# Continuous Random Variable question. Could you help?

• June 6th 2009, 02:24 AM
db5vry
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 http://thestudentroom.co.uk/latexren...7d2b498705.png, so E(1/X) was done by http://thestudentroom.co.uk/latexren...685e79c8ed.png which is 10/17, or 0.58 or something like that. Is this right?

For part b] I had http://thestudentroom.co.uk/latexren...a4d4d10c27.png.

For c] if P(X http://thestudentroom.co.uk/latexren...aca311c641.png 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 :)
• June 6th 2009, 02:45 AM
mr fantastic
Quote:

Originally Posted by db5vry
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 http://thestudentroom.co.uk/latexren...7d2b498705.png, so E(1/X) was done by http://thestudentroom.co.uk/latexren...685e79c8ed.png 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 http://thestudentroom.co.uk/latexren...a4d4d10c27.png. Mr F says: Wrong. If this was correct then it should equal 1 when x = 2. It doesn't.

For c] if P(X http://thestudentroom.co.uk/latexren...aca311c641.png 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.
• June 6th 2009, 03:01 AM
db5vry
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 :]
• June 6th 2009, 03:06 AM
mr fantastic
Quote:

Originally Posted by db5vry
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.
• June 6th 2009, 04:29 AM
db5vry
Quote:

Originally Posted by mr fantastic
$\frac{6}{5} \left[\frac{x^2}{2} - x \right]_1^2 \neq 3$. 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 :)
• June 6th 2009, 04:00 PM
mr fantastic
Quote:

Originally Posted by db5vry
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 ....