The mean and variance of a set of 15 numbers | vedclass

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Question

(D) Let the two sets be $S_1$ and $S_2$ with $n_1 = 15, n_2 = 15$. Given: $\bar{x}_1 = 12, \sigma_1^2 = 14$ and $\bar{x}_2 = 14, \sigma_2^2 = \sigma^2

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Let the two sets be $S_1$ and $S_2$ with $n_1 = 15, n_2 = 15$. Given: $\bar{x}_1 = 12, \sigma_1^2 = 14$ and $\bar{x}_2 = 14, \sigma_2^2 = \sigma^2$. The combined variance $\sigma^2_{comb}$ of $n_1 + n_2$ observations is given by: $\sigma^2_{comb} = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2 (\bar{x}_1 - \bar{x}_2)^2}{(n_1 + n_2)^2}$ Substituting the given values: $13 = \frac{15(14) + 15(\sigma^2)}{15 + 15} + \frac{15 	imes 15 (12 - 14)^2}{(15 + 15)^2}$ $13 = \frac{15(14 + \sigma^2)}{30} + \frac{225(-2)^2}{900}$ $13 = \frac{14 + \sigma^2}{2} + \frac{900}{900}$ $13 = \frac{14 + \sigma^2}{2} + 1$ $12 = \frac{14 + \sigma^2}{2}$ $24 = 14 + \sigma^2$ $\sigma^2 = 10$

The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^2$ respectively. If the variance of all the $30$ numbers in the two sets is $13$,then $\sigma^2$ is equal to $.........$.

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