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  • The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70 . I

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70 . If the correct mean is 16 , then the correct variance is equal to:

Option: 1

10


Option: 2

36


Option: 3

43


Option: 4

60


Answers (1)

best_answer

\mathrm{\begin{aligned} &\bar{x}=15 \\ &r=2 \\ &n=50 \end{aligned} }
let incorrect observation =\mathrm{x_{1}}
correct obsenvation =\mathrm{x_{1}^{*}}

\mathrm{\begin{aligned} &\therefore \quad x_{1}+x_{2}+\ldots-x_{50}=750 \\ &x_{1}^{*}+x_{2}+\ldots x_{50}=800 \\ &x_{1}^{*}-x_{1}=50 \end{aligned} }
also \mathrm{ \quad x_{1}^{*}+x_{1}=70}
\mathrm{\therefore \quad x_{1}=10, \quad x_{1}^{*}=60 }

\mathrm{\therefore \frac{x_{1}^{2}+x_{2}^{2}+--x_{50}^{2}}{50}-225=4 }

\mathrm{\begin{aligned} &x_{1}^{2}+x_{2}^{2}+\cdots-x_{50}^{2}=11450 \\ &\left(x_{1}^{*}\right)^{2}+x_{2}^{2}+\cdots-x_{50}^{2} =11450+3500 \\ &=14950 \end{aligned} }

\mathrm{ \begin{aligned} &=\frac{14950}{50}-256 \\ &=299-256 \\ &=43 \end{aligned} }

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vishal kumar

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