# A poisson distribution question

## 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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vela
Staff Emeritus
Homework Helper
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.

Mark44
Mentor
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.

vela
Staff Emeritus
Homework Helper
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? im assuming probability on the y axis and number of eggs on the x axis by the way

vela
Staff Emeritus