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  • If for a distribution ∑ (x −5)=3, ∑ (x −5)2 = 43 and the total number of item is 18, find the mean and standard deviation.

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

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\\\\ \text{It is given that for a distribution } \Sigma \left( x-5 \right) =3,~ \Sigma \left( x-5 \right) ^{2}=43 \text{ and the total number of item is 18} \\ \text{We have to find the mean and standard deviation} \\ \text{As per given criteria, Number of items, n=18}

\\ \text{~ Mean can be written as }\overline{x}=A+\frac{ \Sigma \left( x-5 \right) }{n}~~ \\ \\ \text{ Here assumed mean is 5, so substituting the corresponding values in above equation we get} \\ \\ ~\overline{x}=5+\frac{3}{18}=\frac{93}{18}=5.17~ \\ \\ \text{And we know that the standard deviation can be written as}, \sigma =\sqrt {\frac{ \Sigma \left( x-5 \right) ^{2}}{n}- \left( \frac{ \Sigma \left( x-5 \right) }{n} \right) ^{2}}~ \\



\\ \text{Substituting the corresponding values, we get }\sigma =\sqrt {\frac{43}{18}- \left( \frac{3}{18} \right) ^{2}}= \sqrt {2.39- \left( 0.166 \right) ^{2}} =\sqrt {2.39-0.027}= \sqrt {2.363}=1.54~ \\\\ \\ \text{ Thus the mean and standard deviation of given items are 5.17 and 1.54 respectively} \\\\

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