## 17.8 Examples using \(z\)-scores

**Example 17.7 (Normal distributions) **Aedo-Ortiz et al. (1997)
simulated mechanized forest harvesting systems
(Devore and Berk 2007).

As part of their study, they assumed that the specific trees in their study would vary in diameter, with

- a normal distribution; with
- a mean of \(\mu=8.8\) inches; and
- a standard deviation of \(\sigma=2.7\) inches.

*greater than*than 6 inches?

Follow the steps identified earlier:

**Draw**a normal curve, and mark on 6 inches (Fig. 17.7, top panel).**Shade**the region corresponding to ‘greater than 6 inches’ (Fig. 17.7, bottom panel).**Compute**the \(z\)-score using Eq. (17.1). Here, \(x=6\), \(\mu=8.8\), \(\sigma=2.7\), so \(\displaystyle z = (6 - 8.8)/2.7 = -2.8/2.7 = -1.04\) to two decimal places.**Use**tables: The probability of a tree diameter*shorter*than 6 inches is \(0.1492\). (The tables always give area*less*than the value of \(z\) that is looked up.)**Compute**the answer: Since the*total*area under the normal distribution is one, the probability of a tree diameter*greater*than 6 inches is \(1 - 0.1492 = 0.8508\), or about 85%.

The normal-distribution tables in
the Appendix
**always** provide area to the **left**
of the \(z\)-scores that is looked up.
Drawing a picture of the situation is important:
it helps visualise how to get the answer from what the table give us.

*total*area under the normal distribution is one.

**Think 17.1 (Drawing diagrams) **Match the diagram in
Fig. 17.8
with the meaning
for the tree-diameter model (recall: \(\mu=8.8\) inches):

- Tree diameters greater than 11 inches.
- Tree diameters
*between*6 and 11 inches. - Tree diameters less than 11 inches.
- Tree diameters between 3 and 6 inches.

**Example 17.8 (Normal distributions)**Using the model for tree diameters in Example 17.7 (Aedo-Ortiz et al. 1997), what is the probability that a tree has a diameter

*between*6 and 11 inches?

First, **draw** the situation, and **shade** ‘between 6 and 10 inches’
(Fig. 17.9).
Then,
**compute** the \(z\)-scores for *both* tree diameters:

\[\begin{align*}
\text{6 inches: } &z = \displaystyle \frac{6 - 8.8}{2.7} = -1.04;\\[6pt]
\text{11 inches: } &z = \displaystyle \frac{11 - 8.8}{2.7} = 0.81.
\end{align*}\]
Table B can then be used
to find the area to the *left* of \(z = -1.04\),
and also
the area to the *left* of \(z = 0.81\).
However,
neither of these provide the area
*between*
\(z = -1.04\) and \(z = 0.81\)
(Fig. 17.10).

Looking carefully at the areas from the tables and the area sought, that area between the two \(z\)-scores is

\[ 0.7910 - 0.1492 = 0.6418; \] see the animation below. The probability that a tree has a diameter between 6 and 11 inches is about 0.6418, or about 64%.

Click on the hotspots in the following image, to see what the areas under the normal curve mean.