Answer:
The answer is below
Step-by-step explanation:
The question is not complete, let me solve a question that is exactly like this one.
A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation 5 mpg. A pizza delivery company buys 43 of these cars. What is the probability that the average mileage of the fleet is greater than 33.5 mpg?
Answer:
Given that the mean (μ) is 34 miles per gallon (mpg) with a standard deviation (σ) 5 mpg. The sample (n) is 43.
The z score is used in statistics to determine by how much the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma} \\for\ a \ sample\ size:\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]
For the average mileage of the fleet is greater than 33.5 mpg (x > 33.5):
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} }\\z=\frac{33.5-34}{5/\sqrt{43} } =-0.66[/tex]
From the normal distribution table, The probability that the average mileage of the fleet is greater than 33.5 mpg = P(x > 33.5) = P(z > -0.66) = 1 - P(z < -0.66) = 1 - 0.2546 = 0.7454 = 74.54%
