Questions

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

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5.13, 1.53

5.17, 1.53

5.14, 1 .55

None of these

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detailed solution

Correct option is B

Given, $Σ (x - 5) = 3 and Σ (x - 5)^{2} = 43, n = 18$

$\begin{array}{l} ∴ Mean = 5 + \frac{\sum (x - 5)}{18} = 5 + \frac{3}{18} = 5 + \frac{1}{6} \\ = \frac{31}{6} = 5 .17 \end{array}$

, $\begin{array}{ll} Now, & Σ (x - 5)^{2} = 43 \\ \Rightarrow & {Σx}^{2} - Σ 10 x + Σ 25 = 43 \\ \Rightarrow & {Σx}^{2} - 10 {Σ (x - 5 + 5)} + 18 \times 25 = 43 \\ \Rightarrow & {Σx}^{2} - 10 Σ (x - 5) - 10 Σ 5 + 450 = 43 \\ \Rightarrow & {Σx}^{2} - 10 (3) - 10 \times 18 \times 5 + 450 = 43 \\ \Rightarrow & {Σx}^{2} - 30 - 900 + 450 = 43 \\ \Rightarrow & {Σx}^{2} = 523 \\ and & Σ (x - 5) = 3 \\ \Rightarrow & Σx - 5 Σ 1 = 3 \\ \Rightarrow & Σx = 3 + 5 \times 18 = 93 \\ ∵ & σ = \sqrt{\frac{{Σx}^{2}}{n} - {(\frac{Σx}{n})}^{2}} = \sqrt{\frac{523}{18} - {(\frac{93}{18})}^{2}} \\ = \sqrt{29 .06 - (5 .17)^{2}} \\ = \sqrt{29 .06 - 26 .73} \\ = \sqrt{2 .33} = 1 .53 \end{array}$