distribution function, calculating $f(y)$ and $F(Y)$ The length of time to failure (in hundreds of hours) for a transistor is a random variable $Y$ with distribution function given by:
$F(y) = 0$ when $y <0$, and $1-e^{-y^2}$ when y greater than or equal to $0$
a.) Show that $F(y)$ has the properties of a distribution function.
$F(-\infty) = 0$
$F(\infty) = 1$
$F(Y_2) - F(Y_1) \ge 0$
b.) find $f(y)$
I just took the derivative of $1-e^{-y^2}$ which is $2e^{-y^2}$
c.) Find the probability that the transistor operates for at least $200$ hours, find $P(Y \ge 2)$ for y in $100$s of hours
$P(Y \ge 2) = 1 - F(1) = 1 -  1-e^{-1^2}$
d.) Find $P(Y > 100 | Y \le 200)$
$P(Y > 100 \cap Y \le 200) / P(Y \le 200) = ?/F(2)$
Are these correct? I am not $100$% sure how to do $P(Y > 100 \cap Y \le 200)$ for part d.
 A: Ok, so your CDF is $$
  F(y) = \begin{cases}
    1 - e^{-y^2} &\text{for $y \geq 0$} \\
    0 &\text{otherwise}
  \end{cases}
$$
A. Show that $F$ is a distribution function
Showing that $\lim_{y\to-\infty} F(y) = 0$ and $\lim_{y\to+\infty} F(y) = 1$ and that $F$ is monotone is a good start. You also need that $F$ is right-continuous, i.e. that for all $y$ you have $\lim_{\epsilon \to 0,\epsilon > 0} F(y+\epsilon) = F(y)$. In your case, $F$ is even continuous - as the combination of continuous functions - except possibly at $0$, since that's where you switch from one case to another. But since $1 - e^{-0^2} = 0$, $F$ is continuous at $0$ also.
B. The idea is sound, but your result is wrong. Check how you compute that derivative...
C. In general, you have $P(Y \geq y) = P(Y = y) + P(Y > y) = P(Y = y) + 1 - P(Y \leq y)$. Now, $P(Y = y) = F(y) - \lim_{\epsilon\to 0, \epsilon > 0} F(y-\epsilon)$. Since your distribution function is not only right-continuous, but fully continuous, that limit is always $0$, and you get $$
  P(Y \geq y) = 1 - P(Y \leq y) = 1 - F(y) \text{.}
$$
D. As André and I already pointed out in the comment, $$
  P(a < X \leq b) = F(b) - F(a) \text{.}
$$
Note that, just as in (c), you may exchange $<$ with $\leq$, and also $>$ with $\geq$ if the distribution function is fully continuous, i.e. also left-continuous and not only  right-continuous as all distribution functions must be. Your distribution function satisfies that property, but of course not all distribution functions do.
