Question

Calculate the sample standard deviation 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.

Lower Bound Upper Bound Frequency
56 64 11
65 73 15
74 82 14
83 91 4
92 100 11

Required:
a. What is the sample standard deviation?
b. What is the sample variance?

Answer 1

`\sigma = 12.5` ---- sample standard deviation

`\sigma^2 = 157.2` ---- sample variance

Explanation:

Solving (a): The sample variance

First, we calculate the midpoint of each class (this is the average of the limits)

So, we have:

`x_1 = (56 + 64)/(2) = 60`

`x_2 = (65 + 73)/(2) = 69`

And so on

So, the table becomes:

`\begin{array}{cc}{x} & {f} & {60} & {11} & {69} & {15} & {78} & {14} & {87} & {4} & {96} & 11 \ \end{array}`

Calculate the mean

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

`\bar x = (60*11+69*15+78*14+87*4+96*11)/(11+15+14+4+11)`

`\bar x = (4191)/(55)`

`\bar x = 76.2`

The variance is:

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

So, we have:

`\sigma^2 = ((60 - 76.2)^2*11+(69 - 76.2)^2*15+(78 - 76.2)^2 *14+(87 - 76.2)^2*4+(96 - 76.2)^2*11)/(11+15+14+4+11-1)`

`\sigma^2 = (8488.8)/(54)`

`\sigma^2 = 157.2`

The sample standard deviation is:

`\sigma = \sqrt{\sigma^2`

`\sigma = \sqrt{157.2`

`\sigma = 12.5`

Solving (b): The sample variance

In (a), we calculate the sample variance to be:

`\sigma^2 = 157.2`