Cumulative distribution function integration

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May 27th 2011, 07:58 AM
isx99

Cumulative distribution function integration

The Question:
The continuous random variable X has probability density function given by:
$f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1$
$\ \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3$

$\ 0\ otherwise$

a) Sketch the graph of f.
b) Explain why the value of n, the median of X, is 1.
c) Show that the value of u, the mean of X, is $\frac{13}{12}$ .
d) Find P(X < 3u - n).

I've done a), b), and c), but am stuck on d).

My Attempt:
F(x) (the cumulative distribution function for X) is
$\int{\frac{3\ -\ x}{4}}\ from\ 1\ to\ x$ , which becomes:
$[\frac{1}{4}(3x\ -\ x^2)]$ with limits 1 and x.

Substituting:
$\frac{3x\ -\ x^2}{4}\ -\ \frac{5}{8}$ .

Now 3u - n = $\frac{9}{4}$ .

Hence $F(\frac{9}{4})$ =:

$\frac{3(\frac{9}{4})\ -\ \(\frac{9}{4})^2}{4}\ - \frac{5}{8}$ .

This works out to $\frac{55}{128}$ .

However, the actual answer is $\frac{119}{128}$ , and the cumulative distribution function F(x) is $1\ -\ \frac{1}{8}(3\ -\ x)^2$ .

Some advice on how to proceed would be appreciated.

isx99
May 29th 2011, 12:29 AM
CaptainBlack

Quote:

Originally Posted by isx99

The Question:
The continuous random variable X has probability density function given by:
$f(x)\ =\ \frac{1}{2}\ for\ x\ between\ 0\ and\ 1$
$\ \frac{3\ -\ x}{4}\ for\ x\ between\ 1\ and\ 3$

$\ 0\ otherwise$

a) Sketch the graph of f.
b) Explain why the value of n, the median of X, is 1.
c) Show that the value of u, the mean of X, is $\frac{13}{12}$ .
d) Find P(X < 3u - n).

$3u-n=2.25$

$P(X< 3u-n)=P(X<2.25)=\int_0^{2.25} f(x) \; dx$

Now as $1<2.25<3$ the integral may be written:

$\begin{array}{lll}P(X<2.25)&=&{\displaystyle \int_0^{1} f(x) \; dx+\int_1^{2.25} f(x) \; dx} \\ \\ &=&{\displaystyle \frac{1}{2}+\int_1^{2.25}\frac{3-x}{4}\;dx} \end{array}$

CB
May 29th 2011, 06:15 AM
isx99

Thank you for your reply

Thank you for your reply. This is a much simpler method than the one I was using, and it is also much easier to understand.

isx99