SOLUTION: the waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes. find the 75th percentile for wait

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Question 1055026: the waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes.
find the 75th percentile for waiting times at this bank.
Find the probability that a customer has to wait more than 6 minutes.
I believe the probability is 0.05 or 5 percent for this one but I'm not sure.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
mean of 3.7 minutes and a standard deviation of 1.4 minutes
z+=blue+%28x+-+mu%29%2Fblue%28sigma%29
0r
blue%28sigma%29%2Az+%2B+mu=+blue+%28x%29
the 75th percentile for waiting times at this bank: z = invNorm(.75) = .6745
blue%281.4%29%2A.6745+%2B+3.7=+blue+%284.6436%29 x = 4.7 always round Up for these
|
P(x > 6) = P(z+=blue+%286.0+-+3.7%29%2Fblue%281.4%29 )
Find z and then find P(z > value found) = normalcdf(value found, 100) 100 a placeholder
z = 1.6429
P(z > 1.6429) = normalcdf(1.6429, 100) = .05 0r 5%