# 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??

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