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  • 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.

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\\ \text{If at the time of calculation of mean and variance of 10 readings, a student wrongly used the reading} \\\\ \\ \text{ 52 instead of the correct reading 25. The mean and variance were obtained as 45 and 16 respectively.} \\ \text{ We have to find the correct mean and the variance.} \\ ~~ \text{As per the given criteria, Number of observations}, n=10 \\\\ \text{Mean of the given observations before correction,} \\\\ \overline{X~}=45 \\ \text{~ But we know, }\overline{X~}=\frac{ \Sigma x_{i}}{n}~~ \\\\
 \\ ~~\text{Substituting the corresponding values, we get }45=\frac{ \Sigma x_{i}}{10}~~ \\\\ \\ ~~ \Sigma x_{i}=45 * 10=450~ \\\\ \\ \text{It~is~given in the question that one reading 25 was taken wrongly as 52 } \\\\ \\ So~~ \Sigma x_{i}=450-52+25=423~ \\\\ \\ \text{ So the correct mean after correction is }\overline{X~}=\frac{ \Sigma x_{i}}{n}=\frac{423}{10}=42.3~~ \\\\
 \\ \text{~Also given that variance of the 10 observations is 16 before correction } \sigma ^{2}=16~ \\\\ \\ \text{But~we know that } \\\\ \\ \sigma ^{2}=\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}~~ \\\\ \\ ~ \text{Substituting the corresponding values we get} 16=\frac{ \Sigma x_{i}^{2}}{10}- \left( 45 \right) ^{2}~~ \\\\ \\ ~\text{Now taking square on both sides, we get } 16=\frac{ \Sigma x_{i}^{2}}{10}-2025~~ \\\\ \\ ~ 16+2025=\frac{ \Sigma x_{i}^{2}}{10}~~ \\\\ \\ ~~ \Sigma x_{i}^{2}=20410~ \\\\
\\ \text{It is said that one reading 25 was wrongly taken as 52. } \\\\ \\ \text{So,~correction is } \Sigma x_{i}^{2}=20410- \left( 52 \right) ^{2}+ \left( 25 \right) ^{2}~~=20410-2704+625=18331 \\\\ \\ \text{~~So, the variance after correction is } \\\\ \\ \sigma ^{2}=\frac{18331}{10}- \left( \frac{423}{10} \right) ^{2} =1833.1- \left( 42.3 \right) ^{2}=1833.1-1789.29~=43.81 \\\\

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