The mean and variance of 10 observations are found to be 10 and 4 respectively. On rechecking it was found that an observation 8 was incorrect. If it is replaced by 18, then the correct variance is
Step-by-Step Solution
Key Concept: Variance calculation with frequency distribution requires updating both the sum and frequency terms when adding new data
Given $\bar{x}_{f(old)} = 10$, $\sum x_{f(old)} = 100$ which gives $\sum f_{(old)} = 100$. Then $\sum x_{f(new)} = 1040$ and $\bar{x}_{f(new)} = \frac{1040}{110} = \frac{104}{11}$. Using the variance formula: $\text{Var}(x_{\text{old}}) = 4 = \frac{\sum x_i^2 f_i}{\sum f_i} - (\bar{x})^2$, we get $\sum x_i^2 f_i = 1040$. For the new data: $\sum x_i^2 f_i^{\text{new}} = 1040 + 64 + 324 = 1428$ and $\text{Var}(x_{\text{new}}) = \frac{1428}{110} - (\frac{104}{11})^2 = 130 - 121 = 9$.
Correct Answer: 9