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Math Help - Continuous Random Variable question. Could you help?

  1. #1
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    Continuous Random Variable question. Could you help?

    The continuous random variable X has probability density function f given by
    f(x) = \frac{6}{5}x (x - 1), for 1 \leq x \leq 2
    f(x) = 0, otherwise.
    a] Evaluate E \frac{1}{X}.
    b] Find an expression for the cumulative distribution function.
    c] Evaluate P(X \leq 1.75).
    d] State, with a reason, whether the median of X is greater or less than 1.75.

    For part a] I calculated E(X) which I found to be , so E(1/X) was done by which is 10/17, or 0.58 or something like that. Is this right?

    For part b] I had .

    For c] if P(X 1.75) is F(1.75), then that should be 0.30625 if part b] is correct,

    and to answer d] you can say that the median of X must be greater than 0.5 because the value of 1.75 is not greater than 0.5.

    Am I right or did something go wrong?
    Thanks for any help
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  2. #2
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    Quote Originally Posted by db5vry View Post
    The continuous random variable X has probability density function f given by
    f(x) = \frac{6}{5}x (x - 1), for 1 \leq x \leq 2
    f(x) = 0, otherwise.
    a] Evaluate E \frac{1}{X}.
    b] Find an expression for the cumulative distribution function.
    c] Evaluate P(X \leq 1.75).
    d] State, with a reason, whether the median of X is greater or less than 1.75.

    For part a] I calculated E(X) which I found to be , so E(1/X) was done by which is 10/17, or 0.58 or something like that. Is this right? Mr F says: No! {\color{red} E \left( \frac{1}{X} \right) \neq \frac{1}{E(X)}}.

    For part b] I had . Mr F says: Wrong. If this was correct then it should equal 1 when x = 2. It doesn't.

    For c] if P(X 1.75) is F(1.75), then that should be 0.30625 if part b] is correct,

    and to answer d] you can say that the median of X must be greater than 0.5 because the value of 1.75 is not greater than 0.5.

    Am I right or did something go wrong?
    Thanks for any help
    a) E \left( \frac{1}{X}\right) = \int_1^2 \frac{1}{x} \cdot \frac{6}{5} x (x - 1) \, dx.


    b) F(x) = \left\{<br />
\begin{array}{ll}<br />
0, & x < 1 \\<br />
& \\<br />
\int_1^x \frac{6}{5} t (t - 1) \, dt = \frac{ax^3 - bx^2 + c}{5}, & 1 \leq x \leq 2 \\<br />
& \\<br />
1, & x > 2 \end{array} \right.

    where I'll let you calculate the values of a, b and c for yourself. Note: The first and third lines of this rule for F(x) are just as important as the second line, by the way.

    c) Substitute x = 1.75 into the correct rule.

    d) You will find that F(1.75) > 0.5.
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    Quote Originally Posted by mr fantastic View Post
    a) E \left( \frac{1}{X}\right) = \int_1^2 \frac{1}{x} \cdot \frac{6}{5} x (x - 1) \, dx.


    b) F(x) = \left\{<br />
\begin{array}{ll}<br />
0, & x < 1 \\<br />
& \\<br />
\int_1^x \frac{6}{5} t (t - 1) \, dt = \frac{ax^3 - bx^2 + c}{5}, & 1 \leq x \leq 2 \\<br />
& \\<br />
1, & x > 2 \end{array} \right.

    where I'll let you calculate the values of a, b and c for yourself. Note: The first and third lines of this rule for F(x) are just as important as the second line, by the way.

    c) Substitute x = 1.75 into the correct rule.

    d) You will find that F(1.75) > 0.5.
    part a] confuses me in terms of the multiplication. When I have
    \frac{1}{X} \frac{6}{5}x^2 - \frac{6}{5}x dx
    I canceled it down to \frac{6}{5}x - \frac{6}{5}
    And integrated it to \frac{\frac{6}{5}x^2}{2} - \frac{6}{5}x between the limits of 2 and 1
    So I ended up with 2\frac{2}{5} - - \frac{3}{5} and got 3 as the value.
    It doesn't seem right. Is this really how it is done?
    Thanks for helping :]
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  4. #4
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    Quote Originally Posted by db5vry View Post
    part a] confuses me in terms of the multiplication. When I have
    \frac{1}{X} \frac{6}{5}x^2 - \frac{6}{5}x dx
    I canceled it down to \frac{6}{5}x - \frac{6}{5}
    And integrated it to \frac{\frac{6}{5}x^2}{2} - \frac{6}{5}x between the limits of 2 and 1
    So I ended up with 2\frac{2}{5} - - \frac{3}{5} and got 3 as the value.
    It doesn't seem right. Is this really how it is done?
    Thanks for helping :]
    \frac{6}{5} \left[\frac{x^2}{2} - x \right]_1^2 \neq 3. Check your arithmetic.
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    Quote Originally Posted by mr fantastic View Post
    \frac{6}{5} \left[\frac{x^2}{2} - x \right]_1^2 \neq 3. Check your arithmetic.
    So you take the \frac{6}{5} out as a constant! If I do that then I get 0 - - \frac{3}{5} and an answer of \frac{3}{5}. Is this correct this time?
    I will get there in the end. Thanks for pointing this out
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  6. #6
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    Quote Originally Posted by db5vry View Post
    So you take the \frac{6}{5} out as a constant! If I do that then I get 0 - - \frac{3}{5} and an answer of \frac{3}{5}. Is this correct this time?
    I will get there in the end. Thanks for pointing this out
    If you take the 6/5 out the arithmetic is simpler. But the answer is no different to if you don't take it out ....

    3/5 is the corerct answer.
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