Question

A random variable X has probability density function given by
f(x) = 2/3 x, 1 ≤ x ≤ 2,
0, otherwise.
(i) Find E(X)
(ii) Find P(X < E(X))
(iii) Hence explain whether the mean of X is less than, equal to or greater than the median of X.

Workings and explanation please

Answer 1

if i recall correctly
\[E(x)=\int_{-\infty}^{\infty}xf(x)dx\] which in your case would be
\[\int_1^2 \frac{2}{3}x^2dx\]
get a second opinion though

Answer 2

Yes bingo. I actually have difficulties in the last question.

Answer 3

Refer to satellite's integral for \(E(X)\). I'll call the result \(k\).

Second part wants the probability that \(X\) is less than the expectation
\[P(X<k)=F(k)=\int_{-\infty}^kf(x)~dx=\int_1^k\frac{2}{3}x~dx\]
where \(f\) denote the probability density function and \(F\) denotes the cumulative distribution function.

The median is the value \(c\) that makes
\[\int_{-\infty}^cf(x)~dx=\int_1^c\frac{2}{3}x~dx=\frac{1}{2}=n\]
Compare \(n\) and \(k\).

Answer 4

as for iii), in the answer scheme it says 115/243 < 1/2

Answer 5

What are you getting for \(E(X)=k\)?

Answer 6

14/9

Answer 7

Okay, good. What's the value of
\[\int_1^{14/9}\frac{2}{3}x~dx~~?\]

Answer 8

115/243

Answer 9

I didnt have any problem in i) and ii). I just need some clarification in iii). why is it 115/243<1/2 and not 14/9<1/2

Answer 10

Right again. So you now have to find \(c\)... I just noticed that I included \(n\) for some reason... Ignore that.

For (iii), you want to compare the values of mean \(k\) and the median \(c\). I don't see how saying \(\dfrac{115}{243}<\dfrac{1}{2}\) tells you anything about what you want to show...

Here's how you compute the median:
\[\int_1^c\frac{2}{3}x~dx=\frac{1}{2}~~\iff~~\left[\frac{x^2}{3}\right]_1^c=\frac{1}{2}~~\iff~~\frac{c^2}{3}-\frac{1^2}{3}=\frac{1}{2}\]
which gives \(c=\sqrt{\dfrac{5}{2}}\approx1.581\) as the median. \(k=\dfrac{14}{9}\approx1.555\), so clearly the median is greater than the mean.

Answer 11

I dont see it also. Anyways, thanks. Ill get back with my tutor. :)