the heights of the trees in a forest are normally distributed, with a mean of 25 meters and a standard…

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.\n\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.50 | 0.6915 |\n| 1.00 | 0.8413 |\n| 2.00 | 0.9772 |\n| 3.00 | 0.9987 |\n\n0.13%\n0.26%\n2.3%\n4.6%

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.\n\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.50 | 0.6915 |\n| 1.00 | 0.8413 |\n| 2.00 | 0.9772 |\n| 3.00 | 0.9987 |\n\n0.13%\n0.26%\n2.3%\n4.6%

Answer

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 37$ (the value we are interested in), $\mu = 25$ (the mean), and $\sigma=6$ (the standard deviation). So, $z=\frac{37 - 25}{6}=\frac{12}{6}=2$.

Step2: Find the cumulative probability for $z = 2$

From the standard - normal table, when $z = 2$, the cumulative probability $P(Z\leq2)=0.9772$.

Step3: Find the probability for $Z\geq2$

We know that $P(Z\geq z)=1 - P(Z\leq z)$. So, $P(Z\geq2)=1 - 0.9772 = 0.0228$. Converting this to a percentage, we get $0.0228\times100% = 2.28%\approx2.3%$.

Answer:

2.3%