Question

1. A forester measured a sample of trees in a tract of land being sold for a lumber harvest. Among 27 trees, she found a mean diameter of 10.4 inches and a standard deviation of 4.7 inches. Suppose her sample gives an accurate representation of the entire tract of land and that the tree diameters follow a normal distribution. Round to 2 decimal places, when applicable.

(a) Sketch a graph of the distribution of tree diameters, labeling the mean and one standard deviation in either direction.

(b) What diameter would you expect the central 95% of trees to be?

(c) What percentage of trees should be less than 1 inch in diameter?

(d) What percentage of trees should be between 10.4 and 19.8 inches in diameter?

(e) What is the minimum diameter a tree can have in order to be placed in the top 5% of all such trees on the tract regarding their size?

(f) Suppose the forester looked at another tract of land and found that those tree’s diameters also followed a normal distribution with a mean diameter of 12.3 inches with a standard deviation of 2.25 inches. Would it be more unusual to find a tree with a diameter of 15 inches in the first tract of land described or in the second one described in this part? Hint: use z-scores. You must support your work mathematically for full credit, simply saying “1^st tract” or “2^nd tract” is not sufficient.

Answer 1

Given = 10.4, = 4.7

To find the probability, we need to find the z scores.

____________________

(a) In a normal (bell curved) distribution, the mean would be at the center and the rest of the distribution would have values to the left and to the right. One standard deviation to the right of the mean = 10.4 + 4.7 = 15.1 inches and one standard deviation to the left would be 10.4 - 4.7 = 5.7 inches. The diagram is given below

__________________

(b) 95% of all diameters centered about the mean.

This means that the remaining 5% will be distributed equally to the right and the left = 0.05/2 = 0.025

So the lower p value = 0.025 and the Upper p value = 1 - 0.025 = 0.975

The Z score at p = 0.025 and 0.975 are -1.96 and +1.96 respectively.

The Lower value: (X - 10.4) / 4.7 = -1.96. Solving for X, X = (-1.96 * 4.7) + 10.4 = 1.19 inches

The Upper value: (X - 10.4) / 4.7 = 1.96. Solving for X, X = (1.96 * 4.7) + 10.4 = 19.61 inches

Therefore 95% will be distributed about 1.19 < X < 19.61

_________________________________________

(c) For P( X < 1)

Z = (1 – 10.4) / 4.7 = -2.00

The required percentage from the normal distribution tables is = 0.0228 * 100 = 2.28%

__________________________________________________________

(d) Between 10.4 and 19.8 = P(10.4 < X < 19.8) = P(X <19.8) - P(X < 10.4)

For P(X < 19.8) ; z = (19.8 - 10.4) / 4.7 = 2. The p value at this score is = 0.9772

For P(X < 10.4) ; z = (10.4 - 10.4) / 4.7 = 0. The p value at this score is = 0.5

Therefore the required percentage is (0.9772 – 0.5) * 100 = 0.4772 * 100 = 47.72%

________________________________________________________

(e) To be in the top 5%, therefore 95% will be below this value.

To find P(X < x) = 0.95

From the standard normal distribution table, the z score at a p value of 0.95 Is 1.645

Therefore 1.645 = (X – 10.4) / 4.7

Solving for X, we get X = (1.645 * 4.7) + 10.4 = 18.13 inches

_______________________

(f) For the 1st Tract, For P(X < 15); z = (15 - 10.4) / 4.7 = 0.98.

For the 2nd Tract, For P(X < 15); z = (15 - 12.3) / 2.25 = 1.2.

Since the z score for Tract 2 is greater (Which means it is further away from the mean by 1.2 SD as compared to 0.98 SD in Tract 1), it is more unusual in tract 2.

_________________________