Answer:
[tex]s^2 = 0.9938[/tex] --- sample variance
[tex]s = 0.9969[/tex] --- sample standard deviation
Step-by-step explanation:
Given
[tex]\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}[/tex]
Required
Calculate the sample variance and sample standard deviation
First, we calculate the midpoint of each class:
[tex]\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}[/tex]
The midpoints are calculated by taking the average of the class intervals.
For instance:
Class 116.1 to 120.0 has a midpoint of
[tex]m = \frac{1}{2}(116.1+120.0)[/tex]
[tex]m = \frac{1}{2}(236.1)[/tex]
[tex]m = 118.05[/tex]
The same approach is applied to other classes.
Next, is to calculate the mean:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
In this case, it is:
[tex]\bar x = \frac{\sum fm}{\sum f}[/tex]
Where
[tex]m = midpoint[/tex]
So, we have:
[tex]\bar x = \frac{118.05*38+122.05*24+126.05*28+130.05*43+134.05*29}{38+24+28+43+29}[/tex]
[tex]\bar x = \frac{20424.10}{162}[/tex]
[tex]\bar x = 126.07[/tex]
The sample variance (s^2) is:
[tex]s^2 = \frac{\sum(m - \bar x)^2}{\sum f -1}[/tex]
This gives:
[tex]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}[/tex]
[tex]s^2 = \frac{160.002}{161}[/tex]
[tex]s^2 = 0.9938[/tex]
The sample standard deviation (s) is:
[tex]s = \sqrt{s^2[/tex]
[tex]s = \sqrt{0.9938[/tex]
[tex]s = 0.9969[/tex]
