The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :

1. Calculate the sum of the original observations: <br/><br/>$\frac{x_1+x_2+\ldots+x_9+50}{10}=20$ <br/><br/>$x_1+x_2+\ldots+x_9=150$ <br/><br/>2. Calculate the sum of the squares of the original observations using the original variance: <br/><br/>$\sigma^2 = 8^2 = 64$ <br/><br/>$64 = \frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10} - 400$ <br/><br/>$x_1^2+x_2^2+\ldots+x_9^2 = 2140$ <br/><br/>3. Calculate the new mean after correcting the error: <br/><br/>$\text{New mean} = \frac{150+40}{10} = 19$ <br/><br/>4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations: <br/><br/>$\text{New } \sigma^2 = \frac{2140+1600}{10} - (19)^2$ <br/><br/>$\Rightarrow$ $\sigma^2 = 13$ <br/><br/>The correct variance after correcting the error is 13 (Option C).