SOLUTION: The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of

Question 438819: The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes.
What proportion of members spend between 80 and 100 minutes at the gym? Use the standard normal distribution tables on p564 of your Wegner textbook.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The manager of a local gym has determined that the length of time members spend at the gym is a normally distributed variable with a mean of 80 minutes and a standard deviation of 20 minutes.
What proportion of members spend between 80 and 100 minutes at the gym?
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z(80) = 0
z(100) = (100-80)/20 = 1
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P(80<= x <= 100) = P(0<= z <=1) = 0.3413
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Cheers,
Stan H.
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