Question

For questions 2 and 3. It is known that the mean diameter of pine trees in a national forest is 9.6 (inches) with a standard deviation 2.4 (inches). Assuming that the tree diameters are distributed normally. If a selected pine tree from this forest is at the 90th percentile, what is its diameter (rounded off to two decimal places) in inches?

Answer 1

Answer:

12.67 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.6, \sigma = 2.4$

If a selected pine tree from this forest is at the 90th percentile, what is its diameter (rounded off to two decimal places) in inches?

90th percentile is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 9.6}{2.4}$

$X - 9.6 = 1.28*2.4$

$X = 12.67$