Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of $\mathrm{A}$ and adding 2 to each element of $\mathrm{B}$. Then the sum of the mean and variance of the elements of $\mathrm{C}$ is ___________.

<p>To solve this problem, let&#39;s break it down step by step :</p> <p><strong>Step 1 :</strong> Determine the mean of set C</p> <p>The mean of set A = $5$ <br/><br/>The mean of set B = $8$</p> <p>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 $8 + 2 = 10$.</p> <p>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:</p> <p>Mean of C = $\frac{5 \times 2 + 5 \times 10}{10} = \frac{10 + 50}{10} = 6$</p> <p><strong>Step 2 :</strong> Determine the variance of set C</p> <p>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.</p> <p>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.</p> <p>Now, when combining variances from two datasets into one : <br/><br/>Variance of C <br/><br/>= $\frac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$</p> <p>Given : <br/><br/>$n_1 = n_2 = 5$ <br/><br/>Variance of A = $12$, Variance of B = $20$ <br/><br/>Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$</p> <p>Plugging in the values, we get : <br/><br/>Variance of C = $\frac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$ <br/><br/>Variance of C = $\frac{60 + 80 + 100 + 80}{10} = \frac{320}{10} = 32$</p> <p><strong>Step 3 :</strong> Sum of the mean and variance of set C <br/><br/>Sum = Mean of C + Variance of C = $6 + 32 = 38$</p> <p>So, the correct answer is : <br/><br/>Option C : 38.</p>