Question

a forester measured 29 of the trees in a large woods that is up for sale. he found a mean diameter of 10.2 inches and a standard deviation of 4.8 inches. suppose that these trees provide an accurate description of the whole forest and that a normal model applies.
a) choose the correct normal model for tree diameters.

b) what size would you expect the central 95% of all tree diameters to be?
using the 68 - 95 - 99.7 rule, the central 95% of the tree diameters are between.6 inches and 19.8 inches.
(do not round. type integers or decimals.)
c) about what percent of the trees should have diameters below 5.4 inches?
using the 68 - 95 - 99.7 rule, about % of the trees should have diameters below 5.4 inches.
(do not round. type an integer or a decimal.)

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule for a normal distribution states that about 95% of the data lies within \(\mu\pm2\sigma\), about 68% lies within \(\mu\pm\sigma\), and about 99.7% lies within \(\mu\pm3\sigma\).

Step2: Calculate the value of \(\mu - 2\sigma\)

Given \(\mu = 10.2\) and \(\sigma=4.8\).
\(\mu - 2\sigma=10.2-2\times4.8=10.2 - 9.6 = 0.6\)
\(\mu + 2\sigma=10.2 + 2\times4.8=10.2+9.6 = 19.8\)

Step3: Calculate the percentage for part c

We want to find the percentage of data less than \(x = 5.4\).
First, note that \(\mu-\sigma=10.2-4.8 = 5.4\).
The total area under the normal curve is 100%. The area to the left of \(\mu-\sigma\) is \(\frac{100 - 68}{2}\%=16\%\)

Answer:

a) Since the mean \(\mu = 10.2\) and standard deviation \(\sigma = 4.8\), the correct normal model is the one where \(\mu = 10.2\) and we can calculate the intervals using \(\mu\pm n\sigma\) (the one with \(\mu = 10.2\) is the correct model, assume option B is the one with \(\mu = 10.2\) as per the given values in the problem - related to the mean and standard deviation provided).
b) The central 95% of tree diameters are between \(0.6\) inches and \(19.8\) inches.
c) About \(16\%\) of the trees should have diameters below \(5.4\) inches.