Question

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15(\mu + \mu^2 + \sigma^2)$ is equal to $\ldots$

Answer 1

Let the incorrect mean be $\mu'$ and standard deviation be $\sigma'$.

We have:
$\mu' = \frac{\sum z_i}{15} = 12 \implies \sum z_i = 15 \times 12 = 180.$

After correcting the value:
$\sum z_i = 180 - 10 + 12 = 182.$

Corrected mean:
$\mu = \frac{182}{15}.$

Also:
$\sigma'^2 = \frac{\sum z_i^2}{15} - \mu'^2.$

Given $\sigma' = 3$:
$\sigma'^2 = 9 \implies \frac{\sum z_i^2}{15} - 9 = 9 \implies \sum z_i^2 = 15 \times 9 + 180^2.$

Corrected variance:
$\sigma^2 = 2339.$

The required value is:
$15 \left(\mu^2 + \sigma^2 + \sigma^2\right) = 2521.$

The Correct answer is: 2521