SOLUTION: A manufacturer of window frames knows from past experience that 10 per cent of the production will
have some type of minor defect that will require adjustment. Suppose 30 windows
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-> SOLUTION: A manufacturer of window frames knows from past experience that 10 per cent of the production will
have some type of minor defect that will require adjustment. Suppose 30 windows
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Question 1056956: A manufacturer of window frames knows from past experience that 10 per cent of the production will
have some type of minor defect that will require adjustment. Suppose 30 windows are selected at
random:
a. How many window frames would you expect to have minor defects?
b. What is the probability that more than 3 window frames will need minor adjustments? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Expected value is 0.10*30=3 window frames x*(p(x))
probability of 0 is 0.90^30=0.0424
probability of 1 is 30*0.9^29*0.10=0.1413
probability of 2 is 435*0.9^28*0.10^2=0.2277
probability of 3 is 4060*0.9^27*0.1^3=0.2361
They add to 0.6051
More than 3 is 1-0.6051=0.3949, using the complement rule.
