SOLUTION: The time that customers take to complete their transaction at a money machine is a random variable with mean of 2 minutes and a standard deviation of 0.6 minutes. Assume the variab

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Question 1031950: The time that customers take to complete their transaction at a money machine is a random variable with mean of 2 minutes and a standard deviation of 0.6 minutes. Assume the variable is normally distributed.
a) What is the probability that a randomly selected customer will take between 1.5 and 2.5 minutes to complete his transaction?
b) Find the probability that a random sample of 50 customers will take more than 112 minutes to complete all their transactions

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The time that customers take to complete their transaction at a money machine is a random variable with mean of 2 minutes and a standard deviation of 0.6 minutes. Assume the variable is normally distributed.
a) What is the probability that a randomly selected customer will take between 1.5 and 2.5 minutes to complete his transaction?
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z(1.5) = (1.5-2)/0.6 = -0.8333
z(2.5) = (2.5-1.5)/0.6 = 1/0.6 = 10/6 = 5/3
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P(1.5< x < 2.5) = P(-0.8333< z < 5/3) = normalcdf(-0.8333,5/3) = 0.7500
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b) Find the probability that a random sample of 50 customers will take more than 112 minutes to complete all their transactions
z(112) = (112-2)/(0.6/sqrt(50)) = 1296
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Note:: I suspect there is a typo in your post.
Cheers,
Stan H.
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