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**bossk171****Member**- Registered: 2007-07-16
- Posts: 305

My buddy sent me this question in an email. Unfortunately I've never take a formal prob and stats class, so I couldn't answer very well. If you could help (with the method, the answer is kinda irrelevant). Much Thanks:

He wrote:

In some small town 7 people die per year out of a population of 760.......

a, what is the average deaths per day

b, what is the probability that there will be no deaths on a given day

c, what is the prob. that there will be 1 death on a given day

d, what is the prob. that there will be 2 death on a given day

I wrote:

a. 7 deaths per year. 365 days in a year: (7/365): 0.019ish

b. About 92.5%

c. A little under 2%

d. 0.04% or less

I didn't use any formulas or anything, went more on intuition, I may be way off (but I don't think I am). Also, deaths aren't random, time of year, death of a loved one, natural disasters, etc. influence the odds. For example, if a house were to burn down, I'd think the odds of 2 deaths in one day would jump considerably. In general, I'd think deaths in small towns would tend to be grouped together in some way.

He wrote:

Well we were going that problem in stat class...over simplified because we were assuming deaths were random.

So I got part a right and comprehended it, so I didn't help there but I didn't know if you would need that part to do the rest.

then my teacher didn't really go over how to figure out the rest.

The back of the book said it would about 98.1%....I couldn't figure it out....my professors just writes on the board 358/365 which is 98.1%...but with no explanation,

then he moved on and said there is something else he needs to teach us...or I think he said that...he mumbles worse than me.

So you weren't horribly off, but I dunno I just don't see it.. the only thing I could think of was that if there's 7 deaths a year, there are 358 days where no body dies....but like that only works if there is a guaranteed one death a day rule...which there isn't...so that's why I'm all hung up.

There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

There can happen 7 deaths in a single day (7), in two (6-1 or 5-2 or 4-3),......, in six (2-1-1-1-1-1) or in seven different days (1-1-1-1-1-1-1). The probabilities are respectively p(1),p(2),...,p(6),p(7).The sum is obviously 1. The calculation is simple but messy and ends up to (appr.): p(1)= 4.2*10^-16,p(2)=9.7*10^-12,p(3)=1.1*10^-8,p(4)=3*10^-6,p(5)=7*10^-4,p(6)=0.055532694 and p(7)=0.943764295. Only p(5),p(6),p(7) are interesting since the others are too small. The answer to (b) is p(5)*360/365+p(6)*359/365+p(7)*358/365=0.981(appr.) or 98.1%. So the teacher was right. (c): a little under 1.9%. (d): the little!

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**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

A. What is the average deaths per day?

B. What is the probability that there will be no deaths on a given day?

C. What is the probability that there will be 1 death on a given day?

D. What is the probability that there will be 2 death on a given day?

Edit to add: I just noticed that the town population is given. This makes it a binomial distribution instead of a Poisson distribution:

We'll let Y be the same random variable as above.

For Part B:

For Part C:

For Part D:

Notice how small the difference is for the results using Poisson versus binomial distribution. The Poisson distribution can be used to approximate the binomial distribution when n is large, p is small, and λ < ~7.

*You can shear a sheep many times but skin him only once.*

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