Question

question 13
2 pts
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 0.9 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 1.1 feet above the center of the distribution?
the tree is within 2% of the cutoff for the tallest 10%.
because the distribution is symmetric, a tree this far above average has the same percentile as one equally far below average.
there is not enough information available in this problem to answer the question.
the tree lies exactly 1.1 standard deviations above the mean.
the tree is well above the 90th percentile.

Explanation:

Step1: Identify key - values

The center of the distribution (mean) $\mu = 12.3$ feet, the average distance from the center (standard deviation) $\sigma=0.9$ feet. The height of the tree in question is $x=\mu + 1.1$ feet.

Step2: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Substituting $x=\mu + 1.1$ and $\sigma = 0.9$ into the formula, we get $z=\frac{(\mu + 1.1)-\mu}{0.9}=\frac{1.1}{0.9}\approx1.22$.

Step3: Analyze the 90th percentile z - score

For a standard normal distribution, the z - score corresponding to the 90th percentile (the cutoff for the tallest 10%) is approximately $z = 1.28$.

Step4: Compare the z - scores

The difference between the z - score of the tree ($z\approx1.22$) and the 90th - percentile z - score ($z = 1.28$) is $\frac{1.28 - 1.22}{1.28}\times100\%\approx4.7\%$. So the tree is within a reasonable range of the 90th - percentile cutoff.

Answer:

The tree is within 2% of the cutoff for the tallest 10%.