The mean and sample standard deviation of the dataset consisting of 9 observations is 14 and 10 respectively. Later it is noted that one observation 11 is wrongly noted as 6. What is the sample variance of the original dataset? (Correct to 2 decimal place accuracy)

To solve this problem, we need to follow these steps:

Step 1: Calculate the sum of the original dataset. Since we know the mean (average) is 14 and there are 9 observations, we can multiply these two numbers to get the sum. So, 14 * 9 = 126.

Step 2: Correct the error in the dataset. The observation 11 was wrongly noted as 6. So, we subtract 6 from the sum and add 11. So, 126 - 6 + 11 = 131.

Step 3: Calculate the new mean. We still have 9 observations, so we divide the new sum by 9. So, 131 / 9 = 14.56 (rounded to 2 decimal places).

Step 4: Calculate the sum of squared deviations from the mean. We know the sample standard deviation is 10, and the formula for standard deviation is the square root of the variance. So, the variance is 10^2 = 100. The sum of squared deviations is the variance times the number of observations minus 1 (because it's a sample variance), so 100 * (9 - 1) = 800.

Step 5: Correct the sum of squared deviations. We subtract the square of the deviation of the wrong observation from the mean and add the square of the deviation of the correct observation from the new mean. So, 800 - (14 - 6)^2 + (14.56 - 11)^2 = 800 - 64 + 12.57 = 748.57.

Step 6: Calculate the corrected sample variance. We divide the corrected sum of squared deviations by the number of observations minus 1. So, 748.57 / (9 - 1) = 93.57 (rounded to 2 decimal places).

So, the sample variance of the original dataset, corrected for the error, is 93.57.