Question 1020836: A company manufactures screws and nuts. For every 1000 pieces they manufacture, there are on an average, 3 defective pieces. a) Find the probability that in a given set of 1000 parts there are at most 5 defective pieces. b) Repeat Problem # 4a) by making an approximation.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
A company manufactures screws and nuts. For every 1000 pieces they manufacture, there are on an average, 3 defective pieces. a) Find the probability that in a given set of 1000 parts there are at most 5 defective pieces. b) Repeat Problem # 4a) by making an approximation.
Solution:
Using binomial distribution:
p=0.003.
n=1000
P(X,n,p)=C(n,X)*p^X*(1-p)^(n-X)
where C(n,X) is the combination of X objects out of n.
P(X<=5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.04956+0.14914+0.22415+0.22438+0.16828+0.10087
=0.91639
(b)
When n is large (>10) and 0.05<=p<=0.95, we can apply the normal approximation to the binomial. Since binomial is discrete, we need to apply the continuity correction, namely include an additional 0.5 units at each end of the range (if possible).
Since p=0.003 << 0.05, the approximation is not expected to be very accurate, but partially compensated by the fact that n is very large.
Accordingly, we have
μ=np=1000*0.003=3, σ=sqrt(npq)=sqrt(1000*0.003*0.997)=1.72945.
Z=(5.5-3)/1.72945=1.4455
P(z<Z)=0.92584 (from normal distribution tables)
with an error of about 1%.
Also,
in order that you get answers as fast as possible, I suggest you post new questions instead of in the messages. You are welcome to alert me to the question in your message if you so wish.
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