If three cars of the same model are randomly selected, what is the probability that each of the three cars have efficiencies exceeding 32 mpg?

Question

The mean of fuel efficiency is 30, with a standard deviation of 1.2 mpg. 

Answer 1

First find the probability that one car has an efficiency exceeding 32 mpg. Assuming a normal distribution, you do this by finding the corresponding z-score:

z=(32-30)/1.2=1.667.

Now use a table or calculator to find the probability P(z≥1.667):

P(z≥1.667)=0.0478.

 

Now for the random selection of three cars you use the binomial distribution B(3,0.0478) - assuming each selection is an independent event so the 0.0478 stays the same.

The probability of three "successes" (three cars exceeding 32 mpg) is 0.0478³=1.09*10^-4.

Answer 2

On the normal curve, about 68% (+/-1 Standard Deviation) of these cars will have fuel efficiency from 28.8 to 31.2. The definition of "normal distribution" tells us that 34% of random selections are within -1 Standard Deviation from the mean, while another 34% of random selections are within +1 Standard Deviation. Thus, mpgs greater than 32 mpg are +2 or more Standard Deviations from the mean.

A simple way of thinking about & estimating your answer is this:

100% (the entire normal curve) minus 84% (the part of the curve that is below the the +2 Standard Deviation) leaves 16%, which represents all random mpgs that are within the +2 Standard Deviation or higher. The lowest part of the +2 Standard Deviation is not actually high enough to exceed 32 mpg - these mpgs are higher than 31.2 - so we're probably talking about approximately 15% or less of the normal curve exceeding 32 mpg.

Using this estimate of 15%, one can assume that you have a 15% chance of randomly selecting one car exceeding 32 mpg efficiency.

Randomly selecting two such cars, however, will be quite unlikely - 15% times 15%, or 2.25%.

Three such random selections will be even more unlikely - 15% times 15% times 15%, or 0.3375%. 

Thus, when making random selections of three cars, 99.6625% of those three-car selections will include at least one car with 32mpg or less. Remember that, because we are selecting randomly from a "normal distribution," about 85% of all these cars will have an mpg of 32 or less.

To compute the exact percentage based upon exactly 32 mpg, I believe you would use the one-tailed t distribution chart. I hope this helps - good luck!