SOLUTION: 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

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Question 1205385: 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?

Answer by ikleyn(52776) About Me  (Show Source):
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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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


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   p%5E6 = 3%2F90 = 1%2F30,
which implies   p = root%286%2C1%2F30%29 = 0.5673   (approximately).


From the other side,  the probability to have 0 successful trials  (no damaged eggs in a cartoon)  is   %281-p%29%5E6 = 52%2F90 = 26%2F45,
which implies   1-p = root%286%2C26%2F45%29 = 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.