Question

seattle, washington, averages μ = 34 inches of annual precipitation. assuming that the distribution of precipitation amounts is approximately normal with a standard - deviation of σ = 6.5 inches, determine whether each of the following represents a fairly typical year, an extremely wet year, or an extremely dry year.
a. annual precipitation of 41.8 inches
b. annual precipitation of 49.6 inches
c. annual precipitation of 28.0 inches

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data point, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 34$ inches and $\sigma=6.5$ inches.
For $x = 41.8$ inches:
$z_1=\frac{41.8 - 34}{6.5}=\frac{7.8}{6.5}=1.2$
For $x = 49.6$ inches:
$z_2=\frac{49.6 - 34}{6.5}=\frac{15.6}{6.5}=2.4$
For $x = 28.0$ inches:
$z_3=\frac{28.0 - 34}{6.5}=\frac{- 6}{6.5}\approx - 0.92$

Step2: Interpret z - scores

In a normal distribution, z - scores between - 2 and 2 are considered fairly typical.
A z - score of 1.2 (for 41.8 inches) is within the range of fairly typical values.
A z - score of 2.4 (for 49.6 inches) is outside the range of fairly typical values, indicating an extremely wet year.
A z - score of approximately - 0.92 (for 28.0 inches) is within the range of fairly typical values.

Answer:

a. An extremely wet year: 49.6 inches
b. A fairly typical year: 41.8 inches and 28.0 inches