Answer to Question #344616 in Statistics and Probability for Kool

Question

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?

Answer 1

$$P(X>120)=1-P(Z\le\dfrac{120-80}{20})$$

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

$2.275\%$