27.03.2023 - 23:54

# Costco installs automobile tires on a first-come first-serve basis. The total time a customer needs to wait for the installation to be completed follows the normal distribution with a mean time of 106.3 minutes and a standard deviation of 18.5 minutes. Wh

Question:

Costco installs automobile tires on a first-come first-serve basis. The total time a customer needs to wait for the installation to be completed follows the normal distribution with a mean time of 106.3 minutes and a standard deviation of 18.5 minutes. What is the probability that a randomly selected customer will wait between 50 and 95 minutes for his or her tires to be installed?

A) 0.3455

B) 0.2960

C) 0.1931

D) 0.1245

• April 7, 2023 в 18:16

To solve this problem, we need to standardize the normal distribution using the z-score formula, which is:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, we want to find the probability that a customer will wait between 50 and 95 minutes, so we need to calculate the z-score for both values:

z1 = (50 - 106.3) / 18.5 = -3.043

z2 = (95 - 106.3) / 18.5 = -0.6108

Then, we use a standard normal distribution table or a calculator to find the area under the curve between these two z-scores. The probability can be calculated as:

P(50 ≤ x ≤ 95) = P(z1 ≤ z ≤ z2) = Φ(z2) - Φ(z1)

where Φ(z) is the cumulative distribution function of the standard normal distribution.

Using a standard normal distribution table, we find Φ(-0.6108) = 0.2700 and Φ(-3.043) = 0.0011. Therefore,

P(50 ≤ x ≤ 95) = Φ(-0.6108) - Φ(-3.043) = 0.2700 - 0.0011 = 0.2689

Therefore, the correct answer is not one of the options provided, but it is closest to option B) 0.2960. However, the closest option is not always the correct one, so it's important to check the calculations and interpret the results. The probability of a randomly selected customer waiting between 50 and 95 minutes for their tires to be installed is about 0.2689, or \approx imately 27%.