.
Eggs are packed into cartons of six. A sample of 90 cartons is randomly
selected and the number of damaged eggs in each carton counted.
Number of damaged eggs
0
1
2
3
4
5
6
Number of Cartons
52
15
8
5
4
3
3
Does the number of damaged eggs in a carton follow a Binomial distribution?
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Let assume for a minute that the given distribution is a binomial with the probability of a successful outcome
(success in this problem is getting a damaged egg) at each individual trial p = const.
Then the probability to have 6 successful trials (six damaged eggs in a cartoon) is = = ,
which implies p = = 0.5673 (approximately).
From the other side, the probability to have 0 successful trials (no damaged eggs in a cartoon) is = = ,
which implies 1-p = = 0.9126 (approximately).
But then we have this contradiction:
the sum of p and ( 1-p), which is 0.5673 + 0.9126 ~ 1.4700, is FAR from to be equal 1 (one).
From this reasoning, my conclusion is that the given distribution IS NOT a binomial.