Answer to Question #323661 in Statistics and Probability for Bene

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.

5.1 What proportion of patrons spends more than 2 hours at the gym? (3)

5.2 What proportion of patrons spend less than 1 hour at the gym? (3)

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

Answer 1

$z_1 = \dfrac{x_1-\mu}{\sigma} = \dfrac{120-80}{20} = +2 \\ P(z \leq +2) = 0.9772$

$z_2 = \dfrac{x_2-\mu}{\sigma} = \dfrac{60-80}{20} = -1 \\ P(z \leq -1) = 0.1587$

Part a.

The proportion of clients who spend more than 120 minutes (2 hours) in the gym is equal to 1 minus the proportion of clients who spend less than 120 minutes (z<2)

$P(z > 2) = 1- P(z \leq +2) = 1-0.9772 = \boxed{0.0228}$

Part b.

The proportion of clients who spend less than 60 minutes (1 hour) in the gym is $P(z \leq -1) = \boxed{0.1587}$

Part c.

Let's calculate the variable z (z3) from the z-table, knowing that the proportion of clients is less than or equal to 60%

$P(z \leq z_3) = 0.60 \to z_3 = 0.26$

Finally, the least amount of time spent by 60% of customers is:

$x_3 = \mu+\sigma\cdot z_3 \\ x_3 = 80+20(0.26) = \boxed{85.2 \,min}$