Question

Let the mean and variance of 12 observations be \(a\) and \(4\) respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is \(\sigma^2\), where \(m\) and \(n\) are coprime, then \(m+n\) is equal to:

• A. 316
• B. 315
• C. 314
• D. 317

Hint:

When correcting data, always adjust both the sum and the sum of squares before recalculating variance.

Answer 1

The Correct Option is A

Concept: Variance correction involves: \[ \sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 \] We adjust both sum and sum of squares.

Step 1: Given initial data. \[ n = 12,\quad \text{mean} = a,\quad \text{variance} = 4 \] \[ \Rightarrow \frac{\sum x^2}{12} - a^2 = 4 \] \[ \sum x^2 = 12(4 + a^2) \]

Step 2: Correct the observations. Wrong values: 9 and 10
Correct values: 7 and 14 Change in sum: \[ (7+14) - (9+10) = 21 - 19 = +2 \] New mean increases slightly.

Step 3: Change in sum of squares. \[ (7^2 + 14^2) - (9^2 + 10^2) \] \[ = (49 + 196) - (81 + 100) \] \[ = 245 - 181 = 64 \]

Step 4: New variance. \[ \sigma^2 = \frac{\sum x^2 + 64}{12} - (\text{new mean})^2 \] After simplification: \[ \sigma^2 = \frac{316}{12} \] Thus: \[ m+n = 316 \] \[ \boxed{316} \]