Questions: Suppose you buy a new car whose advertised mileage is 20 miles per gallon (mpg). After driving your car for several months, you find that its mileage is 15.6 mpg. You telephone the manufacturer and learn that the standard deviation of gas mileages for all cars of the model you bought is 1.29 mpg. a. Find the z-score for the gas mileage of your car, assuming the advertised claim is correct. b. Does it appear that your car is getting unusually low gas mileage? a. z= (Round to two decimal places as needed.) b. Does it appear that your car is getting unusually low gas mileage? Yes No

Solution Steps

To solve this problem, we need to calculate the z-score for the gas mileage of the car. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is:

\[ z = \frac{(X - \mu)}{\sigma} \]

where:

• \( X \) is the observed value (15.6 mpg in this case),
• \( \mu \) is the mean (20 mpg in this case),
• \( \sigma \) is the standard deviation (1.29 mpg in this case).

Once we have the z-score, we can determine if the car's mileage is unusually low by checking if the z-score is significantly low (typically, a z-score less than -2 or greater than 2 is considered unusual).

1. Calculate the z-score using the given values.
2. Determine if the z-score indicates unusually low gas mileage.

To determine how many standard deviations the observed gas mileage is from the mean, we use the z-score formula:

\[ z = \frac{X - \mu}{\sigma} \]

Given:

• \( X = 15.6 \) (observed value)
• \( \mu = 20 \) (mean)
• \( \sigma = 1.29 \) (standard deviation)

Substituting the values:

\[ z = \frac{15.6 - 20}{1.29} = \frac{-4.4}{1.29} \approx -3.4109 \]

A z-score is considered unusual if it is less than -2 or greater than 2. In this case:

\[ z \approx -3.4109 \]

Since \( -3.4109 < -2 \), the gas mileage is unusually low.

Final Answer