Question

According to secret company tests, the highway gas mileage that Urbana trucks manufactured in Germany get is normally distributed with mean 18 m.p.g. and standard deviation of 3.3. The government does not know this, and is testing a single new Urbana truck.

(a) If the test vehicle registers highway gas mileage between 21 and 25 m.p.g., the Urbana will be designated “efficient”. What is the probability that the truck gets this designation? Show your work. (b) You purchased an Urbana truck manufactured in Germany, and the company informed you that this particular truck is, among all German-made Urbanas, at the 90th percentile for gas mileage. What is your Urbana truck’s gas mileage? Show your work. (c) A company purchased one Urbana truck manufactured in Germany and another one manufactured in Austria. Their German-made truck had gas mileage of 27 m.p.g, and the Austrian-made one had gas mileage of 29 m.p.g. However, the distribution of gas mileage for Austrian-made Urbana trucks also differed from the distribution for those made in Germany. For all Urbana trucks manufactured in Austria, the mean was 20 m.p.g. and the standard deviation was 4.4 m.p.g. Which of these two trucks had better gas mileage relative to all Urbana trucks manufactured in its same country? Explain your answer.

Answer 1

Mean, = 18

Standard deviation, = 3.3

(a) The probability that the truck gets the designation 'efficient'

= P(21 ≤ X ≤ 25)

= P{(21 - 18)/3.3 ≤ Z ≤ (25 - 18)/3.3}

= P(0.909 ≤ Z ≤ 2.12)

= 0.1648

(b) Corresponding to 90th percentile, the z score = 1.2845

Thus, the required mileage = 18 + 1.2845*3.3 = 22.239 mpg

(c) Z score corresponding to 27 mpg for German made truck

= (27 - 18)/3.3 = 2.727

Z score corresponding to 29 mpg of Austrian made truck

= (29 - 20)/4.4 = 2.045

Thus, the German truck had greater mileage relative to all Urbana trucks manufactured in the same country. As this mileage is relatively more better than the average