Question

A normal distribution is observed from the number of rentals per week for a certain car rental company. If the mean is 15 rentals and the standard deviation is 3 rentals, what is the probability that on a randomly selected week, the car company rented greater than 24 rentals

Answer 1

The probability that on a randomly selected week, the car company rented greater than 24 rentals  is 0.4987

First, we need to calculate the z-score using the formula;

$z=\frac{x-\mu}{\sigma}$

where:

• $\mu$ is the mean value = 15

• $\sigma$ is the standard deviation = 3

• x = 24

Substitute the given values into the formula to get the z-score

$z=\frac{24-15}{3} \\z=\frac{9}{3} \\z=3.000$

Checking the probability table for P(x > 3.000), the probability that on a randomly selected week, the car company rented greater than 24 rentals

is 0.4987

Learn more on probability here: brainly.com/question/22664861