Answer:
[tex](a)\ \bar x = 7.07[/tex]
[tex](b)\ \sigma^2 = 18.67[/tex]
[tex](c)\ \sigma = 4.32[/tex]
Step-by-step explanation:
Given
[tex]\begin{array}{cc}{Rate (\%)} & {f} & {0-3} & {10} & {3-6} & {9} & {6-9} & {8} & {9-12} & {7} & {12-15} & {8} \ \end{array}[/tex]
Solving (a): The average vacancy rate
First, we calculate the class midpoint (x). This is the mean of the class boundaries.
[tex]x = \frac{0+3}{2} = 1.5[/tex] [tex]x = \frac{3+6}{2} = 4.5[/tex] [tex]x = \frac{6+9}{2} = 7.5[/tex]
When done for the whole class, the frequency table becomes:
[tex]\begin{array}{ccc}{Rate (\%)} & {f} & {x} & {0-3} & {10} &{1.5}& {3-6} & {9} &{4.5}& {6-9} & {8} & {7.5}&{9-12} & {7} &{10.5}& {12-15} & {8} &{13.5}\ \end{array}[/tex]
The average rate is:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
[tex]\bar x = \frac{10*1.5 + 9*4.5+8*7.5+7*10.5+8*13.5}{10+9+8+7+8}[/tex]
[tex]\bar x = \frac{297}{42}[/tex]
[tex]\bar x = 7.07[/tex]
Solving (b): The variance
This is calculated as:
[tex]\sigma^2 = \frac{\sum f(x -\bar x)^2}{\sum f}[/tex]
[tex]\sigma^2 = \frac{10(1.5-7.07)^2 + 9(4.5-7.07)^2+8(7.5-7.07)^2+7(10.5-7.07)^2+8(13.5-7.07)^2}{10+9+8+7+8}[/tex]
[tex]\sigma^2 = \frac{784.2858}{42}[/tex]
[tex]\sigma^2 = 18.67[/tex]
Solving (c): The standard deviation.
This is calculated as:
[tex]\sigma = \sqrt{\sigma^2[/tex]
[tex]\sigma = \sqrt{18.67[/tex]
[tex]\sigma = 4.32[/tex]
