Let the mean and variance of 12 observations be $$\frac{9}{2}$$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$m + n$$ is equal to

We are given 12 observations with mean $$\frac{9}{2}$$ and variance 4. Two observations 9 and 10 were incorrectly recorded instead of 7 and 14. We need to find the correct variance $$\frac{m}{n}$$ (where $$m, n$$ are coprime) and compute $$m + n$$.

Find the original sum and sum of squares.

Mean = $$\frac{9}{2}$$, so $$\sum x_i = 12 \times \frac{9}{2} = 54$$.

Variance = $$\frac{\sum x_i^2}{n} - \bar{x}^2$$, so:

$$4 = \frac{\sum x_i^2}{12} - \left(\frac{9}{2}\right)^2 = \frac{\sum x_i^2}{12} - \frac{81}{4}$$

$$\frac{\sum x_i^2}{12} = 4 + \frac{81}{4} = \frac{97}{4}$$

$$\sum x_i^2 = 12 \times \frac{97}{4} = 291$$

Compute the corrected sum and sum of squares.

Corrected sum: $$54 - 9 - 10 + 7 + 14 = 56$$

Corrected mean: $$\frac{56}{12} = \frac{14}{3}$$

Corrected $$\sum x_i^2 = 291 - 9^2 - 10^2 + 7^2 + 14^2 = 291 - 81 - 100 + 49 + 196 = 355$$

Compute the corrected variance.

$$\text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{355}{12} - \left(\frac{14}{3}\right)^2 = \frac{355}{12} - \frac{196}{9}$$

Finding a common denominator (LCM of 12 and 9 is 36):

$$= \frac{355 \times 3}{36} - \frac{196 \times 4}{36} = \frac{1065 - 784}{36} = \frac{281}{36}$$

Check if the fraction is in lowest terms.

We need $$\gcd(281, 36)$$. Since 281 is not divisible by 2 or 3 (the prime factors of 36), and checking: 281 is prime (not divisible by any prime up to $$\sqrt{281} \approx 16.8$$, i.e., not by 2, 3, 5, 7, 11, or 13).

So $$m = 281$$, $$n = 36$$, and they are coprime.

Compute the answer.

$$m + n = 281 + 36 = 317$$

The answer is Option D: 317.