Question

A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees is 12 inches with a standard deviation of 2.2 inches. Assume the distribution of diameters is roughly mound-shaped. What fraction of the trees will have diameters between 7.6 and 18.6 inches

Answer 1

Answer:

97.6% of the trees will have diameters between 7.6 and 18.6 inches.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12 inches

Standard Deviation, σ = 2.2 inches

We are given that the distribution of diameter of tree is a mound shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(diameters between 7.6 and 18.6 inches)

$P(7.6 \leq x \leq 18.6) = P(\displaystyle\frac{7.6 - 12}{2.2} \leq z \leq \displaystyle\frac{18.6-12}{2.2}) = P(-2 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -2)\\= 0.999 - 0.023 = 0.976= 97.6\%$

$P(7.6 \leq x \leq 18.6) = 97.6\%$

97.6% of the trees will have diameters between 7.6 and 18.6 inches.