Question

Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American adults. If necessary, round to one more decimal place than the largest number of decimal places given in the data. Heart Rates in Beats per Minute Class Frequency 51 - 58 6 59 - 66 3 67 - 74 11 75 - 82 13 83 - 90 4

Answer 1

`Mean = 68.9`

`s^2 =18.1` --- Variance

Explanation:

Given

`\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}`

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

`x = (1)/(2)(51+58) = (1)/(2)(109) = 54.5`

For 59 - 66

`x = (1)/(2)(59+66) = (1)/(2)(125) = 62.5`

For 67 - 74

`x = (1)/(2)(67+74) = (1)/(2)(141) = 70.5`

For 75 - 82

`x = (1)/(2)(75+82) = (1)/(2)(157) = 78.5`

For 83 - 90

`x = (1)/(2)(83+90) = (1)/(2)(173) = 86.5`

So, the table becomes:

`\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}`

The mean is then calculated as:

`Mean = (\sum fx)/(\sum f)`

`Mean = (54.5*4+62.5*3+70.5*11+78.5*13+86.5*4)/(6+3+11+13+4)`

`Mean = (2547.5)/(37)`

`Mean = 68.9` -- approximated

Solving (b): The sample variance:

This is calculated as:

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

So, we have:

`s^2 =((54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2)/(37 - 1)`

`s^2 =(652.8)/(36)`

`s^2 =18.1` -- approximated