Question

an ecologist records the heights of hundreds of trees of a single species. the distribution of these tree heights is bell - shaped and symmetric, with a center of 12.3 feet. the average distance of tree heights from this center is 0.8 feet. a tree is considered \exceptionally tall\ if its height places it among the tallest 10% of all trees of this species. based on this information, which one of the following statements is most accurate for a tree that is 1.3 feet above the center of the distribution?

the tree is below the 90th percentile cutoff.

there is not enough information available in this problem to answer the question.

the tree is exactly 1.3 standard deviations away from the mean.

the tree is at approximately the 90th percentile.

the tree is closer to the 95th percentile than to the 90th percentile.

Explanation:

Step1: Calculate the number of standard deviations

We know that the average distance (standard deviation, \(\sigma\)) is \(0.8\) feet. The tree is \(1.3\) feet above the mean (\(\mu = 12.3\)). The \(z-\)score is calculated as \(z=\frac{x-\mu}{\sigma}\). Here, \(x-\mu = 1.3\) and \(\sigma=0.8\), so \(z=\frac{1.3}{0.8}=1.625\)

Step2: Use the standard normal distribution table

Looking at the standard - normal distribution table (or using a calculator with a normal - distribution function), the \(z-\)score of \(z = 1.28\) corresponds to the \(90^{th}\) percentile (\(P(Z<1.28)=0.90\)) and \(z = 1.645\) corresponds to the \(95^{th}\) percentile (\(P(Z < 1.645)=0.95\))

Since \(1.28<1.625<1.645\), the tree is closer to the \(95^{th}\) percentile than to the \(90^{th}\) percentile

Answer:

The tree is closer to the 95th percentile than to the 90th percentile.