Question

The mean and variance of $20$ observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was $11$. Then the correct variance is :

• A. $4.01$
• B. $3.99$
• C. $3.98$
• D. $4.02$

Answer 1

The Correct Option is B

To find the correct variance after the observation error is corrected, let's go through the necessary calculations step-by-step.

1. The original data consists of 20 observations with a mean \((\bar{x})\) of 10 and a variance \((\sigma^2)\) of 4.
2. We calculate the sum of the original observations using the formula for the mean:
   
   S = n \times \bar{x} = 20 \times 10 = 200
3. The incorrect observation was 9, and it should have been 11. Therefore, the corrected sum of the observations is:
   
   S_{\text{corrected}} = S - 9 + 11 = 200 - 9 + 11 = 202
4. The corrected mean \((\bar{x}_{\text{corrected}})\) is:
   
   \bar{x}_{\text{corrected}} = \frac{S_{\text{corrected}}}{n} = \frac{202}{20} = 10.1
5. To find the variance, we need the sum of the squares of the observations. Using the formula for variance:
   
   \sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - \bar{x}^2
   Rearranging it, we get:
   
   \sum_{i=1}^{n} x_i^2 = n \times (\sigma^2 + \bar{x}^2)
6. Plugging in the known values:
   
   \sum_{i=1}^{20} x_i^2 = 20 \times (4 + 10^2) = 20 \times (4 + 100) = 20 \times 104 = 2080
7. Correcting for the observation change gives:
   
   \sum_{i=1}^{20} x_i^2_{\text{corrected}} = 2080 - 9^2 + 11^2 = 2080 - 81 + 121 = 2120
8. Finally, calculate the corrected variance:
   
   \sigma^2_{\text{corrected}} = \frac{\sum x_i^2_{\text{corrected}}}{n} - \left(\bar{x}_{\text{corrected}}\right)^2
   
   Substituting:
   
   \sigma^2_{\text{corrected}} = \frac{2120}{20} - (10.1)^2 = 106 - 102.01 = 3.99

Therefore, the correct variance is 3.99, which matches the provided correct answer option.