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

We are given a statistical data set \( x_1, x_2, \dots, x_{10} \) with the mean and the sum of squares as follows:

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

The student initially noted two values incorrectly as 4 and 5 instead of 6 and 8. We need to find the variance of the corrected data.

Step-by-Step Solution:

1. Calculate the initial sum of data from the given mean: 

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

1. The incorrect data values are 4 and 5. Their sum is 9.
2. The correct data values are 6 and 8. Their sum is 14.
3. Correct the sum of the data: 

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

1. Calculate the corrected mean: 

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

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

\[\text{Incorrect sum of squares} = 4^2 + 5^2 = 16 + 25 = 41\]

• Add the squares of the correct values: 

\[\text{Correct sum of squares} = 6^2 + 8^2 = 36 + 64 = 100\]

• Calculate the corrected sum of squares: 

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

1. Calculate the corrected variance: 

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

Therefore, the variance of the corrected data is 7.

Answer 2

Given: 

\( \sum x_i = (5.5) \times 10 = 55 \), and \( \sum x_i^2 = 371 \).

Corrected Mean Calculation:

\[ \text{Corrected Mean} = \frac{\sum x_i'}{10} = \frac{55 + 6 + 8 - (4 + 5)}{10} = 6. \]

Sum of Squared Corrected Values:

\[ \sum (x_i')^2 = 371 + 6^2 + 8^2 - (4^2 + 5^2) = 471 - 41 = 430. \]

Variance Calculation:

\[ \text{Variance} = \frac{\sum (x_i')^2}{10} - \left( \frac{\sum x_i'}{10} \right)^2. \] Substituting the values: \[ \text{Variance} = \frac{430}{10} - 36 = 43 - 36 = 7. \]

Final Answer:

The correct option is \( \boxed{7} \).