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 6.01 - 8.00 50 8.01 - 10.00 42 10.01 - 12.00 13 12.01 - 14.00 16 14.01 - 16.00 23

[tex]\begin{array}{cc}{Class} & {Frequency} & 6.01 - 8.00 & 50& 8.01 - 10.00 &42 & 10.01 - 12.00 & 13 & 12.01 - 14.00 & 16 & 14.01 - 16.00 & 23 \ \end{array}[/tex]

Solving (a): The mean

First, we calculate the class midpoint (x); this is the average of each class.

For 6.01 - 8.00

[tex]x = \frac{6.01+8.00}{2} =7.005[/tex]

When done for the whole classes, the frequency table will become:

[tex]\begin{array}{cc}{x} & {Frequency} & 7.005 & 50& 9.005 &42 & 11.005 & 13 & 13.005 & 16 & 15.005 & 23 \ \end{array}[/tex]

The mean is then calculated as:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

[tex]\bar x = \frac{7.005*50+9.005*42+11.005*13+13.005*16+15.005*23}{50+42+13+16+23}[/tex]

[tex]\bar x = \frac{1424.72}{144}[/tex]

[tex]\bar x= 9.894[/tex]

Solving (b): The sample variance

This is calculated using:

[tex]Var = \frac{\sum f(x - \bar x)^2}{\sum f-1}[/tex]

So, we have:

[tex]Var = \frac{(7.005 - 9.894)^2 * 50 + (9.005 - 9.894)^2 * 42 + (11.005 - 9.894)^2 * 13 + (13.005 - 9.894)^2 * 16 + (15.005 - 9.894)^2 *23}{50+42+13+16+23-1}[/tex]

[tex]Var = \frac{1222.222}{143}[/tex]

[tex]Var = 8.547[/tex]