A company manufactures poles. the length of a pole is a random variable X, with mean $\mu_X$ and probability density function f(x).
Poles are cut to obtain an exact length L. If the initial length of the pole is less than L, the entire pole is lost. If it is greater than L, the pole will be cut down to L, and the section left over is lost. We are interested in the random variable Y, defined as the length of each piece lost.
the first part of the question was to:
1) Express E(Y) as a function of f(x) and $\mu_X$. which I did as followed:
E(Y) = $\int$g(x)f(x) = $\int_0^L$xf(x) + $\int_L^\infty$(x-L)f(x)
but I think I may done it wrong because of I didn't really used $\mu_X$ and also because of the next part that I don't know how to answer and probably relays on the previous one:
2)Suppose that X follows a normal distribution with mean $\mu_X$ and variance $σ_X^2$. Show that there exists a value $μ$ of $μ_X$ that minimizes $μ_Y$.
am I missing something in the first part? and what is the logic to answer the second part?
Your work is correct (and indeed you do not need $\mu_X$), but can be written in another way: $$\int_0^L x f(x) \, dx + \int_L^\infty (x-L) f(x) \, dx = \int_0^L x f(x) \, dx - L \int_L^\infty f(x) \, dx = \mu_X - L\int_L^\infty f(x) \, dx.$$