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 :

Statistics PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q27 with Solution

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Correct answer: C

Let the two sets be:
- Set 1: $n_1=15$ , mean $\mu_1=12$ , variance $\sigma_1^2=14$
- Set 2: $n_2=15$ , mean $\mu_2=14$ , variance $\sigma_2^2=\sigma^2$
First find the mean of all $30$ numbers.

\mu=\frac{15\cdot 12+15\cdot 14}{30}=\frac{15(26)}{30}=13

\text{Combined variance}= \frac{n_1\left(\sigma_1^2+(\mu_1-\mu)^2\right)+n_2\left(\sigma_2^2+(\mu_2-\mu)^2\right)}{n_1+n_2}

Given combined variance is $13$ , so

13=\frac{15\left(14+(12-13)^2\right)+15\left(\sigma^2+(14-13)^2\right)}{30}

(12-13)^2=1, \quad (14-13)^2=1

So,

13=\frac{15(14+1)+15(\sigma^2+1)}{30}

13=\frac{15\cdot 15+15(\sigma^2+1)}{30}

13=\frac{225+15\sigma^2+15}{30}

13=\frac{240+15\sigma^2}{30}

390=240+15\sigma^2

15\sigma^2=150

\sigma^2=10

\boxed{\text{C: }10}