Question 363541
I have the answer to this problem and I am sure I am using the correct formula, but I am not getting the correct answer of .227 
The lifetime of a machine component is normally distributed with a mean of 3 and a standard deviation of .16. What is the probability that X is greater than 3.3
X ~ N(3, .16)
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z(3.3) = (3.3-3)/0.16 = 1.875
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P(x > 3.3) = P(z > 1.875) = 0.0304
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You would not be the first person to 
find that the "book" answer was wrong.
Cheers,
Stan H.