Question

Mean of $50$ observations was found to be $80.4$. But later on, it was discovered that $96$ was misread as$69$ at one place. Find the correct mean.

• A. $75.9$
• B. $8.4$
• C. $79$
• D. $80.94$

Answer 1

The correct option is D $80.94$

$\frac{x_1+x_2+...+x_{50}}{50}=80.4$

Say $x_{10}$ (incorrect value) $=69$

${x^{\prime }}_{10}$ (correct value) $=96=x_{10}+27$

$x_1+x_2+...+x_{50}=4020$

Add $27$ on both sides

$x_1+x_2+...+x_{50}+27=4020+27=4047$

$x_1+x_2+...+{x^{\prime }}_{10}+...+x_{50}=4047$

Actual mean $=\frac{x_1+x_2+...+{x^{\prime }}_{10}+...+x_{50}}{50}$

Therefore, $\frac{4047}{50}=80.94$