Question

Let sets A and B have 5 elements each. Let 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 A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is:

• A. 36
• B. 40
• C. 32
• D. 38

Hint:

To find the mean and variance of a new set formed by modifying elements from other sets, remember to apply the changes (such as adding or subtracting) to both the mean and variance accordingly.

Answer 1

The Correct Option is D

Let sets A and B be as follows: \[ A = \{a_1, a_2, a_3, a_4, a_5\}, \quad B = \{b_1, b_2, b_3, b_4, b_5\} \] Given: \[ \text{Mean of A} = 5, \quad \text{Mean of B} = 8 \] and the variances: \[ \text{Variance of A} = 12, \quad \text{Variance of B} = 20 \] First, calculate the sum of the elements of sets A and B: \[ \sum_{i=1}^{5} a_i = 5 \times 5 = 25, \quad \sum_{i=1}^{5} b_i = 5 \times 8 = 40 \] Next, calculate the sum of squares for A and B using the formula for variance: \[ \text{Variance of A} = \frac{\sum_{i=1}^{5} a_i^2}{5} - \left( \frac{\sum_{i=1}^{5} a_i}{5} \right)^2 \] \[ 12 = \frac{\sum_{i=1}^{5} a_i^2}{5} - 5^2 \quad \Rightarrow \quad \sum_{i=1}^{5} a_i^2 = 185 \] Similarly, for set B: \[ \text{Variance of B} = \frac{\sum_{i=1}^{5} b_i^2}{5} - \left( \frac{\sum_{i=1}^{5} b_i}{5} \right)^2 \] \[ 20 = \frac{\sum_{i=1}^{5} b_i^2}{5} - 8^2 \quad \Rightarrow \quad \sum_{i=1}^{5} b_i^2 = 420 \] Now, set C is formed by subtracting 3 from each element of set A and adding 2 to each element of set B: \[ C = \{c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8, c_9, c_{10}\} \] where: \[ c_1 = a_1 - 3, \quad c_2 = a_2 - 3, \quad \dots, \quad c_5 = a_5 - 3 \] and \[ c_6 = b_1 + 2, \quad c_7 = b_2 + 2, \quad \dots, \quad c_{10} = b_5 + 2 \] Now, we calculate the mean of C: \[ \text{Mean of C} = \frac{1}{10} \left( \sum_{i=1}^{5} (a_i - 3) + \sum_{i=1}^{5} (b_i + 2) \right) \] \[ = \frac{1}{10} \left( \sum_{i=1}^{5} a_i - 15 + \sum_{i=1}^{5} b_i + 10 \right) \] \[ = \frac{1}{10} \left( 25 - 15 + 40 + 10 \right) = \frac{1}{10} \times 60 = 6 \] Next, calculate the variance of C: \[ \text{Variance of C} = \frac{1}{10} \left( \sum_{i=1}^{5} (a_i - 3)^2 + \sum_{i=1}^{5} (b_i + 2)^2 \right) \] Using the identity \( (x - 3)^2 = x^2 - 6x + 9 \) and similarly for \( (x + 2)^2 \): \[ \text{Variance of C} = \frac{1}{10} \left( \sum_{i=1}^{5} a_i^2 - 6\sum_{i=1}^{5} a_i + 45 + \sum_{i=1}^{5} b_i^2 + 4\sum_{i=1}^{5} b_i + 20 \right) \] \[ = \frac{1}{10} \left( 185 - 6 \times 25 + 45 + 420 + 4 \times 40 + 20 \right) \] \[ = \frac{1}{10} \left( 185 - 150 + 45 + 420 + 160 + 20 \right) \] \[ = \frac{1}{10} \times 680 = 68 \] Thus, the sum of the mean and variance of C is: \[ 6 + 68 = 38 \]