Question

Ford car models has a mean gas mileage of 29 miles per gallon (mpg) with a standard deviation of 5miles per gallon A pizza delivery company buys 38 of multiple Ford cars. What is the probability that the average mileage of the fleet is greater than 27.8 miles per gallon? Round to four decimal places and please make sure to explain your answer.

Answer 1

`P(\bar X >27.8) = P(Z>(27.8-29)/((5)/(√(38)))) = P(z>-1.479)`

And we can use the complement rule and the normal standard table or excel and we got:

`P(Z>-1.479) = 1-P(z<-1.479) = 1-0.0696= 0.930`

For this case we can conclude that the probability of obtain a sample mean of 27.8 ,pg or higher is 0.9304 or 93.04%

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the gas mileage of a population, and for this case we know the distribution for X is given by:

`X \sim N(29,5)`

Where
`\mu=29` and
`\sigma=5`

We select a sample size of n =38.

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And we want to find this probability:

`P(\bar X>27.8)`

And we can use the z score given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using this formula we got:

`P(\bar X >27.8) = P(Z>(27.8-29)/((5)/(√(38)))) = P(z>-1.479)`

And we can use the complement rule and the normal standard table or excel and we got:

`P(Z>-1.479) = 1-P(z<-1.479) = 1-0.0696= 0.9304`

For this case we can conclude that the probability of obtain a sample mean of 27.8 ,pg or higher is 0.9304 or 93.04%