To determine the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters, follow these steps:
1. Identify the parameters of the normal distribution:
- Mean ([tex]\(\mu\)[/tex]) = 25 meters.
- Standard deviation ([tex]\(\sigma\)[/tex]) = 6 meters.
2. Find the z-score for the height threshold (37 meters):
The z-score is a measure of how many standard deviations an element is from the mean. It is calculated using the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X = 37\)[/tex] meters.
[tex]\[
z = \frac{37 - 25}{6} = \frac{12}{6} = 2.0
\][/tex]
3. Locate the cumulative probability for the z-score of 2.0 in the standard normal distribution table:
According to the given table:
[tex]\[
\begin{array}{|c|c|}
\hline
z & \text{Probability} \\
\hline
2.00 & 0.9772 \\
\hline
\end{array}
\][/tex]
This probability (0.9772) represents the area to the left of [tex]\(z = 2.0\)[/tex].
4. Determine the probability of a tree having a height greater than or equal to 37 meters:
Since we need the probability of a height greater than or equal to 37 meters, we are interested in the right tail of the distribution. The area to the right of [tex]\(z = 2.0\)[/tex] can be found by subtracting the cumulative probability from 1.
[tex]\[
P(X \geq 37) = 1 - P(X < 37) = 1 - 0.9772 = 0.0228
\][/tex]
5. Convert the probability into percentage:
[tex]\[
P(X \geq 37) \times 100 = 0.0228 \times 100 = 2.28\%
\][/tex]
So, the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is approximately [tex]\(2.3\% \)[/tex].
