Question

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

Answer 1

But either the length or the breadth should be given.

Answer 2

First please don't re-post questions. Second, enumerate every possible integer length, compute the area resulting from each, and average them - that gives you the expected area.

Answer 3

Since the perimeter is constrained to 36 meters the length can't be any longer than 17 (since 18 meters would give you a degenerate rectangle)

Answer 4

So we should take the length as 17?

Answer 5

No, you have to compute the area for every possible length from 1 to 17 and average them

Answer 6

Length Width Area

1 17 17
2 16 32
3 15 45
4 14 56
5 13 65
6 12 72
7 11 77
8 10 80
9 9 81
10 8 80
11 7 77
12 6 72
13 5 65
14 4 56
15 3 45
16 2 32
17 1 17

Answer 7

we should find the average of 17 values

Answer 8

1223 isn't replying so i guess he figured out the rest

Answer 9

the average of the areas, but i don't think that's the same as the average of the dimensions

Answer 10

Sorry!! I srsly don't understand the question.

Answer 11

I assume this is a probability class?

Answer 12

yess. Data Management. The answer is 57 m 2. it is Uniform Distribution class

Answer 13

question*

Answer 14

OK so there are a finite number of ways you can build a rectangle with a 36 meter perimeter under the constraint that the length must be an integer. Your job is to find the expected area meaning the average of all those cases.

Answer 15

Okay. How did you get 17?

Answer 16

The perimeter is 2 lengths plus 2 widths. The max perimeter is 36 meters. An 18 meter length, would consume all 36 meters of fence just for the two length sides and leave none for the width sides. Therefore 17 is the largest length.

Answer 17

(Any lengths larger than 18, of course, would not even have enough fence to complete the two length sides, let alone the width.)

Answer 18

Anyway, I made a table of all possible lengths (from 1 to 17), the corresponding width (which is 3(6-2*length)/2), and the corresponding area (which is length * width).

Average all those areas, and I got 57.

Answer 19

Sorry, typo in the width formula - it should have said (36 - 2*length) / 2.

I put the "(" in the wrong place since I was typing too fast.

Answer 20

Okay, I get it. Thank You soooooo much!!!!

Answer 21

np.

Answer 22

One more question. How would i use this formula... E(x) = (x1) P(x1) + (x2) P(x2) +...+ (xn) P(xn)

Answer 23

THat's basically what we did here, but since each outcome was equally likely it ends up being the same as an average.

Answer 24

You could multiply each area by 1/17 and sum them (since each area has a 1 in 17 probability of being the one picked). But since all the probabilies are uniform it's easier just to add them all up and divide by 17 in the end, which is mathematically equivalent.

Answer 25

THANKKK YOU SOO MUCH!

Answer 26

my pleasure