Question 900743
The manager of a gym has determined that the length of time members spend training at the gym is normally distributed with a mean of 80 minutes and a standard deviation of 20 minutes. What percentage of members spend more than 2 hours training at the gym?
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2 hrs = 120 minutes
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z(120) = (120-80)/20 = 2
P(x > 120) = P(z > 2) = normalcdf(2,100) = 0.0228 = 2.28%
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Cheers,
Stan H.
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