Math Help Forum - View Single Post - Continuous Random Variable question. Could you help?

db5vry · #3 June 6th, 2009, 05:01 AM

Quote:

Originally Posted by mr fantastic

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.

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 :]