# Data Management - Prob. Help

• Dec 31st 2008, 02:04 PM
Faisal2007
Data Management - Prob. Help
A computer is programmed to draw rectangles with perimeter or 24 cm. the program randomly chooses integer lengths. what is the expected area of a rectangle.

i really dont get this one...

It would be great if you could point me to the correct direction...

Not begging for the answer only understanding.
• Dec 31st 2008, 02:30 PM
nzmathman
Let the width of the rectangle be x, and the length be y.

Then $2x + 2y = 24$
$x + y = 12$
$y = 12 - x$
And the area $= xy = x(12 - x)$

x can be all integers between and including 1 and 11.
All the outcomes below,

x=1, area=11
x=2, area=20
x=3, area=27
x=4, area=32
x=5, area=35
x=6, area=36
x=7, area=35
x=8, area=32
x=9, area=27
x=10, area=20
x=11, area=11

Now, use $E(X) = n \times p$ to find the solution.
• Dec 31st 2008, 02:49 PM
Faisal2007
We never learnt it using that method...

Is this a binomial distibution ?

I think it goes with P(X) nCx . p^x . q^n-x
p = success
q = fail
n = total trials
x = successes from total trials
• Dec 31st 2008, 03:26 PM
nzmathman
I don't see how this is binomial distribution. The computer has RANDOMLY chosen the x,y values and so each value of X has equal probability.

It is a simple distribution like this:

Area=11, p=2/11
Area=20, p=2/11
Area=27, p=2/11
Area=32, p=2/11
Area=35, p=2/11
Area=36, p=1/11

This allows the expected area to be calculated relatively simply.
• Dec 31st 2008, 03:30 PM
Faisal2007
I am lost...I hate questions like these.
I am actually having trouble in *approaching* the question...
• Dec 31st 2008, 04:11 PM
nzmathman
But do you understand the method to find the answer?
• Dec 31st 2008, 04:31 PM
mr fantastic
Quote:

Originally Posted by nzmathman
But do you understand the method to find the answer?

The more interesting problem is when x is a continuous random variable .... (of course, in practice x is always discrete).