In its Fuel Economy Guide for 2014 model vehicles, the Environmental Protection
ID: 2947195 • Letter: I
Question
In its Fuel Economy Guide for 2014 model vehicles, the Environmental Protection Agency gives data on 1134 vehicles. The combined city and highway gas mileage of these 1134 vehicles is approximately Normal with mean 22.2 miles per gallon (mpg) and standard deviation 5.2 mpg. (a) The 2014 Volkswagen Beetle with a four-cylinder 1.8 L engine and automatic transmission has combined gas milage of 28 mpg. What percent of all vehicles have better gas mileage than the Beetle? (b) How high must a 2014 vehicle’s gas milage be to fall in the top 5% of all vehicles? (c) The quintiles of any distribution are the values with cumulative proportions 0.20, 0.40, and 0.80. What are the quintiles of the distribution go gas mileage?
Explanation / Answer
Solution
Let X = The combined city and highway gas mileage (mpg) of the 1134 vehicles covered by the guide. We are given X ~ N(22.2, 5.22)………………………………………………. (1)
Back-up Theory
If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then, pdf of X, f(x) = {1/??(2?)}e^-[(1/2){(x - µ)/?}2] …………………………….(A)
Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(2)
P(X ? or ? t) = P[{(X - µ)/?} ? or ? {(t - µ)/?}] = P[Z ? or ? {(t - µ)/?}] .………(3)
Probability values for the Standard Normal Variable, Z, can be directly read off from
Standard Normal Tables or can be found using Excel Function NORMDIST(x,Mean,Standard_dev,Cumulative)……………………..(4)
Now, to work out the answer,
Part (a)
Percent of all vehicles that have better gas mileage than the Beetle
= 100 x P(a vehicle has gas mileage greater than )
[Given ‘The 2014 Volkswagen Beetle with a four-cylinder 1.8 L engine and automatic
transmission has combined gas milage of 28 mpg.’]
= 100 x P(X > 28)
= 100 x P[Z > {(28 – 22.2)/5.2}] [vide (1) and (3)]
= 100 x P(Z > 1.1154)
= 100 x 0.13234 [vide (4)]
= 13.23% ANSWER
Part (b)
Let t = gas milage of a 2014 vehicle for it to be in the top 5% of all vehicles. Then, we should have:
P(X > t) = 0.05 or
P[Z > {t – 22.2)/5.2}] = 0.05 [vide (1) and (3)]
=> {t – 22.2)/5.2} = 1.645 [vide (4)]
Or t = 30.754.
So, the required mileage is at least 30.75 mpg ANSWER
Part (c)
Let q1, q2 and q3 be the quintiles of the distribution of gas mileage. Then, by definition,
P(X < q1) = 0.20…………………………………………………………………..(5),
P(X < q2) = 0.40…………………………………………………………………..(6), and
P(X < q3) = 0.80…………………………………………………………………..(7)
(1), (3) and (5) => P[Z < {q1 – 22.2)/5.2}] = 0.2
=> {q1 – 22.2)/5.2} = - 0.84162 [vide (4)]
Or q1 = 17.82. ANSWER 1
(1), (3) and (6) => P[Z < {q2 – 22.2)/5.2}] = 0.4
=> {q2 – 22.2)/5.2} = - 0.25355 [vide (4)]
Or q2 = 20.46. ANSWER 2
(1), (3) and (7) => P[Z < {q3 – 22.2)/5.2}] = 0.8
=> {q3 – 22.2)/5.2} = 0.841621 [vide (4)]
Or q3 = 26.58. ANSWER 3
DONE