Question

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

Short answer

a) The probability is 0.7967. b) It would be very surprising because the probability is almost 0. c) The probability is 0.0001. d) The efficiency is 28.026 mpg.

Step-by-step solution

Find the Z-Score

Calculate the Z-score for both values (29 and 31). For 29, the Z score is calculated as \( z = \frac{29-30}{1.2} = -0.833 \). For 31, the Z score is calculated as \( z = \frac{31-30}{1.2} = 0.833 \).

Returning the Probability

The probability that the fuel efficiency for a randomly selected car of this model is between 29 and 31 mpg is given by the area between these two Z-scores. Using the Standard Normal Table, we find the probability as 0.7967.

Find the Z-Score for Efficiency Less Than 25 mpg

Calculate the Z-score for the value 25. The Z-score for 25 is \( z = \frac{25-30}{1.2} = -4.167 \).

Find the Probability of Efficiency Less Than 25 mpg

Using the Standard Normal Table again, we find the probability as being almost 0. This indicates that it would be highly surprising to find an efficiency less than 25 mpg.

Find the Probability of Three Cars Having Efficiency Over 32 mpg

First, find the Z-score for 32, which is \( z = \frac{32-30}{1.2} = 1.667 \). The probability of one car having efficiency over 32 mpg is about 0.0485. However, we need the probability of three cars each having efficiency more than 32 mpg, which is \( 0.0485^3 = 0.0001 \).

Calculate x* for Efficiency Exceeding 95% of Cars

The Z score for the 5th percentile is approximately -1.645. By using the formula to convert z-score to x-score, \( x = x^{*} = \mu + \sigma * Z = 30 + 1.2 * -1.645 = 28.026 \), it means 95% of all cars of this model have efficiencies exceeding about \( 28.026 \) mpg