Question

A forester measured 27 of the trees in a large woods that is up for sale. He found a mean diameter of \(10.4\) inches and a standard deviation of \(4.7\) inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies. a) Draw the Normal model for tree diameters. b) What size would you expect the central \(95 \%\) of all trees to be? c) About what percent of the trees should be less than an inch in diameter? d) About what percent of the trees should be between \(5.7\) and \(10.4\) inches in diameter? e) About what percent of the trees should be over 15 inches in diameter?

Short answer

a) Sketch a Normal curve centered at 10.4 with intervals of 4.7. b) 1 to 19.8 inches. c) About 2.28%. d) About 34.13%. e) About 16.45%.

Step-by-step solution

Understanding the Normal Model

The Normal distribution is described by its mean (\( \mu \)) and standard deviation (\( \sigma \)). Here, the mean diameter is \( 10.4 \) inches, and the standard deviation is \( 4.7 \) inches. Hence, our Normal model for tree diameters can be denoted as \( N(10.4, 4.7) \).

Drawing the Normal Model

To draw the Normal model, sketch a symmetric bell-shaped curve centered at the mean (\( 10.4 \) inches). Mark one standard deviation (\( 4.7 \)) away both on the left and right from the mean. These points are \( 10.4 - 4.7 = 5.7 \) and \( 10.4 + 4.7 = 15.1 \). Continue marking points for 2 standard deviations (\( 10.4 \pm 2 \times 4.7 \)) and 3 standard deviations (\( 10.4 \pm 3 \times 4.7 \)).

Calculating Central 95% of Trees

For a Normal distribution, approximately 95% of data falls within two standard deviations from the mean. Therefore, the central 95% of tree diameters are between \( 10.4 - 2 \times 4.7 = 1 \) inch and \( 10.4 + 2 \times 4.7 = 19.8 \) inches.

Calculating Percent of Trees Less than 1 Inch

For values less than 1 inch, calculate the Z-score: \( Z = \frac{1 - 10.4}{4.7} = -2 \). The percentage of trees with diameters less than this Z-score can be found using a Z-table or calculator, which gives about 2.28%.

Calculating Percent of Trees between 5.7 and 10.4 inches

First, find Z-scores for 5.7 and 10.4:- For 5.7 inches: \( Z = \frac{5.7 - 10.4}{4.7} = -1 \).- For 10.4 inches: \( Z = \frac{10.4 - 10.4}{4.7} = 0 \).The cumulative probability associated with \( Z = -1 \) is about 15.87%, and for \( Z = 0 \) is 50%. The percent of trees between these Z-scores is \( 50\% - 15.87\% = 34.13\% \).

Calculating Percent of Trees Over 15 inches

Calculate the Z-score for 15 inches: \( Z = \frac{15 - 10.4}{4.7} = 0.98 \). Use a Z-table or calculator to find the cumulative probability up to this Z-score (about 83.55%). Thus, the percent of trees over 15 inches is \( 100\% - 83.55\% = 16.45\% \).

Key concepts

Standard Deviation

The standard deviation is an essential statistical concept used to measure the amount of variation or dispersion in a set of values. In simple terms, it tells us how much the values of a dataset deviate from the mean (average) of the dataset.
If the standard deviation is small, the values tend to be close to the mean. If it's large, the values are spread out over a wider range.

For example, the forester measured a standard deviation of \(4.7\) inches for the tree diameters. This means that most of the tree diameters tend to differ by about \(4.7\) inches from the mean diameter of \(10.4\) inches.
Understanding the standard deviation helps in interpreting data variability and making predictions, especially when considering a Normal distribution.

Mean Diameter

The mean diameter is simply the average diameter of a set of trees. It is calculated by adding up all the tree diameters and dividing by the number of trees measured.
The mean provides a central value that represents an entire dataset. In our exercise, the mean diameter of the trees in the woods is \(10.4\) inches, which means that if all the tree diameters were roughly the same, they would each be \(10.4\) inches.
This value is central to drawing and understanding the Normal distribution of the data, as the Normal model is symmetric around the mean. Thus, in a Normal model, most of the tree diameters are clustered around this mean.

Z-score

A Z-score is a measurement that describes a value's position relative to the mean of a group of values, expressed in terms of standard deviations.
Using the formula:
\[ Z = \frac{x - \mu}{\sigma} \]
where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. The Z-score helps us understand how far away a given value is from the mean in standard deviation units.
For example, a Z-score of \(-2\) means the value is 2 standard deviations below the mean. In our exercise, trees less than an inch in diameter have a Z-score of \(-2\), indicating they are quite rare compared to the mean of \(10.4\) inches.
Z-scores are crucial for determining probabilities and percentiles within a data set.

Percentile Calculation

Percentile calculation involves determining the relative standing of a particular value within a dataset. It tells us the percentage of values in the data fall below a particular value.
Percentiles are crucial when evaluating the position of a particular measurement within a distribution. To calculate percentiles associated with tree diameters, we use Z-scores and a Z-table or mathematical tools.
For example, the calculation shows that about 2.28% of trees have a diameter less than 1 inch. Additionally, roughly 16.45% of the trees have a diameter over 15 inches. These calculations help in understanding and predicting the variability and expectation within a dataset.