Question 853814
The scores of students in the NAT is normally distributed with a mean of 60 and a standard deviation of 8. If 50 students took the test, how many got scores above 36? 
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z(36) = (36-60)/8 = -24/8 = -3
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P(x > 36) = P(z >  -3) = normalcdf(-3,100) = 0.9987
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# who scored above 36 = 0.9987*50 = 49.9 (approximately all 50)
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Cheers,
Stan H.