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

We have a set of $$n = 20$$ observations whose (original) mean is given to be $$\bar{x}=10$$ and whose variance is given to be $$\sigma^{2}=4$$.

First, from the mean we find the total (aggregate) of all the observations. Using the relation $$\bar{x}=\dfrac{\sum x_{i}}{n}$$ we get $$\sum x_{i}=n\bar{x}=20 \times 10 = 200.$$

Next, we use the formula that connects variance with the sum of squares of the observations:

$$\sigma^{2}=\dfrac{\sum x_{i}^{2}}{n}-\bar{x}^{2}.$$

Substituting the known values,

$$4 = \dfrac{\sum x_{i}^{2}}{20} - (10)^{2}.$$

Rearranging,

$$\dfrac{\sum x_{i}^{2}}{20} = 4 + 100 = 104.$$ Hence, $$\sum x_{i}^{2} = 104 \times 20 = 2080.$$

Now the re-checking shows that one observation was wrongly recorded as $$9$$; the correct value is $$11$$. We therefore make the following corrections:

1. Corrected sum of observations

Old total $$=200$$. We remove the wrong value and insert the right value:

$$\sum x_{i}^{(\text{new})}=200 - 9 + 11 = 202.$$

2. Corrected mean

$$\bar{x}_{\text{new}} = \dfrac{202}{20} = 10.1.$$

3. Corrected sum of squares of observations

Old sum of squares $$=2080$$. Replace $$9^{2}$$ by $$11^{2}$$:

$$\sum x_{i}^{2\;(\text{new})}=2080 - 9^{2} + 11^{2} = 2080 - 81 + 121 = 2080 + 40 = 2120.$$

4. Corrected variance

Again applying the variance formula,

$$\sigma_{\text{new}}^{2}= \dfrac{\sum x_{i}^{2\;(\text{new})}}{n} - \bigl(\bar{x}_{\text{new}}\bigr)^{2}.$$

Substituting,

$$\sigma_{\text{new}}^{2}= \dfrac{2120}{20} - (10.1)^{2}.$$

We calculate each term: $$\dfrac{2120}{20}=106, \qquad (10.1)^{2}=102.01.$$

Therefore,

$$\sigma_{\text{new}}^{2}=106 - 102.01 = 3.99.$$

Hence, the correct answer is Option A.