Answered

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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

Sagot :

abidemiokin

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;

[tex]z=\frac{x-\mu}{\sigma} \\[/tex]

where:

  • [tex]\mu[/tex] is the mean value = 15
  • [tex]\sigma[/tex] is the standard deviation = 3
  • x = 24

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

[tex]z=\frac{24-15}{3} \\z=\frac{9}{3} \\z=3.000[/tex]

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: https://brainly.com/question/22664861

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