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

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

`s = 0.9969` --- sample standard deviation

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 = (1)/(2)(116.1+120.0)`

`m = (1)/(2)(236.1)`

`m = 118.05`

The same approach is applied to other classes.

Next, is to calculate the mean:

`\bar x = (\sum fx)/(\sum f)`

In this case, it is:

`\bar x = (\sum fm)/(\sum f)`

Where

`m = midpoint`

So, we have:

`\bar x = (118.05*38+122.05*24+126.05*28+130.05*43+134.05*29)/(38+24+28+43+29)`

`\bar x = (20424.10)/(162)`

`\bar x = 126.07`

The sample variance (s^2) is:

`s^2 = (\sum(m - \bar x)^2)/(\sum f -1)`

This gives:

`s^2 = ((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 = (160.002)/(161)`

`s^2 = 0.9938`

The sample standard deviation (s) is:

`s = \sqrt{s^2`

`s = \sqrt{0.9938`

`s = 0.9969`