To determine the percentage of automobiles that have between 18 and 21 miles per gallon, we will follow these steps:

Step 1: Convert 18 and 21 miles per gallon to z-scores

Given:

• Mean $\mu = 27.0$ mpg
• Standard deviation $\sigma = 3.8$ mpg

The z-score formula is: $z = \frac{X - \mu}{\sigma}$ where $X$ is the value we are converting.

For $X = 18$:

$z = \frac{18 - 27.0}{3.8} = \frac{-9.0}{3.8} \approx -2.37$

For $X = 21$:

$z = \frac{21 - 27.0}{3.8} = \frac{-6.0}{3.8} \approx -1.58$

Step 2: Find the Probability for Each Z-Score

Using a z-table or a standard normal distribution calculator, we can find the probabilities for $z = -2.37$ and $z = -1.58$:

• The probability for $z = -2.37$ is approximately 0.0089.
• The probability for $z = -1.58$ is approximately 0.0571.

Step 3: Calculate the Percentage between the Two Z-Scores

To find the percentage of automobiles with in-city mileage between 18 and 21 miles per gallon, we subtract these probabilities: $0.0571 - 0.0089 = 0.0482$

Step 4: Convert to a Percentage

$0.0482 \times 100 \approx 4.82$

Final Answer

The percentage of automobiles that have between 18 and 21 miles per gallon is 4.82.

Would you like further details or have any questions?

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Further Exploration Questions:

1. What is the probability that a car has more than 30 miles per gallon?
2. What percentage of cars have less than 24 miles per gallon?
3. How would the answer change if the mean was 30 mpg with the same standard deviation?
4. What percentage of cars have between 25 and 29 miles per gallon?
5. How would the standard deviation affect the probability range?

Tip: When working with normal distributions, sketching the bell curve can help visualize the z-scores and probability areas.