end{array}}{12}\\ \overline{x}=\frac{147450.92}{12}\\ \overline{x}\approx 12287.58\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
12800.18	12287.58	512.6	262758.76
12606.3	12287.58	318.72	101582.44
12099.3	12287.58	-188.28	35449.36
12207.17	12287.58	-80.41	6465.77
12743.19	12287.58	455.61	207580.47
12182.13	12287.58	-105.45	11119.7
12348.21	12287.58	60.63	3676
12381.02	12287.58	93.44	8731.03
12266.39	12287.58	-21.19	449.02
11893.69	12287.58	-393.89	155149.33
11951.09	12287.58	-336.49	113225.52
11972.25	12287.58	-315.33	99433.01
			1005620.4

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{1005620.4}{12}\\ {\sigma }^2=83801.7\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{83801.7}\\ \sigma =289.49\end{array}$

Conclusion:

The mean and standard deviation for the given data set are $12287.58,289.49$ respectively.