Question

Ch 3 Homework Check Syntax 6. (3.35) In its Fuel Economy Guide for 2013 model vehicles, the Environmental Protection Agency gives data on 1108 vehicles. There are a outliers, however, the combined city and number of high outliers, mainly hybrid gas-electric vehicles. If we ignore the vehicles identified as highway deviation 5 mpo tos vehidles i approximately homal with mean 2.sga) gas mileage of the other 1093 is approximately Normal with mean 21.7 2013 Vollswagen Beetle with a five-cylinder 2.5t engine and automatic transmission has combined gas What percent (:0.1) of all vehicles have better gas mileage than the Beetle (use software)? Check Syntax (3.42) The heights of people of the same sex and similar ages f The first and third quartiles are 126.3 pounds and 181.2 pounds. What can you say about the shape of follow a Normal distribution reasonably closely. Weights, on the other hand 149.4 pounds O Distribution is left-skewed Remaining: 1:48:00 Start: 9:00 PM

Answer 1

Question 6

Here mean combined city and highway gas mileage = $\mu$ = 21.7 mpg

standard deviation of combined city and highway gas mileage =$\sigma$ =5 mpg

Here volkswagen beetle has combined gas mileage = 25 mpg.

so here as we are given that distribution is normal here we will find the z value of the beetle car.

Z = (x - $\mu$)/$\sigma$ = (25 - 21.7)/5 = 0.66

so here as we have to find the probability of number of vehicles who have better mileage than beetle.

Pr(x > 25 mpg) = Pr(Z > 0.66) = 1 - Pr(Z < 0.66)

Now seeing z table for value related to Z = 0.66 or using calculator normalcdf function

Pr(Z < 0.66) = 0.7454

Pr(x > 25 mpg) = Pr(Z > 0.66) = 1 - Pr(Z < 0.66) = 1- 0.7454 = 0.2546

so there are 25.46 % of all vehicles have better mileage than beetle. or in one decimal place

THe answer is 25.5%