Chapter 11.7, Problem 23PPSE

Solution

To find: a) The effect of adding 10 to every value in a data set have on the standard deviation is to be determined.

b) The effect of multiplying 10 to every value in a data set have on the standard deviation is to be determined.

a) Adding 10 to every value does not change the standard deviation.

b) Multiplying 10 to every value increase the standard deviation 10 times.

Given information:

Given data sets are:

  

The range, the mean, and the standard deviation for the given data set of Men are 18, 22.44 and 3.58 respectively.

Calculation:

The mean of the data after adding 10 to each value is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}29+28+31+32+32+33+33+33+33\\ +32+46+32+34+31+31+31+31+32\end{array}}{18}\\ \overline{x}=\frac{584}{18}\\ \overline{x}=32.444\approx 32.44\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
29	32.44	-3.44	11.83
28	32.44	-4.44	19.71
31	32.44	-1.44	2.07
32	32.44	-0.44	0.19
32	32.44	-0.44	0.19
33	32.44	0.56	0.31
33	32.44	0.56	0.31
33	32.44	0.56	0.31
33	32.44	0.56	0.31
32	32.44	-0.44	0.19
46	32.44	13.56	183.87
32	32.44	-0.44	0.19
34	32.44	1.56	2.43
31	32.44	-1.44	2.07
31	32.44	-1.44	2.07
31	32.44	-1.44	2.07
31	32.44	-1.44	2.07
32	32.44	-0.44	0.19
		Sum	230.38

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{230.38}{18}\\ {\sigma }^2=12.8\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{12.8}\\ \sigma =3.58\end{array}$

Hence adding 10 to every value does not change the standard deviation.

The mean of the data after multiplying 10 to each value is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}190+180+210+220+220+230+230+230+230\\ +220+360+220+240+210+210+210+210+220\end{array}}{18}\\ \overline{x}=\frac{4040}{18}\\ \overline{x}=224.444\approx 224.44\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
190	224.44	-34.44	1186.11
180	224.44	-44.44	1974.91
210	224.44	-14.44	208.51
220	224.44	-4.44	19.71
220	224.44	-4.44	19.71
230	224.44	5.56	30.91
230	224.44	5.56	30.91
230	224.44	5.56	30.91
230	224.44	5.56	30.91
220	224.44	-4.44	19.71
360	224.44	135.56	18376.51
220	224.44	-4.44	19.71
240	224.44	15.56	242.11
210	224.44	-14.44	208.51
210	224.44	-14.44	208.51
210	224.44	-14.44	208.51
210	224.44	-14.44	208.51
220	224.44	-4.44	19.71
		Sum	23044.38

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{23044.38}{18}\\ {\sigma }^2=1280.24\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{1280.24}\\ \sigma =35.78\end{array}$

Hence multiplying 10 to every value increase the standard deviation 10 times.

Conclusion:

Adding 10 to every value does not change the standard deviation.

Multiplying 10 to every value increase the standard deviation 10 times.