Question 1062464: 4. Among all the water pumps produced by a certain factory, 6 percent are defective. A sample of 400 water pumps is selected for inspection.
What is the probability that this sample contains between 20 and 25 defective water pumps (including 20 and 25)?
Suppose that each of 40 inspectors collects a sample of 400 pumps. What is the probability that at least 8 inspectors will find between 20 and 25 defective water Pumps in their samples).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a binomial problem where n=400 p=0.06 and 1-p=0.94
Can use the normal approximation as well, where the mean is 24 (np) and the variance is np(1-p)=24(0.94)=22.56 and the sqrt (22.56)=sd=4.75
z=(x-mean)/sd
(20-24)/4.75=-4/4.75=-0.84
(25-24)/4.75=1/4.75=0.21
probability that (-0.84 < z < 0.21)=0.3827.
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Each of the inspectors has an independent (assumed) probability of 0.3827 of getting this sample and 0.6173 chance of not. Need probabilities from 0 to 7
40Cn(0.3827)^n(0.6173)^(40-n)
n=0, too small to consider
n=1, 40*(.3827)^1(0.6173)^39= about 10^-7
n=2 780*(.3827)^2(0.6173^38) is about 10^-6.
n=3 is about 10^-5
n=4 is about 5 x 10^-5
n=5 is 0.0003
n=6 is 0.0009
n=7 is 0.0027
They add to 0.00395 (using the 5 x 10^-5) or 0.004
The probability of at least 8 is therefore 0.996.
A check is the expected value of this is about 40*3/8 or 15, since the probability is about 3/8
the variance is 15*5/8=75/8, and the sd the sqrt of that, or 3.06
8 falls nearly 3 sd s below the expected value, so the likelihood of at least 8 inspectors will be very small.
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