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Cumulative distribution function

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    The continuous random variable X has cumulative distribution function given by
    [tex]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.[/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)

    2. Relevant equations



    3. 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}
    {\frac{{2x}}{3}} & {0 \le x \le 1} \\
    { - \frac{x}{3} + 1} & {1 \le x \le 3} \\
    0 & {otherwise} \\
    \end{array}} \right.[/tex]
    However, I have no idea how to do the rest. Any feedback is appreciated, thanks
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2

    jbunniii

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    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"?
     
  4. Feb 28, 2012 #3
    Thank you for pointing that out to me. Is the median of X equal to that of [tex]\sqrt X [/tex]?
     
  5. Feb 28, 2012 #4

    jbunniii

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    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)?
     
  6. Mar 1, 2012 #5
    That definitely helps. Thanks!
     
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