The sum of 100 observations and the sum of their squares are 400 & 2475, respectively. Later on, three observations 3, 4 & 5 were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is

We are told that the original set contains 100 observations whose sum is $$\sum x_i = 400$$ and whose sum of squares is $$\sum x_i^2 = 2475$$. Later it is discovered that the entries $$3,\,4,\,5$$ were wrong and must be removed.

Removing these three numbers changes the count of observations from $$100$$ to $$100-3 = 97$$, so the revised number of data points is $$n = 97$$.

Their total also changes. The contribution of the wrong numbers to the total was

$$3 + 4 + 5 = 12.$$

Therefore the corrected sum is

$$\sum x_i' = 400 - 12 = 388.$$

In the same way, the contribution of the wrong numbers to the sum of squares was

$$3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50.$$

Thus the corrected sum of squares is

$$\sum {x_i'}^{\,2} = 2475 - 50 = 2425.$$

For the remaining $$n = 97$$ observations, the arithmetic mean is

$$\bar x = \frac{\sum x_i'}{n} = \frac{388}{97}.$$

Noticing that $$97 \times 4 = 388$$, we find

$$\bar x = 4.$$

The variance of the remaining observations is computed with the formula

$$\sigma^2 = \frac{\sum {x_i'}^{\,2}}{n} - \left(\frac{\sum x_i'}{n}\right)^2.$$

First term:

$$\frac{\sum {x_i'}^{\,2}}{n} = \frac{2425}{97}.$$

Because $$97 \times 25 = 2425$$, this simplifies to

$$\frac{2425}{97} = 25.$$

Second term:

$$\left(\frac{\sum x_i'}{n}\right)^2 = \bar x^{\,2} = 4^2 = 16.$$

Putting these together gives

$$\sigma^2 = 25 - 16 = 9.$$

Therefore the variance of the 97 correct observations equals $$9$$.

Hence, the correct answer is Option C.