Question 1182923: 11. A manufacturer of window frames knows from past experience that 8 per cent of the
production will have some type of minor defect that will require adjustment while the
remaining 92 percent will not have any defect. Suppose 20 windows are selected at
random:
a. Is this a binomial distribution? Explain.
b. How many window frames would you expect to have minor defects (show your work
for full marks)?
c. Calculate the probability that no more than 3 window frames will need minor
adjustments (show your work for full marks).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Yes
two outcomes
defined probability
defined n
independent.
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E(X)=np=20*0.92=1.84
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no more than 3 is p(0, 1,2,3)
probability of 0 is 0.92^20=0.1887
probability of 1 is 20*0.92^19*0.08^1=0.3282
probability of 2 is 20C2*0.92^18*0.08^2=0.2711
prob of 3 is 20C3*0.92^17*0.08^3=0.1414
That sum is 0.9294
check with binomcdf (20,0.08,3)=0.9294
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