Binomial Distribution: If hens of a cetrain breed lay eggs on 4 days a week, on the average, find how many...?

Question

If hens of a certain breed lay eggs on 4 days a week, on the average, find how many days during a season of 200 days a poultry keeper with eight hens of this breed will expect to receive at least six eggs.

So, n = 200. Isn't that number a little too big unless I make it a normal approximation or something along those lines? But, that aside, I'm not sure how to interpret the laying eggs 4 days per week to obtain a value for probability (P = 4/7?), and the 200 days is confusing.

Any help'd be very much appreciated!

The answer's 52 days. :)

Answer 1

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.


First start by finding the prob that six or more birds lay eggs on one day.

Let B be the number of birds who lay eggs on one day.
B ~ Binomial(8, 4/7)
P(B ≥ 6) = P(B = 6) + P(B = 7) + P(B = 8)
= 0.17905076 + 0.06820981 + 0.01136830
= 0.2586289

Let X be the number of days that you will have six or more birds laying eggs.

X ~ Binomial( 200, 0.2586289)

the expected value for a binomial random variable is n * p
= 200 * 0.2586289 = 51.72578

the farmer should expect to get 51 days with six birds laying eggs out of the 200 days.

Answer 2

Let
4/7 = probability of one hen laying an egg on a given day
p = probability of six or more hens laying an egg on a given day

Since p is distributed Binomially we have:

p = (8C6)[(4/7)^6](3/7)² + (8C7)[(4/7)^7](3/7) + (8C8)[(4/7)^8]
p = 28[(4/7)^6](3/7)² + 8[(4/7)^7](3/7) + (4/7)^8
p ≈ 0.258628875

The expected number of days out of 200 in which six or more hens lay eggs is:

200p ≈ 51.7257751

This rounds to 52 days.
___________

By the way, the Poisson approximation to the binomial gives an answer of 53 days. So there is close agreement.

Answer 3

No chicks wont come out , it is a process where hen lays eggs and sits on their eggs for some days to provide proper heat so that chicks come out of it

Answer 4

Looks like a Poisson process to me.
You have an egg-laying machine that, on a particular day, has a Poisson parameter = (32/7)/day = 4.57
P(x > 6) = 1 - e^(-(32/7))[1/0! + 4.57^1/1! + 4.57^2/2! +4.57^3/3! + 4.57^4/4! +4.57^5/5!]
= 1 - 0.0103[1 +4.57 + 10.44 + 15.91 + 18.17 + 16.61] = 0.3093
This is the probability that there will be at least 6 eggs/day, so out of 200 days, there will be 0.3093(200) ≈ 62 days with at least 6 eggs.

Answer 5

each hen has a probability of not laying any egg that day of 3/7

you have a lot of days

what are the chances on any day that more than two hens will not lay?

Answer 6

I would say 3/7 days a week the farm could expect 6 eggs or about 86 days.