Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.25 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are at most 0.23 inches? Round your answer to at least four decimal places.

To find the proportion of woodlice with antenna lengths at most 0.23 inches, we'll use the standard normal distribution (Z-distribution) to calculate this.

Given:
- Mean (\(\mu\)) = 0.25 inches
- Standard deviation (\(\sigma\)) = 0.05 inches
- Antenna length of interest = 0.23 inches

### Step 1: Calculate the Z-score
The Z-score is calculated using the formula:

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

Where:
- \(X\) is the value of interest (0.23 inches),
- \(\mu\) is the mean (0.25 inches),
- \(\sigma\) is the standard deviation (0.05 inches).

Substituting the values:

\[
Z = \frac{0.23 - 0.25}{0.05} = \frac{-0.02}{0.05} = -0.4
\]

### Step 2: Find the proportion corresponding to the Z-score
We now need to find the proportion of the standard normal distribution that lies to the left of \(Z = -0.4\).

Using standard normal distribution tables or a calculator:

\[
P(Z \leq -0.4) \approx 0.3446
\]

### Final Answer
The proportion of woodlice with antenna lengths at most 0.23 inches is approximately **0.3446**.

Would you like more details or have any questions?

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Here are some related questions:
1. What is the Z-score for an antenna length of 0.3 inches?
2. How would you calculate the proportion of woodlice with antenna lengths greater than 0.23 inches?
3. What proportion of woodlice have antenna lengths between 0.20 inches and 0.30 inches?
4. How does changing the standard deviation to 0.03 inches affect the proportion of woodlice with antenna lengths at most 0.23 inches?
5. What is the probability that a woodlouse has an antenna length between 0.23 and 0.27 inches?
6. How would you calculate the proportion for an antenna length less than 0.20 inches?
7. What assumptions are made when using the normal distribution in this context?
8. Can antenna lengths be negative? What would a Z-score close to 0 indicate?

**Tip:** Understanding Z-scores is crucial in statistics, as they help standardize different data points for comparison.

Learn how to calculate the proportion of woodlice with antenna lengths at most 0.23 inches using the normal distribution and Z-score formula. Given a mean of 0.25 inches and a standard deviation of 0.05 inches, discover the step-by-step solution and implications. Suitable for advanced high school students and beyond.

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