Question

For two groups of 15 sizes each, mean and variance of first group is 12, 14 respectively, and second group has mean 14 and variance of σ². If combined variance is 13 then find variance of second group?

• A. 9
• B. 11
• C. 10
• D. 12

Answer 1

The Correct Option is C

To find the variance of the second group, let's use the concept of combined variance. The formula for the combined variance \(\sigma^2_c\) of two groups with variances \(\sigma^2_1\) and \(\sigma^2_2\) is given by:

\(\sigma^2_c = \frac{n_1\sigma^2_1 + n_2\sigma^2_2 + \frac{n_1n_2}{n_1+n_2}(M_1 - M_2)^2}{n_1 + n_2}\)

where:

• \(n_1\) and \(n_2\) are the sizes of the first and second groups respectively.
• \(M_1\) and \(M_2\) are the means of the first and second groups respectively.
• \(\sigma^2_1\) and \(\sigma^2_2\) are the variances of the first and second groups respectively.

Given:

• \(n_1 = 15\), \(n_2 = 15\) (both groups have 15 members)
• \(M_1 = 12\), \(\sigma^2_1 = 14\)
• \(M_2 = 14\), \(\sigma^2_2\) (unknown, needs to be found)
• Combined variance, \(\sigma^2_c = 13\)

Substitute these into the formula:

\(13 = \frac{15 \times 14 + 15 \times \sigma^2_2 + \frac{15 \times 15}{15 + 15}(12 - 14)^2}{15 + 15}\)

Simplifying:

1. Calculate \((M_1 - M_2)^2 = (12 - 14)^2 = 4\).
2. Calculate the term \(\frac{n_1 n_2}{n_1 + n_2}(M_1 - M_2)^2 = \frac{15 \times 15}{30} \times 4 = 30\).
3. Substitute the values back: \(13 = \frac{15 \times 14 + 15 \times \sigma^2_2 + 30}{30}\).
4. Multiply through by 30 to eliminate the fraction: \(390 = 210 + 15\sigma^2_2 + 30\).
5. Simplify the equation: \(390 = 240 + 15\sigma^2_2\).
6. Rearrange to solve for \(\sigma^2_2\): \(150 = 15\sigma^2_2\).
7. Divide both sides by 15: \(\sigma^2_2 = 10\).

Therefore, the variance of the second group is 10.