Homework Answers
We have, n= 1000 and p= 0.02
Let X= no of defectives
P(X=x) = 1000Cx*(0.02)^x*(0.98)^1000-x
We want, P(X<=25) = P(X=0) + P(X=1) +.....+P(X=25) = 0.8907
So the required probability is 0.8907
(By putting these values of x in the binomial function above and adding the probabilities)
Mean number of defectives = 1000*0.02= 20
Standard deviation = √1000*0.02*0.98= 4.427
From the standard normal distribution, we know, 20+ 4.427*3= 34 (is the outer boundary ) so defectives >=34 would be significantly high (as it should lie between 20 +-3*SD)
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