Cumulative distribution function

[drawar]

Homework Statement

The continuous random variable X has cumulative distribution function given by
$$F(x) = \left\{ {\begin{array}{*{20}c} 0 & {x \le 0} \\ {\frac{{x^2 }}{k}} & {0 \le x \le 1} \\ { - \frac{{x^2 }}{6} + x - \frac{1}{2}} & {1 \le x \le 3} \\ 1 & {x \ge 3} \\ \end{array}} \right.$$
(i) Find the value of k
(ii) Find the probability density function of X and sketch its graph
(iii) Find the median of $$\sqrt X$$
(iv) 10 independent observations of X are taken. Find the probability that eight of them are less than 2.
(v) Let A be the event X > 1 and B be the event X > 2. Find P(B|A)

Homework Equations

The Attempt at a Solution

I'm able to do the first 2 questions
For (i), by substitution I get k=3
For (ii), I take the derivative of F(x), then $$f(x) = \left\{ {\begin{array}{*{20}c} {\frac{{2x}}{3}} & {0 \le x \le 1} \\ { - \frac{x}{3} + 1} & {1 \le x \le 3} \\ 0 & {otherwise} \\ \end{array}} \right.$$
However, I have no idea how to do the rest. Any feedback is appreciated, thanks

 

[jbunniii]

For (iii), the median is the value M such that the random variable is equally likely to be below or above M:

$$\int_0^M f(x) dx = 0.5$$
(Solve for M.)

For (iv), start by finding the probability that a single observation is less than 2.

For (v), start by writing down the definition of P(B|A).

P.S. Is this really "precalculus mathematics"?

 

[drawar]

Thank you for pointing that out to me. Is the median of X equal to that of $$\sqrt X$$?

 

[jbunniii]

Oh, sorry, I overlooked the $\sqrt{x}$ in the question. No, in general, it won't be the same as the median of X.

You are looking for the value of M such that

$$P(\sqrt{X} < M) = 0.5$$

Now, since X is non-negative, it follows that $\sqrt{X} < M$ if and only if $X < M^2$.

So

$$P(\sqrt{X} < M) = P(X < M^2)$$

So what is $P(X < M^2)$ in terms of an integral expression involving f(x)?

 

[drawar]

That definitely helps. Thanks!