Let X be a random variable uniformly distributed on [0,1] and let Y := [6X] + 1. Determine the distribution of Y and compute E(Y).
edit: It was pointed out that I neglected to account for the integer function part of the problem. Please fact check my work as if the problem stated $Y$ as $Y = 6X + 1$.
My attempt at solving:
$F_Y(y) = P(Y\le y) = P(X\le (y-1)/6) = 1$ if $y \gt 7, y/6 - 1/6$ if $y \in [1,7], 0$ if $y\lt 1$. Thus, $f_Y(y) = F_X'(y/6 - 1/6)(y/6 - 1/6)' = (1/6)*F_X(y/6-1/6) = 1/6 * 1/6$ if $y\in[1,7]$ and $0$ otherwise. But this is not a density.
I think that probably the correct answer is that Y~Uniform[1,7], but I would like to know where in my reasoning I went astray.
For $E[Y], E[Y] = E[6X + 1] = 6E[X] + 1 = 6(.5) + 1 = 4.$
Would like to know if my answer is correct and if not, why. Thank you