Consider two independent random variables: $X ∼ N(5, 4)$ and $Y~N(3, 2)$. If $(2X + 3Y)~N(\mu, \sigma ^2)$, then the values of mean ($\mu$) and variance ($\sigma ^2$) are
- $\mu$ = 19 and $\sigma ^2$ = 34
- $\mu$ = 8 and $\sigma ^2$ = 14
- $\mu$ = 19 and $\sigma ^2$ = 14
- $\mu$ = 8 and $\sigma ^2$ = 34
The Correct Option is A
Solution and Explanation
Mean and Variance of a Linear Combination of Independent Random Variables
For two independent random variables X and Y, the mean and variance of the linear combination 2X + 3Y are calculated as follows:
Step 1: Calculating the Mean
The expected value (mean) follows the linearity property:
E(2X + 3Y) = 2E(X) + 3E(Y)
Given:
- E(X) = 5
- E(Y) = 3
Substituting values:
E(2X + 3Y) = 2 × 5 + 3 × 3 = 10 + 9 = 19
Step 2: Calculating the Variance
For independent random variables, the variance of a linear combination is given by:
Var(aX + bY) = a² Var(X) + b² Var(Y)
Given:
- Var(X) = 4
- Var(Y) = 2
Applying the formula:
Var(2X + 3Y) = 2² × Var(X) + 3² × Var(Y)
= 4 × 4 + 9 × 2
= 16 + 18 = 34
Final Answer:
- Mean (E(2X + 3Y)) = 19
- Variance (Var(2X + 3Y)) = 34
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