#### 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.

#### Solution

Given:

Number of observations, *n* = 10

Mean,

\[\bar {x}\] = 45 Variance,
Now,

Incorrect mean,

\[\sigma^2\] = 16

Incorrect mean,

\[\bar {x}\] = 45

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

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

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

∴ 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

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

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

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

∴ Correct

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

Now,

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

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

Is there an error in this question or solution?

Solution 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. Concept: Variance and Standard Deviation - Standard Deviation of a Discrete Frequency Distribution.