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  • The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively, Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. The

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively, Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is ______.

Option: 1

269


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \text { Mean }=\frac{\Sigma x_i}{10} \\ & \Rightarrow 50=\frac{\Sigma x_i}{10} \\ & \Rightarrow \Sigma x_i=500 \\ & \text { correct } \Sigma x_i=500-45-50+20+25=450 \\ & \sigma^2=\frac{\Sigma x_i^2}{10}=(\overline{\mathrm{x}})^2 \\ & \Rightarrow 144=\frac{\Sigma x_i^2}{10}-2500 \end{aligned}

\begin{aligned} & \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=26440 \\ & \text { correct } \Sigma \mathrm{x}_{\mathrm{i}}^2=26440-(45)^2-(50)^2+(20)^2+(25)^2 \\ & =26440-2025-2500+400+625 \\ & =22940 \\ & \sigma^2=\frac{\text { correct } \Sigma \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\operatorname{correct} \Sigma \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & =\frac{22940}{10}-\left(\frac{450}{10}\right)^2=2294-2025 \\ & =269 \end{aligned}

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Irshad Anwar

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