Answer to Question #344614 in Statistics and Probability for Kool

Question

Q: 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.

a)What proportion of patrons spend more than two hours at the gym?

b)What proportion of patrons spend less than one hour at the gym?

c)What is the least amount of time spent by 60% of patrons at the gym?

Answer 1

a)

$$P(X>120)=1-P(Z\le\dfrac{120-80}{20})$$ $$=1-P(Z\le2)\approx0.02275$$

$2.275\%$

b)

$$P(X<60)=1(Z<\dfrac{60-80}{20})$$ $$=P(Z<-1)\approx0.15866$$

$15.866\%$

c)

$$P(X>x)=1-P(Z\le\dfrac{x-80}{20})=0.6$$ $$\dfrac{x-80}{20}=-0.25335$$

$$x=80-20(0.25335)$$

$$x=75\ minutes$$