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.
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 16% of the trees should have diameters below 5.4 inches.
(do not round. type an integer or a decimal.)
d) about what percent of the trees should have diameters between 15.0 and 19.8 inches?
using the 68 - 95 - 99.7 rule, about 13.5% of the trees should have diameters between 15.0 and 19.8 inches.
(do not round. type an integer or a decimal.)
e) about what percent of the trees should have diameters over 15.0 inches?
using the 68 - 95 - 99.7 rule, about % of the trees should have diameters over 15.0 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 68% of the data lies within \( \mu\pm\sigma\), about 95% lies within \( \mu\pm2\sigma\), and about 99.7% lies within \( \mu\pm3\sigma\). The total percentage of data in a normal distribution is 100%.

Step2: Analyze part (c)

We know that \( \mu = 10.2\) and \( \sigma=4.8\). For diameters below \(x = 5.4\), we calculate \( \mu-\sigma=10.2 - 4.8=5.4\). According to the 68 - 95 - 99.7 rule, the percentage of data within \( \mu\pm\sigma\) is 68%. The percentage of data below \( \mu-\sigma\) is \(\frac{100 - 68}{2}=16\%\)

Step3: Analyze part (d)

We have \(x_1 = 15.0\) and \(x_2 = 19.8\). First, find \( \mu+\sigma=10.2 + 4.8 = 15.0\) and \( \mu + 2\sigma=10.2+2\times4.8=10.2 + 9.6=19.8\). The percentage of data within \( \mu+\sigma\) and \( \mu + 2\sigma\) is \(\frac{95 - 68}{2}=13.5\%\)

Step4: Analyze part (e)

We know \(x = 15.0=\mu+\sigma\). The percentage of data above \( \mu+\sigma\) is \(\frac{100 - 68}{2}=16\%\)

Answer:

c) \(16\%\)
d) \(13.5\%\)
e) \(16\%\)