Chapter 11.7, Problem 14PPSE

Solution

To find: The standard deviation for each data set is to be determined and are to be used to compare each pair of data set.

The standard deviations for the given data sets are $14.55,52.34$ respectively.

Since the second data set has a larger standard deviation, the data set is much more spread out than the data in the first data set.

Given information:

The given data sets are:

Fastest recorded speeds of various large wild cats (miles per hour):

  $70,50,30,40,35,30,30,40,15$

Fastest recorded speeds of various birds in flight (miles per hour):

  $217,106,95,56,65,37,50,31,53,25,25,25$

Calculation:

The mean of the first data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{70+50+30+40+35+30+30+40+15}{9}\\ \overline{x}=\frac{340}{9}\\ \overline{x}=37.78\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
70	37.78	32.22	1038.13
50	37.78	12.22	149.33
30	37.78	-7.78	60.53
40	37.78	2.22	4.93
35	37.78	-2.78	7.73
30	37.78	-7.78	60.53
30	37.78	-7.78	60.53
40	37.78	2.22	4.93
15	37.78	-22.78	518.93
			1905.57

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{1905.57}{9}\\ {\sigma }^2=211.73\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{211.73}\\ \sigma =14.55\end{array}$

The mean and standard deviation for the first data set are $37.78,14.55$ respectively.

The mean of the second data is given by:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x_n}}{n}\\ \overline{x}=\frac{217+106+95+56+65+37+50+31+53+25+25+25}{12}\\ \overline{x}=\frac{785}{12}\\ \overline{x}=65.42\end{array}$

To determine the variance, the following table is used:

$x$	$\overline{x}$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
217	65.42	151.58	22976.5
106	65.42	40.58	1646.74
95	65.42	29.58	874.98
56	65.42	-9.42	88.74
65	65.42	-0.42	0.18
37	65.42	-28.42	807.7
50	65.42	-15.42	237.78
31	65.42	-34.42	1184.74
53	65.42	-12.42	154.26
25	65.42	-40.42	1633.78
25	65.42	-40.42	1633.78
25	65.42	-40.42	1633.78
			32872.96

Hence,

  $\begin{array}{l}{\sigma }^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}\\ {\sigma }^2=\frac{32872.96}{12}\\ {\sigma }^2=2739.41\end{array}$

The standard deviation is given by:

  $\begin{array}{l}\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{2739.41}\\ \sigma =52.34\end{array}$

The mean and standard deviation for the first data set are $65.42,52.34$ respectively.

Conclusion:

The standard deviations for the given data sets are $14.55,52.34$ respectively.

Since the second data set has a larger standard deviation, the data set is much more spread out than the data in the first data set.