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Home » While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
CBSE | Class 11 | Maths | NCERT Exemplar | Statistics
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Solution:

Given that n=10, \bar{x}=45 and \sigma^{2}=16
\begin{array}{l} \bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\ \Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \Sigma x_{i}=450 \end{array}

Corrected \Sigma x_{i}=450-52+25=423

\therefore \quad Corrected mean, \bar{x}=\frac{423}{10}=42.3

\Rightarrow \quad \sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}

\Rightarrow \quad 16=\frac{\Sigma x_{i}^{2}}{10}-(45)^{2}

\Rightarrow \quad \Sigma x_{i}^{2}=20410

\therefore \quad Corrected \Sigma x_{i}^{2}=20410-(53)^{2}+(25)^{2}=18331

Corrected \sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81

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