Chapter 3.2, Problem 26E

To determine

To find: The variance and standard deviation for each league.

To compare: The results.

Answer to Problem 26E

The variance and standard deviation for NL teamis 0.00005 and 0.007.

The variance and standard deviation for AL team0.00005 and 0.007.

The results are same for both the teams.

Explanation of Solution

Given info:

The data represents the Team batting averages for major league baseball in 2005.

NL		AL
0.252-0.256	4	0.256-0.261	2
0.257-0.261	6	0.262-0.267	5
0.262-0.266	1	0.268-0.273	4
0.267-0.271	4	0.274-0.279	2
0.272-0.276	1	0.280-0.285	1

Calculation:

For NL Team:

The midpoints for each class are calculated below,

NL	Frequency $f$	Midpoint $X_m$
0.252-0.256	4	$\frac{0.252+0.256}{2}=0.254$
0.257-0.261	6	$\frac{0.257+0.261}{2}=0.259$
0.262-0.266	1	$\frac{0.262+0.266}{2}=0.264$
0.267-0.271	4	$\frac{0.267+0.271}{2}=0.269$
0.272-0.276	1	$\frac{0.272+0.276}{2}=0.274$
	${\displaystyle \sum f}=16$

Mean of the frequency distribution:

The formula is,

$\overline{X}=\frac{{\displaystyle \sum f\cdot X_m}}{n}$

${\displaystyle \sum f\cdot X_m}$ is obtained by multiplying the midpoint with the corresponding frequency value as shown below:

Midpoint $X_m$	Frequency $f$	$f\cdot X_m$
0.254	4	$0.254\times 4=1.016$
0.259	6	$0.259\times 6=1.554$
0.264	1	$0.264\times 1=0.264$
0.269	4	$0.269\times 4=1.076$
0.274	1	$0.274\times 1=0.274$
	${\displaystyle \sum f}=16$	${\displaystyle \sum f\cdot X_m}=4.184$

Substitute n as 16 and ${\displaystyle \sum f\cdot X_m}$ as 4.184 in the formula.

$\begin{array}{c}\overline{X}=\frac{4.184}{16}\\ =0.2615\\ \approx 0.26\end{array}$

Thus, the mean for frequency distribution is 0.26.

Variance for grouped data:

The formula of sample variance for grouped data is,

$\begin{array}{c}\text{Sample}\text{Variance}=s^2\\ =\frac{n\left({\displaystyle \sum f\cdot X_m^2}\right)-{\left({\displaystyle \sum f\cdot X_m}\right)}^2}{n\left(n-1\right)}\end{array}$

Midpoint $X_m$	$X_m^2$	Frequency $f$	$f\cdot X_m^2$
0.254	0.064516	4	$\begin{array}{c}0.064516\end{array}\times 4=0.258064$
0.259	0.067081	6	$\begin{array}{c}0.067081\end{array}\times 6=0.402486$
0.264	0.069696	1	$\begin{array}{c}0.069696\end{array}\times 1=\begin{array}{c}0.069696\end{array}$
0.269	0.072361	4	$\begin{array}{c}0.072361\end{array}\times 4=0.289444$
0.274	0.075076	1	$\begin{array}{c}\begin{array}{c}0.075076\end{array}\end{array}\times 1=\begin{array}{c}0.075076\end{array}$
		${\displaystyle \sum f}=16$	${\displaystyle \sum f\cdot X_m^2}=1.094766$

Subtitiute ${\displaystyle \sum f\cdot X_m^2}$ as 1.094766, n as 16 and ${\displaystyle \sum f\cdot X_m}$ as 4.184

$\begin{array}{c}{s}^2=\frac{16\left(1.094766\right)-{\left(4.184\right)}^2}{16\left(16-1\right)}\\ =\frac{17.516256-17.505856}{16\left(15\right)}\\ =\frac{0.0104}{240}\\ =0.0000433\end{array}$

$\approx 0.00005$

Thus, the variance for the data set is approximately 0.00005.

Sample standard deviation for grouped data:

The square root of the sample variance is termed as sample standard deviation.

$\begin{array}{c}\text{Sample}\text{standard}\text{deviation}=s\\ =\sqrt{s^2}\\ =\sqrt{\frac{n\left({\displaystyle \sum f\cdot X_m^2}\right)-{\left({\displaystyle \sum f\cdot X_m}\right)}^2}{n\left(n-1\right)}}\end{array}$

Substitute s² as 0.0000433 in the formula,

$\begin{array}{c}s=\sqrt{0.0000433}\\ =0.00658\\ \approx 0.007\end{array}$

Thus, the standard deviation for the data set is approximately 0.007.

For AL team:

The midpoints for each class are calculated below,

AL	Frequency $f$	Midpoint $X_m$
0.256-0.261	2	$\frac{0.256+0.261}{2}=0.2585$
0.262-0.267	5	$\frac{0.262+0.267}{2}=0.2645$
0.268-0.273	4	$\frac{0.268+0.273}{2}=0.2705$
0.274-0.279	2	$\frac{0.274+0.279}{2}=0.2765$
0.280-0.285	1	$\frac{0.280+0.285}{2}=0.2825$
	${\displaystyle \sum f}=14$

Mean of the frequency distribution:

The formula is,

$\overline{X}=\frac{{\displaystyle \sum f\cdot X_m}}{n}$

${\displaystyle \sum f\cdot X_m}$ is obtained by multiplying the midpoint with the corresponding frequency value as shown below:

Midpoint $X_m$	Frequency $f$	$f\cdot X_m$
0.2585	2	$0.2585\times 2=0.517$
0.2645	5	$0.2645\times 5=1.3225$
0.2705	4	$0.2705\times 4=1.082$
0.2765	2	$0.2765\times 2=0.553$
0.2825	1	$0.2825\times 1=0.2825$
	${\displaystyle \sum f}=14$	${\displaystyle \sum f\cdot X_m}=3.757$

Substitute n as 14 and ${\displaystyle \sum f\cdot X_m}$ as 3.757 in the formula.

$\begin{array}{c}\overline{X}=\frac{3.757}{14}\\ =0.2684\\ \approx 0.27\end{array}$

Thus, the mean for frequency distribution is 0.27.

Variance for grouped data:

The formula of sample variance for grouped data is,

$\begin{array}{c}\text{Sample}\text{Variance}=s^2\\ =\frac{n\left({\displaystyle \sum f\cdot X_m^2}\right)-{\left({\displaystyle \sum f\cdot X_m}\right)}^2}{n\left(n-1\right)}\end{array}$

Midpoint $X_m$	$X_m^2$	Frequency $f$	$f\cdot X_m^2$
0.2585	0.06682225	2	$\begin{array}{c}0.06682225\end{array}\times 2=0.1336445$
0.2645	0.06996025	5	$\begin{array}{c}0.06996025\end{array}\times 5=0.3498013$
0.2705	0.07317025	4	$\begin{array}{c}0.07317025\end{array}\times 4=0.292681$
0.2765	0.07645225	2	$\begin{array}{c}0.07645225\end{array}\times 2=0.1529045$
0.2825	0.07980625	1	$\begin{array}{c}0.07980625\end{array}\times 1=\begin{array}{c}0.07980625\end{array}$
		${\displaystyle \sum f}=14$	${\displaystyle \sum f\cdot X_m^2}=1.0088375$

Subtitiute ${\displaystyle \sum f\cdot X_m^2}$ as 1.0088375, n as 16 and ${\displaystyle \sum f\cdot X_m}$ as 3.757

$\begin{array}{c}{s}^2=\frac{14\left(1.0088375\right)-{\left(3.757\right)}^2}{14\left(14-1\right)}\\ =\frac{14.123725-14.115049}{14\left(13\right)}\\ =\frac{0.008676}{182}\\ =0.0000477\end{array}$

$\approx 0.00005$

Thus, the variance for the data set is approximately 0.00005.

Sample standard deviation for grouped data:

The square root of the sample variance is termed as sample standard deviation.

$\begin{array}{c}\text{Sample}\text{standard}\text{deviation}=s\\ =\sqrt{s^2}\\ =\sqrt{\frac{n\left({\displaystyle \sum f\cdot X_m^2}\right)-{\left({\displaystyle \sum f\cdot X_m}\right)}^2}{n\left(n-1\right)}}\end{array}$

Substitute s² as 0.0000477 in the formula,

$\begin{array}{c}s=\sqrt{0.0000477}\\ =0.00690\\ \approx 0.007\end{array}$

Thus, the standard deviation for the data set is approximately 0.007.

From the results the variance and standard deviation for both the teams is same.