# Thread: Data Management - Prob. Help

1. ## 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.

2. Let the width of the rectangle be x, and the length be y.

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

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 $\displaystyle E(X) = n \times p$ to find the solution.

3. 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

4. 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.

5. I am lost...I hate questions like these.
I am actually having trouble in *approaching* the question...

6. But do you understand the method to find the answer?

7. 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).

### 11 rectangles of 24 cm perimeter algebra forum

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