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.