Answer:
[tex]\bar x = 53.5[/tex]
[tex]SD = 4.21[/tex]
Step-by-step explanation:
Given
[tex]\begin{array}{cc}{'\$\ Million}&{Companies}&25-35&5&35-45&10&45-55&21&55-65&16&65-75&8&Total&60\end{array}[/tex]
Solving (a): The mean
First, we calculate the class midpoint by calculating the average of the intervals.
For instance:
The midpoint of 25 -35 is: [tex]x = \frac{1}{2}(25 +35) = \frac{1}{2} * 60 = 30[/tex]
So, the table becomes:
[tex]\begin{array}{ccc}{'\$\ Million}&{f}&x&25-35&5&{30}&35-45&10&40&45-55&21&50&55-65&16&60&65-75&8&70&Total&60\end{array}[/tex]
The mean is then calculated as:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
[tex]\bar x = \frac{30 * 5 + 40 *10 + 50 * 21 + 60 * 16 + 70 * 8}{60}[/tex]
[tex]\bar x = \frac{3210}{60}[/tex]
[tex]\bar x = 53.5[/tex]
Solving (b): The standard deviation
This is calculated using:
[tex]SD = \sqrt{\frac{\sum (x - \bar x)^2}{\sum f}}[/tex]
So, we have:
[tex]SD = \sqrt{\frac{(30-53.5)^2+(40-53.5)^2+(50-53.5)^2+(60-53.5)^2+(70-53.5)^2}{60}[/tex]
[tex]SD = \sqrt{\frac{1061.25}{60}[/tex]
[tex]SD = \sqrt{17.6875[/tex]
[tex]SD = 4.20565096032[/tex]
[tex]SD = 4.21[/tex]
