Question

For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. 
The variance of the corrected data is:

• A. 7
• B. 4
• C. 9
• D. 5

Hint:

To find the variance after correcting some values, adjust the sum of squares and the sum of the values accordingly, then apply the formula for variance.

Answer 1

The Correct Option is A

Step 1: Calculate the corrected sum of the data. Initially, the mean of the 10 values was 5.5, so the sum of the values was \( 10 \times 5.5 = 55 \).
The incorrect values were 4 and 5, and the correct values are 6 and 8. The corrected sum is obtained by subtracting the incorrect values and adding the correct values: \[ \text{Corrected Sum} = 55 - 4 - 5 + 6 + 8 = 55 + 5 = 60. \] Step 2: Calculate the corrected mean. The corrected mean is the corrected sum divided by the number of values, which is 10: \[ \text{Corrected Mean} = \frac{60}{10} = 6. \] Step 3: Calculate the corrected sum of squares. The initial sum of squares was given as \( \sum_{i=1}^{10} x_i^2 = 371 \). We need to correct this value by subtracting the squares of the incorrect values and adding the squares of the correct values: \[ \text{Corrected Sum of Squares} = 371 - 4^2 - 5^2 + 6^2 + 8^2 = 371 - 16 - 25 + 36 + 64 = 371 + 59 = 430. \] Step 4: Calculate the variance of the corrected data. The variance is given by the formula: \[ \text{Variance} = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\text{Mean})^2. \] In this case, \( n = 10 \), the corrected sum of squares is 430, and the corrected mean is 6. So, \[ \text{Variance} = \frac{1}{10} (430) - (6)^2 = 43 - 36 = 7. \] Final Answer: The variance of the corrected data is 7.