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

Given a statistical data set \( x_1, x_2, \dots, x_{10} \) with the following mean and sum of squares:

• Mean of the data: 5.5
• Sum of squares: \(\sum_{i=1}^{10} x_i^2 = 371\)

Two values were initially recorded incorrectly as 4 and 5, instead of their correct values of 6 and 8. The objective is to determine the variance of the corrected data.

Step-by-Step Solution:

1. Compute the initial sum of the data using the given mean:

\[\sum_{i=1}^{10} x_i = 10 \times 5.5 = 55\]

1. The sum of the incorrectly recorded values (4 and 5) is 9.
2. The sum of the correct values (6 and 8) is 14.
3. Adjust the sum of the data to reflect the corrections:

\[\text{Corrected sum} = 55 - 9 + 14 = 60\]

1. Calculate the corrected mean:

\[\text{Corrected mean} = \frac{60}{10} = 6\]

1. Modify the sum of squares:
   • Subtract the sum of squares of the incorrect values:

\[\text{Sum of squares of incorrect values} = 4^2 + 5^2 = 16 + 25 = 41\]

• Add the sum of squares of the correct values:

\[\text{Sum of squares of correct values} = 6^2 + 8^2 = 36 + 64 = 100\]

• Determine the corrected sum of squares:

\[\sum_{i=1}^{10} {x_i^2(\text{corrected})} = 371 - 41 + 100 = 430\]

1. Calculate the corrected variance using the formula:

\[\text{Variance} = \frac{\sum_{i=1}^{10} {x_i^2} - \left(\frac{\left(\sum_{i=1}^{10} {x_i}\right)^2}{n}\right)}{n} \] \[ = \frac{430 - \frac{60^2}{10}}{10} = \frac{430 - 360}{10} = \frac{70}{10} = 7\]

Consequently, the variance of the corrected data set is 7.