So can someone help me with this problem

2.)The average density of trees in an acre of forest is 80 trees per acre. In a quarter
acre plot
(a) What is the probability that you observe at least 15 trees?
(b) What is the probability that you observe at most 25 trees?
(c) What is the probability that you observe exactly 20 trees?


[Edited on February 7, 2008 at 9:21 PM. Reason : i know its some easy formula but I dont know it]

2/7/2008 9:20:06 PM

50/50

2/7/2008 9:21:19 PM

are you fucking serious

2/7/2008 9:24:15 PM

yes

2/7/2008 9:24:40 PM

a) 20.98980%
b) 15.009333%
c) zero

2/7/2008 9:24:52 PM

liar

2/7/2008 9:25:28 PM

I don't see how it's possible without knowing the distribution of the # of trees per acre

every acre could have exactly 80 trees
or they could alternately have 0 and 160 trees

2/7/2008 9:26:35 PM

oopss, I meant

b) 1.500933% (damn decimals)

2/7/2008 9:27:03 PM

can we assume even distribution?

2/7/2008 9:27:59 PM

I'd think that part a) would be more than 20%

How'd you get those numbers?

2/7/2008 9:28:15 PM

First you have to assume the acre is a sphere.

2/7/2008 9:28:37 PM

I bet she knows a famous mathematician.

2/7/2008 9:29:11 PM

^hahaha

[Edited on February 7, 2008 at 9:30 PM. Reason : .]

2/7/2008 9:30:15 PM

there's more to this problem than the question offers... what class is it for?

2/7/2008 9:30:20 PM

you need more information in order to solve it.

2/7/2008 9:31:18 PM

the plane won't take off

2/7/2008 9:31:50 PM

DAT 190% ARBOR DRANK

2/7/2008 9:32:09 PM

Quote :

"there's more to this problem than the question offers... what class is it for?"

ST 361

This is the first problem from the homeworks where I've actually had to open the book

2/7/2008 9:34:24 PM

I'd assume it's normally distributed then and yeah... pretty basic formula, if nobody solves it in a few minutes i'll do it

2/7/2008 9:37:25 PM

depends on how wide your peripherals is....

80 per acre / 4 = 20 per quarter acre. 15 / 20 = 3 / 5, so a 3 in 5 chance. now you take that 3/5 chance and you flip it. you plant more trees and increase your odds. chances are, one of them will fall and make a sound even if nobody is around to hear it.

[Edited on February 7, 2008 at 9:41 PM. Reason : ]

2/7/2008 9:38:33 PM

assume normal distr

2/7/2008 9:41:06 PM

v if it doesn't tell you a distr. i'd assume normal

[Edited on February 7, 2008 at 9:42 PM. Reason : .]

2/7/2008 9:41:42 PM

you shouldn't assume it's a normal distribution for a few reasons. firstly, you can't observe a negative number of trees, and a normal distribution wouldn't cut off at zero. secondly, a tree is a discrete object; normal distributions are continuous (can't observe 16.7324943 trees).

2/7/2008 9:41:42 PM

and yes, a normal distribution would probably be a good approximation regardless of the actual distribution. but a normal distribution centered at 20 would give the same answer for a) and b), but if you consider that you can't observe a negative number of trees, it becomes pretty clear that a) and b) should have different answers.

2/7/2008 9:43:16 PM

hard to believe a statistics book wouldnt tell the distribution... is this section on a certain distribution?

2/7/2008 9:44:44 PM

You should have gone to class.

2/7/2008 9:48:28 PM

a.) write down/perform all calculations you can
b.) take said writings to teacher/TA and ask how to continue
c.) profit

2/7/2008 9:51:43 PM

I missed one class and apparently that was the class he taught this

it's an 8:00 am class so that's ^ out of question

[Edited on February 7, 2008 at 9:54 PM. Reason : it's due at 8]

2/7/2008 9:53:08 PM

teachers will answer questions in 8:00 am classes too

2/7/2008 9:54:08 PM

I only showed up for the tests when I took stats, so I don't really know any of the equations but I understand the concepts decently (I still passed the class...).

But, after skimming these 2 pages:
http://stattrek.com/Lesson2/Normal.aspx
http://www.stat.yale.edu/Courses/1997-98/101/normal.htm

and using the calculator here:
http://stattrek.com/Tables/Normal.aspx

I think you'd assume it's a normal distribution with a normal standard deviation (68% falling within 1 sd of mean).

So that means that your mean for a 1/4 acre is 20 trees, and the sd is .68/2 * 20=6.8

So for seeing AT LEAST 15 trees, you put those numbers in to the calculator, leave the "cumulative probabilty" blank, and put 15 as the normal random variable, which results in a probability of a space having LESS THAN 15 trees of 23.1%. But since you want a space to have AT LEAST 15 trees, you have to do 1-23.1% which is 76.9%.

b is also 76.9%
and c is 50%

[Edited on February 7, 2008 at 10:05 PM. Reason : ]

2/7/2008 10:02:26 PM

sounds reasonable

2/7/2008 10:21:52 PM

The probability that you will get laid with threads like this = 0




[Edited on February 7, 2008 at 10:23 PM. Reason : ]

2/7/2008 10:22:54 PM

that's a big 10-4 moron

2/7/2008 10:23:53 PM

any thoughts on moron's ideas?

2/7/2008 10:24:56 PM

I agree... seriously, that's basically what I was going to say but kept drinking and couldn't coordinate the mouse enough to get to the right websites to check

[Edited on February 7, 2008 at 10:27 PM. Reason : and I have a statistics minor---again, seriously]

2/7/2008 10:27:21 PM

he did a lot of good work and seems mostly correct

but the standard deviation should be 13.6, because if one standard deviation in both directions will capture 68% of the data, then one standard deviation in the negative direction will capture 68% of the data that is below the mean, ie .68*20=13.6

on this basis the numbers are

a. .643
b. .643 again (these numbers should definitely be identical...he's right)
c. technically 0, since the normal distribution is continuous so the probability of finding any given number is 0
but the odds of observing between 19.5 and 20.5 trees is .0293


[Edited on February 7, 2008 at 10:39 PM. Reason : /]

2/7/2008 10:34:27 PM

is the quarter acre plot on a treadmill?

2/7/2008 10:44:11 PM

^^ That's the one thing I was not sure about.

The site said 68% fall within 1 sd OF the mean, which to me meant 1sd represented 34%.

Also, I don't see how realistically your chances of seeing an exact amount of trees would be 0. I'm not saying my number is correct, I just don't see how it'd be 0.

2/7/2008 10:58:32 PM

i haven't looked at any distribution tables or anything, but moron's suggestion that c) is 50% is definitely wrong. JeffreyBSG's reasoning that you would take the probability of being between 19.5 and 20.5 trees on a normal curve is the way it should be done, so I'd assume he's got the right answer for c).

2/7/2008 11:00:38 PM

your first reply was the correct answer

2/7/2008 11:02:56 PM

The "technically 0" part is because we're applying a normal distribution to a discrete situation. A normal distribution is continuous, so any particular number is one of an infinite number of possibilities, so any particular number would have a zero probability. Since we only have whole numbers of trees, it's not actually zero, but that's why he said "technically."

2/7/2008 11:03:09 PM

it looks like a Poisson distribution to me

a)0.8951
b)0.8878
c)0.0888

[Edited on February 7, 2008 at 11:11 PM. Reason : ]

2/7/2008 11:04:39 PM

^^ Yeah, i get it's calculated by a discrete integral, so it's technically 0.

And I see that 50% is wrong, because that's seeing AT MOST 20 trees.

But to figure out a point probability, you'd have to just find the height of the bell curve of the normal distribution.

I stand by the sd being 6.8 though because the 13.6 is 2sd, which is 1sd each way from the center.

I change my c) to 5.9% chance

[Edited on February 7, 2008 at 11:26 PM. Reason : ]

2/7/2008 11:08:08 PM

I'll post the answers when I get them

[Edited on February 7, 2008 at 11:34 PM. Reason : thanks for the help]

2/7/2008 11:33:13 PM

^ set x=20 as in the graph there

^^^ I doubt they'd be doing a poisson this early in the semester.


[Edited on February 7, 2008 at 11:35 PM. Reason : ]

2/7/2008 11:34:06 PM

I got those answers....because I pulled them out of my ass.

2/7/2008 11:38:23 PM

wait...how big are these trees? are any trunks skinnier than the others? that needs to be taken into consideration. also, is it the winter (all leaves fallen off) or the spring (very leaf-full branches)? that would block your vision of other trees.

[Edited on February 8, 2008 at 7:45 PM. Reason : ]

2/8/2008 7:44:16 PM

Quote :

"it looks like a Poisson distribution to me a)0.8951 b)0.8878 c)0.0888"

Wolfman Tim FTW

he was dead on with the Poisson distribution

2/14/2008 5:31:45 PM

boy that sure is a practical application of probabilities

wow what is the probability that i'll see a certain number of trees

sounds like one of those things where the student would say "is this going to help me in real life" and the teacher/professor is like "no but its going to be on the test"

2/14/2008 5:34:45 PM

yea it is poisson

2/14/2008 5:44:29 PM