Question

A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys 54 of these cars. What is the probability that the average mileage of the fleet is between 33.3 and 34.3 mpg?

Answer 1

Answer:

$z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028$

$z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441$

An we can use the normal standard table and the following difference and we got this result:

$P(-1.028$

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

$\mu = 34$ represent the mean

$\sigma= 5$ represent the deviation

We have a sample size of n = 54 and we want to find this probability:

$P(33.3 < \bar X< 34.3)$

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

And we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028$

$z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441$

An we can use the normal standard table and the following difference and we got this result:

$P(-1.028$