Answer:
[tex]\bar x = 50.25[/tex]
[tex]\sigma = 3.32[/tex]
Step-by-step explanation:
Given
The sumarized data
Solving (a): The mean
First, we calculate the class mid-points (x); This is the mean of the class intervals.
For 40 - 44, x = (40 + 44)/2 = 42. This is applied to all other classes.
So:
x 42 47 52 57 62
f 1 4 11 6 2
The mean is:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
[tex]\bar x = \frac{42 * 1 + 47 * 4 + 52 * 11 + 57 * 6 + 62 * 1}{1 + 4 + 11 + 6 + 2}[/tex]
[tex]\bar x = \frac{1206}{24}[/tex]
[tex]\bar x = 50.25[/tex]
Compared to the actual mean (53.3), the computed mean is lower
Solving (b): The standard deviation
This is calculated using:
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{\sum f}}[/tex]
So, we have:
[tex]\sigma = \sqrt{\frac{(42 - 50.25)^2+(47-50.25)^2+(52-50.25)^2+(57-50.25)^2)+(62-50.25)^2}{24}}[/tex]
[tex]\sigma = \sqrt{\frac{265.3125}{24}}[/tex]
[tex]\sigma = \sqrt{11.055}[/tex]
[tex]\sigma = 3.32[/tex]
