X= # of spoiled milk cartons out of 250
P= P(success)=P(carton being spoiled)=0.01
We wish to calculate:
Where P(5)=( 250)(0.01) (0.99) 245
Np=250(0.01)=2.5 since this is less than 5, we can use the poissom distribution to approx. the
p(x>5) with U=np+2.5
P(x) =(𝑒 2.5 )/𝑥!
Thus P(x>5) =1-P(x<5) =10.89%
This happen when you sample without replacement and you have a small population.
You can use the hypergeomentric distribution
N=100 n=15 X=# of defectives out of n=15
P 2 : conditioned on fact that 1 item was defective.
P 3 : given 1 2 items were defective
We should not us Binomial distribution. (put chosen sample back in the box)
1. to use the HD, you MUST know N
2. it is known that r of them are of a certain kind and N-r of them are of another kind
3. KEY We take a random sample of n items, drawn without replacement, BUT p = P(success) is
not the same from trial
( ) 𝑒 𝑒
𝑥 𝑒 𝑒 𝑒