Question 1026357
The time required to verify and fill a common prescription at a neighborhood pharmacy is normally distributed with a mean of 10 minutes and a standard deviation of 3 minutes. 
What is the probability that a customer will have to wait more than 15 minutes for her prescription to be verified and filled?
z(15) = (15-10)/3 = 5/3
P(x < 15) = P(z > 5/3) = normalcdf(5/3,100) = 0.0478
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What is the probability that a customer will have to wait less than 8 minutes for her prescription to be verified and filled?
z(3) = (3-10)/3 = -7/3
P(x < 3) = P(z < -7/3) = normalcdf(-100,-7/3) = 0.0098
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The time required to verify and fill a common prescription at a neighborhood pharmacy is normally distributed with a mean of 10 minutes and a standard deviation of 3 minutes. 
Determine the wait time for which 80% of all prescriptions will be verified and filled.
Find the z-value with a left tail of 0.8
invNorm(0.8) = 0.8416
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Solve for "x" when x = z*s + u
x = 0.8416*3 + 10 = 12.52 minutes
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Cheers,
Stan H.
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