Question

Consider two sets A and B. Set A has 5 elements whose mean & variance are 5 and 8 respectively. Set B has also 5 elements whose mean & variance are 12 & 20 respectively. A new set C is formed by subtracting 3 from each element of set A and by adding 2 to each element of set B. The sum of mean & variance of the set C is

Answer 1

Solution:
We are given two sets, A and B, with the following properties:

- Set A: 5 elements, mean = 5, variance = 8
- Set B: 5 elements, mean = 12, variance = 20

A new set C is formed by:
- Subtracting 3 from each element of Set A.
- Adding 2 to each element of Set B.
We need to find the sum of the mean and variance of set C.

Step 1: Mean of Set C
The mean of a set is affected by constant additions or subtractions to each element of the set. Specifically:
- When you subtract a constant from each element of a set, the mean of the set decreases by that constant.
- When you add a constant to each element of a set, the mean of the set increases by that constant.

Therefore:
- For Set A, the new mean will be:
\[ \mu_C(A) = \mu_A - 3 = 5 - 3 = 2. \]
- For Set B, the new mean will be:
\[ \mu_C(B) = \mu_B + 2 = 12 + 2 = 14. \]

The mean of Set C is the sum of the new means of Set A and Set B:
\[ \mu_C = \mu_C(A) + \mu_C(B) = 2 + 14 = 16. \]

Step 2: Variance of Set C
The variance of a set is only affected by the scaling of the elements, but not by constant additions or subtractions. Since only additions or subtractions are involved here, the variance of each set will remain unchanged:
- For Set A, the variance of Set C will be the same as the variance of Set A:
\[ \sigma_C^2(A) = \sigma_A^2 = 8. \]
- For Set B, the variance of Set C will be the same as the variance of Set B:
\[ \sigma_C^2(B) = \sigma_B^2 = 20. \]

The variance of Set C is the sum of the variances of Set A and Set B:
\[ \sigma_C^2 = \sigma_C^2(A) + \sigma_C^2(B) = 8 + 20 = 28. \]

Step 3: Sum of the mean and variance of Set C
\[ \text{Sum} = \mu_C + \sigma_C^2 = 16 + 28 = 44. \]

Final Answer: 44