Question

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively Later, the marks of one of the students is increased from 8 to 12 If the new mean of the marks is $10.2$, then their new variance is equal to :

• A. $3.92$
• B. $3.96$
• C. $4.04$
• D. $4.08$

Answer 1

The Correct Option is B

1. Let the total number of students be \(n\). The initial mean is given as: \[ \text{Mean} = 10 \implies \frac{\text{Sum of marks}}{n} = 10. \] Thus, the sum of marks is: \[ \text{Sum of marks} = 10n. \] 2. After increasing one student’s marks from \(8\) to \(12\), the sum of marks becomes: \[ 10n - 8 + 12 = 10n + 4. \] 3. The new mean is given as \(10.2\): \[ \frac{10n + 4}{n} = 10.2. \] Simplify to find \(n\): \[ 10n + 4 = 10.2n \implies 0.2n = 4 \implies n = 20. \] 4. The variance formula is: \[ \text{Variance} = \frac{\sum x_i^2}{n} - (\text{Mean})^2. \] 5. Initially, the variance is \(4\): \[ \frac{\sum x_i^2}{20} - 10^2 = 4 \implies \frac{\sum x_i^2}{20} = 104 \implies \sum x_i^2 = 2080. \] 6. After the change, the updated \(\sum x_i^2\) is: \[ \sum x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160. \] 7. The new variance is: \[ \text{New Variance} = \frac{\sum x_i^2}{20} - (\text{New Mean})^2 = \frac{2160}{20} - 10.2^2. \] Simplify: \[ \text{New Variance} = 108 - 104.04 = 3.96. \] Thus, the new variance is \(3.96\). The variance changes when marks are adjusted because it depends on both the sum of squares of individual data points and the square of the mean. Adjust both terms carefully to compute the new variance.