JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 5)
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 :
12
11
10
9
Explanation
We know that if $n_1, n_2$ are the sizes, $\bar{X}_1, \bar{X}_2$ are the means and $\sigma_1, \sigma_2$ are the standard deviation of the series, then the combine variance of the series.
$$ \begin{array}{ll} & \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}\left(\bar{X}_1-\bar{X}_2\right)^2 \\\\ &\Rightarrow 13=\frac{15 \times 14+15 \times \sigma^2}{15+15}+\frac{15 \times 15}{(15+15)^2}(12-14)^2 \\\\ &\Rightarrow 13=\frac{14+\sigma^2}{2}+\frac{1}{4} \times 4 \\\\ &\Rightarrow 14+\sigma^2=2 \times 12 \\\\ &\Rightarrow \sigma^2=10 \end{array} $$
$$ \begin{array}{ll} & \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}\left(\bar{X}_1-\bar{X}_2\right)^2 \\\\ &\Rightarrow 13=\frac{15 \times 14+15 \times \sigma^2}{15+15}+\frac{15 \times 15}{(15+15)^2}(12-14)^2 \\\\ &\Rightarrow 13=\frac{14+\sigma^2}{2}+\frac{1}{4} \times 4 \\\\ &\Rightarrow 14+\sigma^2=2 \times 12 \\\\ &\Rightarrow \sigma^2=10 \end{array} $$
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