Question

Calculate the sample variance and sample standard deviation for the following frequency distribution of heights in centimeters for a sample of 8-year-old boys. If necessary, round to one more decimal place than the largest number of decimal places given in the data. Heights in Centimeters Class Frequency 116.1 - 120.0 38 120.1 - 124.0 24 124.1 - 128.0 28 128.1 - 132.0 43 132.1 - 136.0 29

Answer 1

Answer:

$s^2 = 0.9938$ --- sample variance

$s = 0.9969$ --- sample standard deviation

Step-by-step explanation:

Given

$\begin{array}{cc}{Heights} & {Frequency} & {116.1-120.0} & {38} & {120.1-124.0} & {24} &{124.1-128.0} & {28} & {128.1-132.0} & {43} & {132.1-136.0} & {29} \ \end{array}$

Required

Calculate the sample variance and sample standard deviation

First, we calculate the midpoint of each class:

$\begin{array}{ccc}{Heights} & {Frequency} & {m} &{116.1-120.0} & {38} & {118.05}& {120.1-124.0} & {24}& {122.05} &{124.1-128.0} & {28} &{126.05}& {128.1-132.0} & {43} & {130.05} &{132.1-136.0} & {29} &{134.05}\ \end{array}$

The midpoints are calculated by taking the average of the class intervals.

For instance:

Class 116.1 to 120.0 has a midpoint of

$m = \frac{1}{2}(116.1+120.0)$

$m = \frac{1}{2}(236.1)$

$m = 118.05$

The same approach is applied to other classes.

Next, is to calculate the mean:

$\bar x = \frac{\sum fx}{\sum f}$

In this case, it is:

$\bar x = \frac{\sum fm}{\sum f}$

Where

$m = midpoint$

So, we have:

$\bar x = \frac{118.05*38+122.05*24+126.05*28+130.05*43+134.05*29}{38+24+28+43+29}$

$\bar x = \frac{20424.10}{162}$

$\bar x = 126.07$

The sample variance (s^2) is:

$s^2 = \frac{\sum(m - \bar x)^2}{\sum f -1}$

This gives:

$s^2 = \frac{(118.05-126.07)^2+(122.05-126.07)^2+(126.05-126.07)^2+(130.05-126.07)^2+(134.05-126.07)^2}{38+24+28+43+29-1}$

$s^2 = \frac{160.002}{161}$

$s^2 = 0.9938$

The sample standard deviation (s) is:

$s = \sqrt{s^2$

$s = \sqrt{0.9938$

$s = 0.9969$