Question

Calculate the sample mean and sample variance for the following frequency distribution of hourly wages for a sample of pharmacy assistants. If necessary, round to one more decimal place than the largest number of decimal places given in the data. Hourly Wages (in Dollars) Class Frequency 10.01 - 11.50 44 11.51 - 13.00 27 13.01 - 14.50 38 14.51 - 16.00 33 16.01 - 17.50 40

Answer 1

`\bar x = 13.739`

`\sigma^2 = 4.923`

Explanation:

Given

`\begin{array}{cc}{Class} & {Frequency} & 10.01 - 11.50 & 44 & 11.51 - 13.00 & 27 & 13.01 - 14.50 & 38 & 14.51 - 16.00 & 33 & 16.01 - 17.50 & 40 \ \end{array}`

Required

The sample mean and the sample variance

First, calculate the midpoints

`x_1 = (10.01 + 11.50)/(2) = 10.755`

`x_2 = (11.51 + 13.00)/(2) = 12.255`

And so on...

So, the table becomes:

`\begin{array}{ccc}{Class} & {Frequency} & {x} & 10.01 - 11.50 & 44 & 10.755 & 11.51 - 13.00 & 27 & 12.255 & 13.01 - 14.50 & 38 & 13.755 & 14.51 - 16.00 & 33 & 15.255 & 16.01 - 17.50 & 40 & 16.755 \ \end{array}`

So, the sample mean is:

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

`\bar x = (44 * 10.755 + 27 * 12.255 + 38 * 13.755 + 33 * 15.255 + 40 * 16.755)/(44 + 27 + 38 + 33 + 40)`

`\bar x = (2500.41)/(182)`

`\bar x = 13.739`

The sample variance is:

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

`\sigma^2 = (44 * (10.755 - 13.739)^2 + 27 * (12.255 - 13.739)^2+ 38 * (13.755 - 13.739)^2 + 33 * (15.255 - 13.739)^2+ 40 * (16.755- 13.739)^2)/(44 + 27 + 38 + 33 + 40-1)`

`\sigma^2 = (890.950592)/(181)`

`\sigma^2 = 4.923`