Cumulative distribution function

[drawar]

Homework Statement

The continuous random variable X has cumulative distribution function given by
F(x) = \left\{ {\begin{array}{*{20}c}<br /> 0 &amp; {x \le 0} \\<br /> {\frac{{x^2 }}{k}} &amp; {0 \le x \le 1} \\<br /> { - \frac{{x^2 }}{6} + x - \frac{1}{2}} &amp; {1 \le x \le 3} \\<br /> 1 &amp; {x \ge 3} \\<br /> \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}<br /> {\frac{{2x}}{3}} &amp; {0 \le x \le 1} \\<br /> { - \frac{x}{3} + 1} &amp; {1 \le x \le 3} \\<br /> 0 &amp; {otherwise} \\<br /> \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} &lt; M) = 0.5

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

So

P(\sqrt{X} &lt; M) = P(X &lt; M^2)

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

 

[drawar]

That definitely helps. Thanks!