Question 667615
A distribution of values is normal with a mean of 231.6 and a standard deviation of 62.4. 
Find the probability that a randomly selected value is less than 312.7. 
P(X < 312.7) = 
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z(312.7) = (312.7-231.6)/62.4 = 1.3000
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P(x < 312.7) = P(z < 1.3000) = 0.9031
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Cheers,
Stan H.
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