3) The length of time to failure (in hundreds of hours) for a transistor is a random variable Y with distribution function give by F(y) = 0, y < 0 1 − e −y 2 , y ≥ 0 a) Show that F(y) has the properties of a distribution function. b) Find f(y). c) Find the probability that the transistor operates at least 200 hours.

[tex]F(y)\to0[/tex] as [tex]y\to-\infty[/tex];
[tex]F(y)\to1[/tex] as [tex]y\to+\infty[/tex];
[tex]F(y)[/tex] is non-decreasing; and
[tex]F(y)[/tex] is "right-continuous", meaning that the limit of any "piece" of [tex]F(y)[/tex] as [tex]y\to c[/tex] from the right exists.

The first two conditions are met: we have

[tex]\displaystyle\lim_{y\to-\infty}F(y)=0[/tex]

[tex]\displaystyle\lim_{y\to+\infty}F(y)=\lim_{y\to+\infty}(1-e^{-y^2})=1[/tex]

The second condition is also met:

[tex]F'(y)=\begin{cases}0&\text{for }y<0\\2ye^{-y^2}&\text{for }y>0\end{cases}[/tex]

both of which are non-negative over their respective domains.

Finally, the fourth condition is met because [tex]F(y)[/tex] is continuous everywhere.

b. We already found [tex]f(y)[/tex] above when checking for the monotonicity of [tex]F(y)[/tex]; the probability density [tex]f(y)[/tex] is simply the derivative of the distribution function [tex]F(y)[/tex].

[tex]f(y)=F'(y)=\begin{cases}0&\text{for }y<0\\2ye^{-y^2}&\text{for }y\ge0\end{cases}[/tex]

(Notice that we include [tex]y=0[/tex] so that the [tex]P(Y=0)[/tex] exists.)

c. By definition of the distribution function, we have

[tex]P(Y\ge200)=1-P(Y<200)=1-F(200)=e^{-40,000}[/tex]

which is value very close to 0.