Question: Performance of stock screeners.Refer to the American Association of Individual Investors (AAII) statistics on stock screeners, Exercise 2.44 (p. 95). Annualized percentage return on investment (as compared to the Standard & Poor’s 500 Index) for 13 randomly selected stock screeners are reproduced in the table.

(9.0, -.1, -1.6, 14.6, 16.0, 7.7, 19.9, 9.8, 3.2, 24.8, 17.6, 10.7, 9.1)

a.Find the range of the data for the 13 stock screeners. Give the units of measurement for the range.

b.Find the variance of the data for the 13 stock screeners. If possible, give the units of measurement for the variance.

c.Find the standard deviation of the data for the 13 stock screeners. Give the units of measurement for the standard deviation

Step by step solution

01

Finding the range

Minimum Value = -1.6%

Maximum Value = 24.8%

$$\begin{gathered} \mathbf{Range}\mathbf{=}\mathbf{Maximum}\mathbf{-}\mathbf{Minimum} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{8}\mathbf{-}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{8}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{6} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{26}\mathbf{.}\mathbf{4} \end{gathered}$$

Therefore, Range = 26.4%.

02

Calculating the Variance

$$\begin{gathered} \mathbf{Mean}\mathbf{=}\frac{\mathbf{sumofallvalues}}{\mathbf{No}\mathbf{.}\mathbf{ofvalues}} \\ \mathbf{Mean}\mathbf{=}\frac{\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{+}\left(-.1\right)\mathbf{+}\left(-1.6\right)\mathbf{+}\mathbf{14}\mathbf{.}\mathbf{6}\mathbf{+}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{19}\mathbf{.}\mathbf{9}\mathbf{+}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{24}\mathbf{.}\mathbf{8}\mathbf{+}\mathbf{17}\mathbf{.}\mathbf{6}\mathbf{+}\mathbf{10}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{9}\mathbf{.}\mathbf{1}}{\mathbf{13}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{140}\mathbf{.}\mathbf{7}}{\mathbf{13}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{82} \end{gathered}$$

x	localid="1651140122911" $\mathrm{(}\mathrm{x}\mathrm{-}\overline{\mathrm{x}}\mathrm{)}$	${\left(\mathrm{x}-\overline{\mathrm{x}}\right)}^2$
9.0	-1.82	3.31
-.1	-10.92	119.25
-1.6	-12.42	154.26
14.6	3.78	14.29
16.0	5.18	26.83
7.7	-3.12	9.73
19.9	9.08	82.45
9.8	-1.02	1.04
3.2	-7.62	58.06
24.8	13.98	195.44
17.6	6.78	45.97
10.7	-0.12	0.0144
9.1	-1.72	2.96
Sum	0	713.60

$$\begin{gathered} \mathbf{Variance}\mathbf{=}\frac{\mathbf{\sum }{\mathbf{(}\mathbf{x}\mathbf{-}\overline{\mathbf{x}}\mathbf{)}}^{\mathbf{2}}}{\mathbf{n}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{713}\mathbf{.}\mathbf{60}}{\mathbf{12}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{59}\mathbf{.}\mathbf{466} \end{gathered}$$

Therefore, Variance = 59.466.

03

Estimating the standard deviation

$$\begin{gathered} \mathbf{StandardDeviation}\mathbf{=}\sqrt{\mathbf{Variance}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{\mathbf{59}\mathbf{.}\mathbf{466}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{71} \end{gathered}$$

Therefore, Standard Deviation = 7.71%.

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