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A poisson distribution question

  • Thread starter 8614smith
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  • #1
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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



[tex]P_{\mu}(N)=\frac{e^{-\mu}{\mu}N}{N!}[/tex]


The Attempt at a Solution



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

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

so P_[tex]\mu[/tex](N) = 0.329 using the numbers given.

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

Is this correct??
 

Answers and Replies

  • #2
vela
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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.
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
vela
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Oh, your formula is wrong. I thought it was just a typo. It should be

[tex]P(N)=e^{-\mu}\frac{\mu^N}{N!}[/tex]
 
  • #6
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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? im assuming probability on the y axis and number of eggs on the x axis by the way
 
  • #7
vela
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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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