Answer to Question #344616 in Statistics and Probability for Kool

Question #344616

The manager of a local gym has determined that the length of time patrons 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 patrons spend more than two hours at the gym?




1
Expert's answer
2022-05-25T14:34:17-0400
"P(X>120)=1-P(Z\\le\\dfrac{120-80}{20})"

"=1-P(Z\\le2)\\approx0.02275"

"2.275\\%"


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