A poisson distribution question

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Homework Help Overview

The problem involves a Poisson distribution related to the egg-laying frequency of hens. The original poster seeks to determine the average number of eggs found during daily checks and the most probable number of eggs encountered.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the average number of eggs and the interpretation of the variable N in the context of the Poisson distribution. Questions arise about the probability of finding zero eggs and the correct application of the Poisson formula.

Discussion Status

There is an ongoing exploration of the correct interpretation of the Poisson distribution formula and its application to the problem. Some participants are questioning assumptions about the egg-laying rate and discussing how to visualize the distribution.

Contextual Notes

Participants note discrepancies in the understanding of the Poisson distribution and its parameters. There is also mention of varying experiences with hen productivity, which may influence assumptions about the average egg production rate.

8614smith
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Homework Statement


On average, each of the 18 hens in my henhouse lays 1 egg every 30 days. If I check the hens once per day and remove any eggs that have been laid, what is the average number, μ, of eggs that I find on my daily visits? What is the most probable (whole) number of eggs that I find on each visit? HINT: if in doubt sketch the distribution of P(N) in this case.


Homework Equations



P_{\mu}(N)=\frac{e^{-\mu}{\mu}N}{N!}


The Attempt at a Solution



18 hens lay 1 egg every 30 days so average, \mu, is 0.6 eggs a day.

Number of occurences, N, is 1 as it is checked once a day.

so P_\mu(N) = 0.329 using the numbers given.

And the most probable whole number of eggs found on each visit is 0.

Is this correct??
 
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No, you made a mistake regarding what N stands for. N is the number of eggs you could find on a visit. P(0) is the probability of finding no eggs; P(1) is the probability of finding one egg; and so on.
 
8614smith said:
On average, each of the 18 hens in my henhouse lays 1 egg every 30 days.

These are some remarkably unproductive hens! In my experience, a more typical egg production rate would be closer to an egg per day for each hen.
 
so how do i find the probability of finding 0 eggs because if i put zero in the top line it will be 0, and that 0.329 is the probability of finding one egg then.
 
Oh, your formula is wrong. I thought it was just a typo. It should be

P(N)=e^{-\mu}\frac{\mu^N}{N!}
 
ok great thanks makes sense now, but how do i draw the poisson distribution? can it be 0.549 probability at zero? for it to be continuous it would have to have a negative distribution - is that possible? I am assuming probability on the y-axis and number of eggs on the x-axis by the way
 
Look up the Poisson distribution in your textbook. It should answer your questions, like what the possible values of N are.
 

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