91Ó°ÊÓ

Fuel economy (2.2) In its recent Fuel Economy Guide, the Environmental Protection Agency (EPA) gives data on 1152 vehicles. There are a number of outliers, main vehicles with very poor gas mileage or hybrids with very good gas mileage. If we ignore the outliers, however, the combined city and highway gas mileage of the other 1120 or so vehicles is approximately Normal with a mean of 18.7 miles per gallon (mpg) and a standard deviation of 4.3 mpg.

a. The Chevrolet Malibu with a four-cylinder engine has a combined gas mileage of 25 mpg. What percent of the 1120 vehicles have worse gas mileage than the Malibu?

b. How high must a vehicle’s gas mileage be in order to fall in the top 10% of the 1120 vehicles?

Step by step solution

01

Part (a) Step 1: Given information

$$\begin{gathered} \mathrm{μ}=18.7 \\ \mathrm{σ}=4.3 \\ x=25 \end{gathered}$$

02

Part (a) Step 2: Concept

The formula used:$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

03

Part (a) Step 3: Explanation

Thus the z -score will be as:

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{25−18.7}{4.3} \\ =1.47 \end{gathered}$$

Thus the corresponding probability using the normal probability table will be as:

$$\begin{gathered} P(X<25)=P(Z<1.47) \\ =0.9292 \\ =92.92\% \end{gathered}$$

Thus, about $92.92\%$of the $1120$vehicles have worse gas mileage than the Malibu.

04

Part (b) Step 1: Explanation

The top ten percentile's cutoff value is the same as the bottom ninety percentile's cutoff value.

Thus the $z-$score with the probability of $90\%$will be as:

$z=1.28$

The $z-$score is as:

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

$=\frac{x−18.7}{4.3}$

Thus, we will find the value of $x$by computing as:

$$\begin{gathered} 1.28=\frac{x−18.7}{4.3} \\ ⇒x−18.7=1.28(4.3) \\ ⇒x=18.7+1.28(4.3) \\ ⇒x=24.204 \end{gathered}$$

Thus the top $10\%$ of the $1120$ vehicles have a gas mileage above $24.204$ mpg.

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