The mean and variance of 20 observations are found to be 10 and 4 respectively. On rechecking, it was found that an observation 8 is incorrect. If the wrong observation is omitted, then the correct variance is
Step-by-Step Solution
Key Concept: Recalculating variance after correcting an erroneous data point by updating $\sum x_i^2$ and recalculating the mean
Given $n = 20$, $\bar{x}_{old} = 10$, and $\text{Var}(x) = 4$. We calculate $\bar{x} = \frac{200}{20} = 10$ and $\sum x_i = 200$. Using the variance formula, $\text{Var}(x_{old}) = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we get $\sum x_i^2 = 2080$. For the new data after correction: $\sum x_i^{\text{new}} = 2080 - 64 = 2016$ and $\text{Var}(x_{\text{new}}) = \frac{2016}{20} - (\bar{x}_{\text{new}})^2 = 100.8 - (\frac{1116}{20})^2 = \frac{1180}{64}$.
Correct Answer: 1180/64