The DBHs (Diameters at Breast Height) of the trees in Question 1 (Douglas-firs and Hemlocks taken together) are normally distributed with an average of 20 cm and a standard deviation of 5 cm. What is the probability that the researcher will select a tree that is less 15 cm in diameter?

To find the probability that a randomly selected tree has a DBH less than 15 cm, we use the normal distribution with:

- Mean (\(\mu\)) = 20 cm
- Standard deviation (\(\sigma\)) = 5 cm

We seek the probability \(P(X < 15)\), where \(X\) represents the DBH of the trees.

### Step-by-Step Solution

1. **Convert the DBH of 15 cm to a z-score**:
   \[
   z = \frac{X - \mu}{\sigma} = \frac{15 - 20}{5} = \frac{-5}{5} = -1
   \]

2. **Find the probability corresponding to \(z = -1\)**:
   From standard normal distribution tables or using a calculator, we find:
   \[
   P(Z < -1) \approx 0.1587
   \]

### Conclusion

The probability that the researcher will select a tree with a DBH less than 15 cm is approximately **0.1587**, or **15.87%**.

Would you like a more detailed breakdown or have further questions?

Here are some related questions you might find useful:

1. What is the probability of selecting a tree with a DBH greater than 25 cm?
2. What is the probability of selecting a tree with a DBH between 15 cm and 25 cm?
3. How would the probability change if the standard deviation was 3 cm instead of 5 cm?
4. What DBH value corresponds to the top 10% of tree sizes in this distribution?
5. How can we calculate the range in which 95% of tree DBHs fall?

**Tip:** For normal distributions, converting to a z-score is a useful method for finding probabilities for values not directly given in tables.

Learn how to calculate the probability of selecting a tree with a DBH less than 15 cm from a normal distribution. This problem covers concepts of normal distribution, z-scores, and probability calculation. The solution shows that the probability is approximately 15.87%. Suitable for high school students in Grades 9-12.

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