Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Answer 1

Given:

Number of observations, n = 10

Mean,

\[\bar {x}\]  = 45  Variance,

\[\sigma^2\]   = 16

Now,
Incorrect mean,

\[\bar {x}\]  = 45 

\[\Rightarrow \frac{\text{ Incorrect } \sum_{} x_i}{10} = 45\]
\[ \Rightarrow \text{ Incorrect }  \sum_{} x_i = 450\]

∴ Correct \[\sum_{} x_i = 450 - 52 + 25 = 423\]

⇒ Correct mean = \[\frac{\text{ Correct } \sum_{} x_i}{10} = \frac{423}{10} = 42 . 3\]

Incorrect variance,

\[\sigma^2\]  = 16

\[\Rightarrow 16 = \frac{\text{ Incorrect }  \sum_{} x_i^2}{10} - \left( 45 \right)^2 \]
\[ \Rightarrow \text{ Incorrect } \sum_{} x_i^2 = 10\left( 16 + 2025 \right) = 20410\]

∴ Correct 

\[\sum_{} x_i^2 = 20410 - \left( 52 \right)^2 + \left( 25 \right)^2 = 20410 - 2704 + 625 = 18331\] 

Now,
Correct variance = \[\frac{18331}{10} - \left( 42 . 3 \right)^2 = 1833 . 1 - 1789 . 29 = 43 . 81\]