Chapter 3.2, Problem 37E

To determine

To find:

The approximate sample variance of gas price corresponding to the given frequency distribution.

Answer to Problem 37E

Solution:

The approximate sample variance of gas price corresponding to the given frequency distribution is 0.004.

Explanation of Solution

Given:

The following frequency distribution gives the gas price frequency distribution.

Gas Prices
Price in Dollars per Gallon	Frequency
$3.55-3.59$	1
$3.60-3.64$	3
$3.65-3.69$	5
$3.70-3.74$	6
$3.75-3.79$	2
$3.80-3.84$	1

Formula used:

$s=\sqrt{\frac{n\left[{\displaystyle \sum \left(f_i\cdot x_i{}^2\right)}\right]-{\left[{\displaystyle \sum \left(f_i\cdot x_i\right)}\right]}^2}{n\left(n-1\right)}}$,

$\begin{array}{l}\text{where}n=s\text{ample}\text{size},\\ f_i=\text{frequency}\text{of}\text{class}i,\text{and}\\ x_i=\text{midpoint}\text{of}\text{class}i.\end{array}$

Class mid-point $=\frac{\text{upper}\text{limit}\text{of}\text{class}+\text{lower}\text{limit}\text{of}\text{class}}{2}$

$\text{Sample size}={{\displaystyle \sum f}}_i=\text{sum}\text{of}\text{frequencies}$.

$\text{Variance}=s^2$

Calculation:

Add the frequencies to calculate $n$.

$\begin{array}{l}n\\ ={\displaystyle \sum f_i}\\ =1+3+5+6+2+1\\ =18\end{array}$

Construct the following table to compute the standard deviation corresponding to the given frequency distribution. Use formula of class midpoint to calculate $x_i$ in the table.

Class	$x_i$	$f_i$	$x_i{}^2$	$f_i\times x_i$	$f_i\times x_i{}^2$
$3.55-3.59$	$\begin{array}{l}\frac{3.55+3.59}{2}\\ =3.57\end{array}$	1	${3.57}^2=12.7449$	$1\times 3.57=3.57$	$1\times 12.7449=12.7449$
$3.60-3.64$	$\begin{array}{l}\frac{3.60+3.64}{2}\\ =3.62\end{array}$	3	${3.62}^2=13.1044$	$3\times 3.62=10.86$	$3\times 13.1044=39.3132$
$3.65-3.69$	$\begin{array}{l}\frac{3.65+3.69}{2}\\ =3.67\end{array}$	5	${3.67}^2=13.4689$	$5\times 3.67=18.35$	$5\times 13.4689=67.3445$
$3.70-3.74$	$\begin{array}{l}\frac{3.70+3.74}{2}\\ =3.72\end{array}$	6	${3.72}^2=13.8384$	$6\times 3.72=22.32$	$6\times 13.8334=83.0304$
$3.75-3.79$	$\begin{array}{l}\frac{3.75+3.79}{2}\\ =3.77\end{array}$	2	${3.77}^2=14.2129$	$2\times 3.77=7.54$	$2\times 14.2129=28.4258$
$3.80-3.84$	$\begin{array}{l}\frac{3.80+3.84}{2}\\ =3.82\end{array}$	1	${3.82}^2=14.5924$	$1\times 3.82=3.82$	$1\times 14.5924=14.5924$
		$n={\displaystyle \sum f_i}=18$		${\displaystyle \sum \left(f_i\cdot x_i\right)}=66.46$	$\begin{array}{l}{\displaystyle \sum \left(f_i\cdot x_i{}^2\right)}\\ =245.4512\end{array}$

Substitute 18 for $n$, 66.46 for ${\displaystyle \sum \left(f_i\cdot x_i\right)}$, and 245.4512 for ${\displaystyle \sum \left(f_i\cdot x_i{}^2\right)}$ in the given formula to calculate the standard deviation.

$\begin{array}{rll}s& =& \sqrt{\frac{n\left[{\displaystyle \sum \left(f_i\cdot x_i{}^2\right)}\right]-{\left[{\displaystyle \sum \left(f_i\cdot x_i\right)}\right]}^2}{n\left(n-1\right)}}\\ & =& \sqrt{\frac{18\times 245.4512-{\left(66.46\right)}^2}{18\times 17}}\\ & =& \sqrt{\frac{1.19}{308}}\\ & \approx & 0.06\end{array}$

Square the value of $s$ to calculate the variance.

$\begin{array}{rll}{s}^2& =& {\left(0.06\right)}^2\\ & =& 0.0036\\ & \approx & 0.04\end{array}$

The approximate sample variance for the given frequency distribution is 0.004.

Conclusion:

Thus, the approximate sample variance of gas price corresponding to the given frequency distribution is 0.004.