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:
Show Hint
- 36
- 40
- 32
- 38
The Correct Option is D
Solution and Explanation
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:
- 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.
- 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: \( 12 \).
- 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: \( 20 \).
- 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):
- \(\sigma_C^2 = \frac{60 + 100}{10} + \frac{5 \cdot 16 + 5 \cdot 16}{10}\)
- \(\sigma_C^2 = 16 + 16 = 32\)
- Calculate the sum of mean and variance of set \( C \):
- Conclusion: The correct answer is 38.
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