The heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard deviation of 6 meters. What is the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters? Use the portion of the standard normal table given to help answer the question.

\begin{tabular}{|c|c|}
\hline[tex]$z$[/tex] & Probability \\
\hline 0.00 & 0.5000 \\
\hline 0.50 & 0.6915 \\
\hline 1.00 & 0.8413 \\
\hline 2.00 & 0.9772 \\
\hline 3.00 & 0.9987 \\
\hline
\end{tabular}

A. [tex]$0.13 \%$[/tex]
B. [tex]$0.26 \%$[/tex]
C. [tex]$2.3 \%$[/tex]
D. [tex]$4.6 \%$[/tex]



Answer :

To find the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters, we can use the properties of the normal distribution and the provided standard normal table.

1. Identify the Given Information:
- Mean height, \(\mu\) = 25 meters
- Standard deviation, \(\sigma\) = 6 meters
- Selected height, \(X = 37\) meters

2. Calculate the Z-Score:
The Z-score represents the number of standard deviations a given value is from the mean. The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
Substituting the given values:
[tex]\[ Z = \frac{37 - 25}{6} = \frac{12}{6} = 2.0 \][/tex]

3. Find the Probability Corresponding to the Z-Score:
Using the standard normal distribution table, we look up the probability for \(Z = 2.0\). According to the provided table, the probability that a standard normal variable is less than or equal to 2.0 is 0.9772.

4. Calculate the Probability for Greater Heights:
The probability that a randomly selected tree has a height greater than or equal to 37 meters is:
[tex]\[ P(X \geq 37) = 1 - P(Z \leq 2.0) \][/tex]
Using the value from the table:
[tex]\[ P(X \geq 37) = 1 - 0.9772 = 0.0228 \][/tex]

Thus, the probability that a randomly selected tree in the forest has a height greater than or equal to 37 meters is \(0.0228\) or \(2.3\%\).

Answer: [tex]\(2.3\%\)[/tex]