The scores on a national achievement exam are normally distributed with a mean of 600 and a standard deviation of 120. If a student is selected at random, find the probability that the student scored above 660
Since "\\mu=600" and "\\sigma =120" we have:
"P (X>660) = P(X-\\mu>660-600) = P(\\dfrac{X-\\mu}{\\sigma}-\\dfrac{660-600}{120})"
Since "\\dfrac{X-\\mu}{\\sigma}" and "\\dfrac{660-600}{120}" = 0.5 we have:
P(X > 660) = P(Z > 0.5)
Use the standard normal table to conclude that:
P(Z > 0.5) = 0.3085
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