JEE Main 2023 — Statistics Question with Solution

To solve this problem, let's break it down step by step :

Step 1 : Determine the mean of set C

The mean of set A = $5$

The mean of set B = $8$

After subtracting 3 from each element in set A, the new mean becomes $5-3=2$. After adding 2 to each element in set B, the new mean becomes $8+2=10$.

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:

Mean of C = $\frac{5\times 2+5\times 10}{10}=\frac{10+50}{10}=6$

Step 2 : Determine the variance of set C

Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one :

Variance of C

= $\frac{n_1\times \text{Variance of A}+n_1\times (\text{Mean of modified A}-\text{Mean of C}{)}^2+n_2\times \text{Variance of B}+n_2\times (\text{Mean of modified B}-\text{Mean of C}{)}^2}{n_1+n_2}$

Given :

$n_1=n_2=5$

Variance of A = $12$, Variance of B = $20$

Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$

Plugging in the values, we get :

Variance of C = $\frac{5\times 12+5\times (2-6{)}^2+5\times 20+5\times (10-6{)}^2}{10}$

Variance of C = $\frac{60+80+100+80}{10}=\frac{320}{10}=32$

Step 3 : Sum of the mean and variance of set C

Sum = Mean of C + Variance of C = $6+32=38$

So, the correct answer is :

Option C : 38.