Question 522706
 Students took an Algebra 2/Trig Test with a mean of 87.5% with a standard deviation of 4.5%.
What is the probability that a student scored between 84 and 100%?
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z(84) = (84-87.5)/4.5 = -0.7778
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z(100) = (100-87.5)/4.5 = 2.7778
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P(84 <= x <=100) = P(-0.7778 <= z <= 2.778) = 0.7789
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Cheers,
Stan H.
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