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Question
If for distribution `sum(x - 5)` = 3, `sum(x - 5)^2` = 43 and total number of items is 18. Find the mean and standard deviation.
Solution
Given that `n = 18, sum(x - 5) = 3, sum(x - 5)^2` = 43
∴ Mean = `A + (sum(x - 5))/n`
= `5 + 3/18`
= `93/18`
= 5.166
= 5.17
And S.D. = `sqrt((sum(x - 5)^2)/N - [(sum(x - 5)^2)/N]^2`
= `sqrt(43/18 - (3/18)^2`
= `sqrt(2.39 - (0.166)^2`
= `sqrt(2.39 - 0.27)`
= 1.54
Hence, the required mean is 5.17 and S.D. = 1.54
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