Question

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.

z	probability
0.50	0.6915
1.00	0.8413
2.00	0.9772
3.00	0.9987

• 0.13%
• 0.26%
• 2.3%
• 4.6%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 37$, $\mu=25$, and $\sigma = 6$.
Substitute the values into the formula: $z=\frac{37 - 25}{6}=\frac{12}{6}=2$.

Step2: Find the probability for $z = 2$

From the standard normal table, when $z = 2.00$, the probability $P(Z\leq2)$ is $0.9772$.

Step3: Calculate the probability for $x\geq37$

We want $P(X\geq37)$, which is equivalent to $P(Z\geq2)$. Since the total probability under the normal curve is $1$, we have $P(Z\geq2)=1 - P(Z\lt2)=1 - 0.9772 = 0.0228$.
To convert this to a percentage, we multiply by $100$: $0.0228\times100 = 2.28\%\approx2.3\%$.

Answer:

2.3%