Question

A certain model of automobile has its gas mileage (in miles per gallon, or mpg) normally distributed, with a mean of 28 mpg and a standard deviation of 4 mpg. Find the probability that a car selected at random has the following gas mileages. (Round your answers to four decimal places.) (a) less than 26 mpg (b) greater than 34 mpg (c) between 22 and 34 mpg

Answer 1

Answer:

Step-by-step explanation:

We are finding the probability, which is a percentage, of each of these intervals on our standard bell curve. In order to find this percentage, we need to find the z-score that provides this percentage. To find the z-score:

$z=\frac{x_i-\bar{x}}{\sigma}$ which is the number in question minus the mean, all divided by the standard deviation. We're first looking for the probability that the gas mileage on a certain model of car is less than 26 mpg.

To find this z-score:

$z=\frac{26-28}{4}=-.5$ Depending upon which table you look at for the z-score determines how you will find it. The z-score that measure from the value and to the left of it is what we need. This decimal is .3085375, or 30.8538%.

Onto b., which is for the percentage of cars that have gas mileage over 34 mpg. Find the z-score, and this time, we look to the right of the value for the percentage:

$z=\frac{34-28}{4}=1.5$ and to the right of 1.5 standard deviations we will find .0668072, or 6.68072%

Then finally c., which wants the probability that the gas mileage on one of these cars is greater than 22 but less than 34 mpg. To do this we have to find the z-scores of each and then do some subtracting. First the z-scores:

$z=\frac{22-28}{4}=-1.5$ The percentage of data that lies to the right of that z-score is .9331927

The z-score for the other value, 34, was already found as 1.5, having .0668072 of the data to the right of that z-score. We subtract the smaller from the larger to determine what's left in-between:

.9331972 - .0668072 = .86639, or as a percentage, 86.639% of the cars fall into this interval for gas mileage.