Cumulative distribution function

drawar
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Homework Statement


The continuous random variable X has cumulative distribution function given by
[tex]F(x) = \left\{ {\begin{array}{*{20}c}<br /> 0 & {x \le 0} \\<br /> {\frac{{x^2 }}{k}} & {0 \le x \le 1} \\<br /> { - \frac{{x^2 }}{6} + x - \frac{1}{2}} & {1 \le x \le 3} \\<br /> 1 & {x \ge 3} \\<br /> \end{array}} \right.[/tex]
(i) Find the value of k
(ii) Find the probability density function of X and sketch its graph
(iii) Find the median of [tex]\sqrt X[/tex]
(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 [tex]f(x) = \left\{ {\begin{array}{*{20}c}<br /> {\frac{{2x}}{3}} & {0 \le x \le 1} \\<br /> { - \frac{x}{3} + 1} & {1 \le x \le 3} \\<br /> 0 & {otherwise} \\<br /> \end{array}} \right.[/tex]
However, I have no idea how to do the rest. Any feedback is appreciated, thanks
 
Last edited:
on Phys.org
For (iii), the median is the value M such that the random variable is equally likely to be below or above M:

[tex]\int_0^M f(x) dx = 0.5[/tex]
(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"?
 
Thank you for pointing that out to me. Is the median of X equal to that of [tex]\sqrt X[/tex]?
 
Oh, sorry, I overlooked the [itex]\sqrt{x}[/itex] 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

[tex]P(\sqrt{X} < M) = 0.5[/tex]

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

So

[tex]P(\sqrt{X} < M) = P(X < M^2)[/tex]

So what is [itex]P(X < M^2)[/itex] in terms of an integral expression involving f(x)?
 
That definitely helps. Thanks!
 

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