Thread: Continuous Random Variable question. Could you help?
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Old June 6th, 2009, 04:45 AM
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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\{ \begin{array}{ll} 0, & x < 1 \\ & \\ \int_1^x \frac{6}{5} t (t - 1) \, dt = \frac{ax^3 - bx^2 + c}{5}, & 1 \leq x \leq 2 \\ & \\ 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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