I've been trying to solve the following problem for days, but I haven't found anything similar to that on the internet:
So, we have a random variable $X$ that follows the Normal distribution with mean $\mu=60$ and standard deviation $\sigma=4$. Also, we have a density function $f(X) = Y = a_1X + a_0$, which in this case represents a cost, so $\min f(X) = 0$. We are looking for the mean $\Bbb E[f(X)]$.
We know that $$\Bbb E[f(X)] = \int_{\min f(X)}^{\max f(X)} xf(x)\,dx.$$ The problem is we don't know the maximum value that $f$ can take. However, we know that the variable $X$ follows the Normal distribution. Would it be correct to assume that the maximum value of $X$ is $\mu+3\sigma$?
If that's the case, then could we also assume that $$\max f(X) = a_1\max X + a_0 = a_1(\mu+3\sigma) + a_0 = 72a_1 + a_0?$$
Thanks in advance.
Best Answer
Note $\Bbb E[f(X)]=\Bbb E[a_1X+a_0]=a_1\Bbb E[X]+a_0=60a_1+a_0$ due to linearity of $\Bbb E$, and also that $f$ has no minimum or maximum as neither does $X$.