Question 251866
The caffeine content of a cup of home-brewed coffee is a normally distributed random variable with a mean of 115 mg with a standard deviation of 20 mg. 
(a) What is the probability that a randomly chosen cup of home-brewed coffee will have more than 130 mg of caffeine? 
Find the z-value of 130
Then P(x > 130) = P(z > that z-value)
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(b) Less than 100 mg?
Find the z-value of 100.
Then P(x<100) = P(z< that z-value)
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(c) A very strong cup of tea has a caffeine content of 91 mg. What is the probability that a cup of coffee will have less caffeine than a very strong cup of tea?
Find the z-value that corresponds to 91 mg using
x = z*sigma + u
91 = z*20+115
z = -1.2
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Find P(z < -1.2)
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Cheers,
Stan H.
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