Question

In its Fuel Economy Guide for 2014 model vehicles, the Environmental Protection Agency gives data on...

In its Fuel Economy Guide for 2014 model vehicles, the Environmental Protection Agency gives data on 1160 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 1134 vehicles is approximately Normal with mean 22.8 miles per gallon (mpg) and standard deviation 5.3 mpg.

The quintiles of any distribution are the values with cumulative proportions 0.20, 0.40, 0.60, and 0.80. What are the quintiles of the distribution of gas mileage?

(Use software)

The first quintile is (±±0.01 mpg)  mpg.

The first quintile is (±±0.01 mpg)  mpg.

The first quintile is (±±0.01 mpg)  mpg.

The first quintile is (±±0.01 mpg)  mpg.

QUESTION 2:

In its Fuel Economy Guide for 2014 model vehicles, the Environmental Protection Agency gives data on 1160 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 1134 vehicles is approximately Normal with mean 22.3 miles per gallon (mpg) and standard deviation 5.1 mpg. 2014 Volkswagen Beetle with a five-cylinder 2.5L engine and automatic transmission has combined gas mileage 25 mpg. What percent (±0.1) of all vehicles have better gas mileage than the Beetle (use software)?

Homework Answers

Answer #1

Solution:-

1)

Mean 22.8, S.D = 5.3

First quintile

p-value for the first  quintile = 0.20

z-score for the p-value = - 0.842

By applying normal distribution:-

x = 22.67

Second quintile

p-value for the first  quintile = 0.40

z-score for the p-value = - 0.253

By applying normal distribution:-

x = 22.76

Third quintile

p-value for the first  quintile = 0.60

z-score for the p-value = 0.253

By applying normal distribution:-

x = 22.84

Fourth quintile

p-value for the fourth quintile = 0.80

z-score for the p-value = 0.842

By applying normal distribution:-

x = 22.93.

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