Why is $\operatorname{Var}(x) = μ^2$ in this problem?

Question

This is the question: "Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3. What is the variance of the total lifetime of two Type A bulbs and one Type B bulb? "

This is the answer: $$\operatorname{Var}(2X+Y)=2^2 \operatorname{Var}(X)+\operatorname{Var}(Y)=4×4+9=25$$

I understand $\operatorname{Var}(2x)=2^2$, Var(x), but why are $\operatorname{Var}(X) = 4$ and $\operatorname{Var}(Y)=9$? I also know that $\operatorname{Var}(x) = σ^2$. However, I also observed that $\operatorname{Var}(x) = μ^2$ in this case.

Answer 1

Your definition for $Var(x) = σ^2$ is taken from the normal distribution. Note, however, that in this question $X$ and $Y$ are stated to be exponential distributions with means of $2$ and $3$ respectively.

For an exponential distribution, the mean is equal to $β$ and the variance is equal to $β^2$. That's the reason you observe that $Var(x)$ equals the square of the mean in this case.

For more information on exponential distribution: https://en.wikipedia.org/wiki/Exponential_distribution

Answer 2

Note that if $X\sim \mathcal{E}xp(\lambda)$, thus $\mathbb{E}[X] = 1/\lambda$. Hence, if $A$'s mean life time is $2$, that is $\mathbb{E}[X] = 2 = 1/\lambda \to \lambda =1/2.$ As such, $\operatorname{var}(X) = 1/\lambda^2 = 1/(1/2)^2 = 4$. Same is true for type $B$.