The mean and the standard deviation of a data of 8 items are 25 and 5 respectively. If two items 15 and 25…

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
  1. 29
  2. 24
  3. 26
  4. $\sqrt{29}$

Solution

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} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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