In its Fuel Economy Guide for $2008$ model vehicles, the Environmental Protection Agency gives data on $1152$vehicles. There are a number of outliers, main vehicles with very poor gas mileage. If we ignore the outliers, however, the combined city and highway gas mileage of the other $1120$

or so vehicles are approximately Normal with a mean of $18.7$miles per gallon (mpg) and a standard deviation of $4.3$ mpg.

The top $10\%(2.2)$How high must a $2008$vehicle’s gas mileage be in order to fall in the top $10\%$ of all vehicles? (The distribution omits a few high outliers, mainly hybrid gas-electric vehicles.)

The mean gas mileage,$\mu =18.7mpg$

The standard deviation of gas mileage,$\sigma =4.3mpg$

A regression line is a straight line that depicts the change in a response variable $y$ when an explanatory variable $x$ changes. By putting this $x$ into the equation of the line, you may use a regression line to predict the value of $y$ for any value of $x$

We are provided that if the random variable $X$ denotes the number of vehicles with gas mileage $X~N(\mu ,{\sigma }^2)$

Where $\mu =18.7mpg$and $\sigma =4.3mpg$

Its goal is to find the vehicle with the best gas mileage among the top $10\%$of all automobiles.

To put it another way, find the $z$' that corresponds to the top $10\%$of the population.

That is, $$\begin{gathered} P(z>z^1)=10\% \\ \Rightarrow P(z>z^1)=0.90 \end{gathered}$$

From standard normal tables, the value of $z$ corresponding to $0.90$cumulative area is $1.28$

That is, $$\begin{gathered} z^1=1.28 \\ \Rightarrow \frac{x-\mu }{\sigma }=1.28 \\ \Rightarrow x=18.7+1.28\times 4.3 \\ =24.204 \end{gathered}$$

As a result, the highest gas mileage that falls into the top $10\%$of all vehicles is $24.204mpg$