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The mean and the standard deviation of a data of 8 items are 25 and 5 respectively. If two items 15 and 25 are added to this data, then the variance of the new data is
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2553 Upvotes
Verified Answer
The correct answer is:
29
We have, $n=8, \bar{x}=25$ and $\sigma=5$
$$
\begin{aligned}
& \bar{x}=\frac{\Sigma x_i}{n} \\
& \Rightarrow \quad \Sigma x_i=n \bar{x}=8 \times 25=200 \\
& \Rightarrow \text { In corrected } \Sigma x_i=200 \\
& \text { and } \quad \sigma=5 \\
& \Rightarrow \quad \sigma^2=25 \\
& \Rightarrow \frac{1}{n} \Sigma x_i^2-(\text { mean })^2=25 \\
& \Rightarrow \quad \frac{\Sigma x_i^2}{8}-625=25 \\
& \Rightarrow \quad \Sigma x_i^2=5200 \\
& \text { Corrected } \Sigma x_i^2=5200+225+625=6050 \\
& \text { and corrected mean }=200+15+25=240 \\
& \therefore \text { Corrected variance }=\frac{6050}{10}-\left(\frac{240}{10}\right)^2 \\
& =605-(24)^2=605-576=29 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \bar{x}=\frac{\Sigma x_i}{n} \\
& \Rightarrow \quad \Sigma x_i=n \bar{x}=8 \times 25=200 \\
& \Rightarrow \text { In corrected } \Sigma x_i=200 \\
& \text { and } \quad \sigma=5 \\
& \Rightarrow \quad \sigma^2=25 \\
& \Rightarrow \frac{1}{n} \Sigma x_i^2-(\text { mean })^2=25 \\
& \Rightarrow \quad \frac{\Sigma x_i^2}{8}-625=25 \\
& \Rightarrow \quad \Sigma x_i^2=5200 \\
& \text { Corrected } \Sigma x_i^2=5200+225+625=6050 \\
& \text { and corrected mean }=200+15+25=240 \\
& \therefore \text { Corrected variance }=\frac{6050}{10}-\left(\frac{240}{10}\right)^2 \\
& =605-(24)^2=605-576=29 \\
&
\end{aligned}
$$
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