# Math Help - [SOLVED] Probability distribution problem :(

1. ## [SOLVED] Probability distribution problem :(

An environmental artist is planning to
construct a rectangle with 36 m of fencing as
part of an outdoor installation. If the length
of the rectangle is a randomly chosen
integral number of metres, what is the
expected area of this enclosure?

No idea how to approach this question

2. Originally Posted by ibetan
If the length
of the rectangle is a randomly chosen
integral number of metres
Can you clarify what this means.

3. Originally Posted by Anonymous1
Can you clarify what this means.
I think it means the length of the side is an integer

4. I'm going to take a quick stab at this, but hopefully you'll post the correct answer if/when you find it.

Okay, we have 36m of fencing to use and the person wants to build a rectangle.

So, the person can rectangles of the following dimensions:

1m width by 17m length
2m width by 16m length
3m width by 15m length
4m width by 14m length
5m width by 13m length
6m width by 12m length
7m width by 11m length
8m width by 10m length

Those are all of the possible dimensions that can be used to create a rectangular enclose whose perimeter is 36m.

We're looking for the expected area of the enclosure, which means you're looking for the average area from all of the possible enclosure sizes listed above. So we calculate the area for each of the dimensions listed above:
$1*17=17$
$2*16=32$
$3*15=45$
$4*14=64$
$5*13=65$
$6*12=72$
$7*11=77$
$8*10=80$

So, the expected or average area would simply be:

$\frac{17+32+45+64+65+72+77+80}{8}=56.5m$

I'm fairly confident that this is the correct way of approaching the problem, but if it is wrong I'd love to see the proper way to do it. Thanks

5. Originally Posted by downthesun01
I'm going to take a quick stab at this, but hopefully you'll post the correct answer if/when you find it.

Okay, we have 36m of fencing to use and the person wants to build a rectangle.

So, the person can rectangles of the following dimensions:

1m width by 17m length
2m width by 16m length
3m width by 15m length
4m width by 14m length
5m width by 13m length
6m width by 12m length
7m width by 11m length
8m width by 10m length

Those are all of the possible dimensions that can be used to create a rectangular enclose whose perimeter is 36m.

We're looking for the expected area of the enclosure, which means you're looking for the average area from all of the possible enclosure sizes listed above. So we calculate the area for each of the dimensions listed above:
$1*17=17$
$2*16=32$
$3*15=45$
$4*14=64$
$5*13=65$
$6*12=72$
$7*11=77$
$8*10=80$

So, the expected or average area would simply be:

$\frac{17+32+45+64+65+72+77+80}{8}=56.5m$

I'm fairly confident that this is the correct way of approaching the problem, but if it is wrong I'd love to see the proper way to do it. Thanks
Thank you! that is the correct answer from the solution