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

To solve this problem, we need to determine the sum of the mean and variance of a new set \( C \) formed from sets \( A \) and \( B \). The steps involved are as follows:

1. Calculate the mean and variance of the new set \( C \):
   • Set \( A \): Mean is 5, Variance is 12.
   • Set \( B \): Mean is 8, Variance is 20.
2. Transform set \( A \):
   • Each element in set \( A \) is decreased by 3. This affects the mean but not the variance.
   • New mean of set \( A \): mean = \( 5 - 3 = 2 \).
   • Variance remains unchanged: \( 12 \).
3. Transform set \( B \):
   • Each element in set \( B \) is increased by 2. This affects the mean but not the variance.
   • New mean of set \( B \): mean = \( 8 + 2 = 10 \).
   • Variance remains unchanged: \( 20 \).
4. Determine the combined set \( C \):
   • Set \( C \) contains 5 elements from transformed set \( A \) and 5 from transformed set \( B \), totalling to 10 elements.
   • Mean of set \( C \) is the weighted average:
   • Variance of set \( C \) (using property of variance with combined sets of equal sizes): 

\[\sigma_C^2 = \frac{5 \cdot 12 + 5 \cdot 20}{10} + \frac{5 \cdot (2 - 6)^2 + 5 \cdot (10 - 6)^2}{10}\]

• \(\sigma_C^2 = \frac{60 + 100}{10} + \frac{5 \cdot 16 + 5 \cdot 16}{10}\)
• \(\sigma_C^2 = 16 + 16 = 32\)

1. Calculate the sum of mean and variance of set \( C \): 

\[\text{Sum} = 6 + 32 = 38\]

1. Conclusion: The correct answer is 38.