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Question
Find the mean and variance for the following frequency distribution.
| Classes | 0 - 30 | 30 - 60 | 60 - 90 | 90 - 120 | 120 - 150 | 150 - 180 | 180 - 210 |
| Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Solution
Let, assumed mean A = 105, class interval h = 30
`y_i = (x_i - A)/h = (x_i - 10)/30`
| Class | Mid-point xi | Frequency | yi | fiyi | yi2 | fiyi2 |
| 0 - 30 | 1 | 2 | −3 | −6 | 9 | 18 |
| 30 - 60 | 4 | 3 | −2 | −6 | 4 | 12 |
| 60 - 90 | 75 | 5 | −1 | −5 | 1 | 5 |
| 90 - 120 | 105 | 10 | 0 | 0 | 0 | 0 |
| 120 - 150 | 135 | 3 | 1 | 3 | 1 | 3 |
| 150 - 180 | 165 | 5 | 2 | 10 | 4 | 20 |
| 180 - 210 | 195 | 2 | 3 | 6 | 9 | 18 |
| Sum | - | 30 | - | 2 | - | 76 |
Mean `overline x = A + ((sumf_iy_i)/N) xx h`
= `105 + 2/30 xx 30`
= 107
Variance σ2 = `h^2/N^2 [N sumf_iy_i^2 - (sumf_iy_i)^2]`
= `(30 xx 30)/(30 xx 30)[30 xx 76 - 2^2]`
= 2280 − 4
= 2276
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