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The problem in the image asks for the proportion of trucks that can be expected to travel between 15 and 30 thousand miles in a year, assuming the distances are normally distributed with a mean of 30 thousand miles and a standard deviation of 12 thousand miles.

To solve this, we need to use the **z-score formula** to find the z-scores for both 15 thousand miles and 30 thousand miles:

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

Where:
- \( x \) is the value we're converting to a z-score,
- \( \mu = 30 \) thousand miles (mean),
- \( \sigma = 12 \) thousand miles (standard deviation).

We will find the z-scores for both 15 thousand miles and 30 thousand miles.

### Step 1: Find the z-score for 15 thousand miles:
\[
z_{15} = \frac{15 - 30}{12} = \frac{-15}{12} = -1.25
\]

### Step 2: Find the z-score for 30 thousand miles:
\[
z_{30} = \frac{30 - 30}{12} = 0
\]

### Step 3: Use the standard normal distribution table or a calculator to find the cumulative probabilities:
- For \( z = -1.25 \), the cumulative probability is approximately **0.1056**.
- For \( z = 0 \), the cumulative probability is **0.5**.

### Step 4: Find the proportion of trucks between 15 and 30 thousand miles:
\[
P(15 < X < 30) = P(0) - P(-1.25) = 0.5 - 0.1056 = 0.3944
\]

Thus, the proportion of trucks that can be expected to travel between 15 and 30 thousand miles in a year is approximately **0.3944** (or 39.44%).

Would you like further explanation on any of the steps or details on normal distribution?

Here are five related questions for you to consider:
1. How can you calculate the probability of a truck traveling more than 30 thousand miles?
2. What is the probability of a truck traveling less than 15 thousand miles?
3. How does changing the standard deviation affect the proportion of trucks within certain mileage ranges?
4. Can you find the mileage range for the middle 50% of trucks?
5. How would you interpret the z-scores in real-world terms?

**Tip:** When working with normally distributed data, z-scores help standardize different values so you can easily use statistical tables or software to find probabilities.

A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 30 thousand miles and a standard deviation of 12 thousand miles. What proportion of trucks can be expected to travel between 15 and 30 thousand miles in a year? Round to four decimal places as needed.

Learn how to calculate the proportion of trucks traveling between 15 and 30 thousand miles in a year, given a normal distribution with a mean of 30 thousand miles and a standard deviation of 12 thousand miles. This solution involves using z-scores and cumulative probabilities.

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