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To solve the problem shown in your image, we need to find the probability that a randomly chosen item has a length less than 18.5 inches. The problem states that the lengths are normally distributed with:

- Mean (\(\mu\)) = 18.7 inches
- Standard deviation (\(\sigma\)) = 0.6 inches

We are asked to find \(P(X < 18.5)\), where \(X\) is the length of a randomly selected item.

### Steps:

1. **Standardize the variable:**
   We convert the length to a standard normal variable using the z-score formula:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   where \(X = 18.5\), \(\mu = 18.7\), and \(\sigma = 0.6\).

2. **Find the z-score:**
   \[
   z = \frac{18.5 - 18.7}{0.6} = \frac{-0.2}{0.6} \approx -0.333
   \]

3. **Use a z-table or calculator to find the probability:**
   We now need to find the probability corresponding to \(z = -0.333\). Looking it up in a z-table or using a calculator, we find:
   \[
   P(Z < -0.333) \approx 0.3707
   \]

Thus, the probability that the item is less than 18.5 inches long is approximately **0.3707**.

Would you like more details on any part of this solution?

### Related questions:
1. What is the probability that the length is greater than 19 inches?
2. How do we calculate probabilities for any normal distribution?
3. What is the probability that the length is between 18.0 and 19.0 inches?
4. How does changing the standard deviation affect the probabilities?
5. What is the cumulative probability for a length greater than 18.7 inches?

### Tip:
Always ensure you are using a z-table or a calculator that provides accurate cumulative probabilities for the normal distribution when finding probabilities from z-scores.

A manufacturer knows that their items have a normally distributed length, with a mean of 18.7 inches, and standard deviation of 0.6 inches. If one item is chosen at random, what is the probability that it is less than 18.5 inches long?

Find the probability that a randomly chosen item from a normally distributed set has a length less than 18.5 inches, given a mean of 18.7 inches and standard deviation of 0.6 inches. Learn how to calculate using the z-score formula and standard normal distribution.

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