SOLUTION: The lengths of lumber a machine cuts are normally distributed with a mean of 93 inches and a standard deviation of 0.6 inch.
(a) What is the probability that a randomly selecte
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Question 1137935: The lengths of lumber a machine cuts are normally distributed with a mean of 93 inches and a standard deviation of 0.6 inch.
(a) What is the probability that a randomly selected board cut by the machine has a length greater than 93.24 inches?
(b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 93.24 inches?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
>(93.24-93)/0.6=0.4
probability z>0.4=0.3446
for the sample, z>(xbar-mu)/sigma/sqrt(n)
this is 0.24*sqrt(44)/0.6, invert and multiply when dividing
z>2.65
probability is 0.0040
Far less likely 44 boards will deviate by the same amount compared to one.
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