Chapter 11.7, Problem 22PPSE

Solution

To find: a) The range, the mean, and the standard deviation of the ages for each team given.

b) For the two data sets, it is to be determined if the set with the greater range necessarily have the greater standard deviation.

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

The range, the mean, and the standard deviation for the given data set of Women are 13, 25.78 and 4.04 respectively.

b) The data with greater range need not necessarily have the greater standard deviation.

Given information:

Given data sets are:

  

Calculation:

The least age for Men is 18 and highest is 36.

Hence the range is $36-18=18$

The mean of the data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}19+18+21+22+22+23+23+23+23\\ +22+36+22+24+21+21+21+21+22\end{array}}{18}\\ \overline{x}=\frac{404}{18}\\ \overline{x}=22.444\approx 22.44\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
19	22.44	-3.44	11.83
18	22.44	-4.44	19.71
21	22.44	-1.44	2.07
22	22.44	-0.44	0.19
22	22.44	-0.44	0.19
23	22.44	0.56	0.31
23	22.44	0.56	0.31
23	22.44	0.56	0.31
23	22.44	0.56	0.31
22	22.44	-0.44	0.19
36	22.44	13.56	183.87
22	22.44	-0.44	0.19
24	22.44	1.56	2.43
21	22.44	-1.44	2.07
21	22.44	-1.44	2.07
21	22.44	-1.44	2.07
21	22.44	-1.44	2.07
22	22.44	-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:

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

The least age for Women is 20 and highest is 33.

Hence the range is $33-20=13$

The mean of the data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{\begin{array}{l}26+31+22+24+20+22+20+30+25\\ +26+32+30+23+33+21+27+24+28\end{array}}{18}\\ \overline{x}=\frac{464}{18}\\ \overline{x}=25.777\approx 25.78\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
26	25.78	0.22	0.05
31	25.78	5.22	27.25
22	25.78	-3.78	14.29
24	25.78	-1.78	3.17
20	25.78	-5.78	33.41
22	25.78	-3.78	14.29
20	25.78	-5.78	33.41
30	25.78	4.22	17.81
25	25.78	-0.78	0.61
26	25.78	0.22	0.05
32	25.78	6.22	38.69
30	25.78	4.22	17.81
23	25.78	-2.78	7.73
33	25.78	7.22	52.13
21	25.78	-4.78	22.85
27	25.78	1.22	1.49
24	25.78	-1.78	3.17
28	25.78	2.22	4.93
		Sum	293.14

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{293.14}{18}\\ {\sigma }^2=16.29\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{16.29}\\ \sigma =4.04\end{array}$

b) The range of data for Men is 18 and the standard deviation is 3.58

The range of data for Women is 13 and the standard deviation is 4.04

Hence the data with greater range need not necessarily have the greater standard deviation.

Conclusion:

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

The range, the mean, and the standard deviation for the given data set of Women are 13, 25.78 and 4.04 respectively.

The data with greater range need not necessarily have the greater standard deviation.