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?

Answer 1

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