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

\(\bar{x}\) = 12, \(\sigma _{1}^{2}\) = 14, \(\bar{y}\) = 12, \(\sigma _{2}^{2}\) = \(\sigma ^{2}\), n₁ = n₂ = 15
\(\sigma _{1}^{2}\) = 14 = \(\sum \frac{x_{i}^{2}}{15}-(12)^{2}\Rightarrow \sum x_{i}^{2}=2370, \sum x_{i}=180\)
\(\sigma _{2}^{2}=\sum \frac{y_{i}^{2}}{15}-(14)^{2}, \sum y_{i}=210\)
13 = \(\frac{\sum x_{i}^{2}\sum y_{i}^{2}}{30}-(\frac{15\bar{x}+15\bar{y}}{30})^{2}\)
\(\frac{2370+\sum y_{i}^{2}}{30}-(13)^{2}\)
\(\sum y_{i}^{2}=3090\Rightarrow \sigma _{2}^{2}=\frac{3090}{15}-(14)^{2}=10\)

Answer 2

Combined Variance Calculation 

Let $n_1$ and $n_2$ be the number of elements in the first and second sets, respectively.
Let $m_1$ and $m_2$ be the means of the first and second sets, respectively.
Let $\sigma_1^2$ and $\sigma_2^2$ be the variances of the first and second sets, respectively. We are given $n_1 = n_2 = 15$, $m_1 = 12$, $m_2 = 14$, $\sigma_1^2 = 14$, and $\sigma_2^2 = \sigma^2$.

The combined variance of the two sets is given by: \[ \sigma_c^2 = \frac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1 + n_2} + \frac{n_1n_2(m_1 - m_2)^2}{(n_1 + n_2)^2}. \]

We are given that the combined variance $\sigma_c^2 = 13$. Plugging in the given values: \[ 13 = \frac{15(14) + 15\sigma^2}{15 + 15} + \frac{15 \cdot 15(12 - 14)^2}{(15 + 15)^2} \] \[ 13 = \frac{210 + 15\sigma^2}{30} + \frac{225(4)}{900} \] \[ 13 = \frac{210 + 15\sigma^2}{30} + \frac{900}{900} \] \[ 13 = \frac{210 + 15\sigma^2}{30} + 1 \] \[ 12 = \frac{210 + 15\sigma^2}{30} \] \[ 360 = 210 + 15\sigma^2\] \[ 150 = 15\sigma^2\] \[ 10 = \sigma^2. \] Therefore, $\sigma^2 = 10$.