To determine the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters, we will make use of the properties of the normal distribution and the standard normal distribution table.
### Step-by-Step Solution:
1. Identify the given parameters:
- Mean height ([tex]\(\mu\)[/tex]): 25 meters
- Standard deviation ([tex]\(\sigma\)[/tex]): 6 meters
- Specific height ([tex]\(X\)[/tex]): 37 meters
2. Calculate the z-score for the specific height:
The z-score formula is given by:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
Substituting the known values:
[tex]\[
z = \frac{37 - 25}{6} = \frac{12}{6} = 2.0
\][/tex]
3. Use the standard normal distribution table for the computed z-score:
From the given table:
[tex]\[
z = 2.00 \implies \text{Probability} = 0.9772
\][/tex]
This probability represents the area under the curve to the left of [tex]\( z = 2.00 \)[/tex].
4. Calculate the probability of a tree being greater than or equal to 37 meters:
Since we need the probability of the height being greater than or equal to 37 meters, we need the area to the right of [tex]\( z = 2.00 \)[/tex]. This is given by:
[tex]\[
P(X \geq 37) = 1 - P(X < 37)
\][/tex]
Using the value from the standard normal distribution table:
[tex]\[
P(X \geq 37) = 1 - 0.9772 = 0.0228
\][/tex]
5. Convert this probability to a percentage:
[tex]\[
\text{Percentage} = 0.0228 \times 100 = 2.28\%
\][/tex]
Therefore, the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is [tex]\(2.3\%\)[/tex] (rounded to the nearest tenth of a percent).
### Final Answer:
[tex]\(\boxed{2.3\%}\)[/tex]
