Question

In its Fuel Economy Guide for 2016 model vehicles, the Environmental Protection Agency gives data on 1170 vehicles. There are a number of high outliers, mainly hybrid gas‑electric vehicles. If we ignore the vehicles identified as outliers, however, the combined city and highway gas mileage of the other 1146 vehicles is approximately Normal with mean 23.0 miles per gallon (mpg) and standard deviation 4.9 mpg. The 2016 Volkswagen Beetle with a four‑cylinder 1.8 ‑L engine and automatic transmission has combined gas mileage of 28 mpg. Find the percentage of 2016 vehicles that have better gas mileage than the Beetle's gas mileage. (Enter your answer rounded to two decimal places.)

Answer 1

Answer:

15.39%

Step-by-step explanation:

Step 1

Find the z-value associated to the given datum x, in this case x = 28 mpg

z is given by the formula

$z=\frac{x-\mu }{\sigma }$

where $\mu$ is the mean of the distribution and $\sigma$ is the standard deviation. In this case,

$z=\frac{x-\mu }{\sigma }=\frac{28-23}{4.9}=1.02$

Step 2

Locate the z-value in the table of z-values of the normal distribution.

(See table attached) and read the number corresponding to this z-value. In this case is highlighted in the table and its value is 0.34614.

This numbers gives us the % of vehicles with a mpg between 0 and 28.

Step 3

As we know that the total area under the curve  for z > 0  is 0.5, then we have to subtract 0.34614 from 0.5 to obtain the % of cars with  a mpg greater than 28 mpg.

0.5 - 0.34614 = 0.15386  

so, the % of 2016 vehicles that have better gas mileage than the Beetle's gas mileage is 15.39 % (rounded to two decimals)