Question

If for a distribution \( \sum (x-5) = 3, \sum (x-5)^{2} = 43 \) and the total number of items is 18, find the mean and standard deviation.

Answer 1

Here,
\( n =18, \sum (x-5)=3 ~ and ~\sum (x-5)^{2} = 43 \)

\( \Rightarrow Mean = A + \dfrac{\sum (x-A)}{n} \)

Here, A = 5

\( \Rightarrow Mean = 5+ \dfrac{3}{18} = 5+0.1666 = 5.17 \)

SD \( = \sqrt{\dfrac{\sum (x-5)^{2}}{n}-\left( \dfrac{\sum (x-5)}{n} \right )^{2}} \)

SD \( = \sqrt{\dfrac{43}{18}-\left( \dfrac{3}{18} \right )^{2}} \)

SD \( = \sqrt{2.3889-0.0277} \)

SD = 1.53

Hence, mean = 5.17 and SD = 1.53