22. Seattle, Washington, averages µ = 34 inches of annual μ precipitation. Assuming that the distribution of precipi- tation amounts is approximately normal with a stan- dard deviation of o= 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
22. Seattle, Washington, averages µ = 34 inches of annual μ precipitation. Assuming that the distribution of precipi- tation amounts is approximately normal with a stan- dard deviation of o= 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
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![**Problem Statement:**
Seattle, Washington, averages \( \mu = 34 \) inches of annual precipitation. Assuming that the distribution of precipitation amounts is approximately normal with a standard deviation of \( \sigma = 6.5 \) 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
**Analysis:**
This exercise requires understanding of a normal distribution where:
- The mean (\( \mu \)) is 34 inches.
- The standard deviation (\( \sigma \)) is 6.5 inches.
To classify the years, we can calculate the z-scores for each amount of precipitation and determine their position relative to the mean.
1. **Z-score Calculation:**
\[
Z = \frac{X - \mu}{\sigma}
\]
- For 41.8 inches:
\[
Z = \frac{41.8 - 34}{6.5}
\]
- For 49.6 inches:
\[
Z = \frac{49.6 - 34}{6.5}
\]
- For 28.0 inches:
\[
Z = \frac{28.0 - 34}{6.5}
\]
2. **Interpretation:**
- A typical year falls within one standard deviation from the mean.
- An extremely wet year is significantly above the mean (greater than two standard deviations).
- An extremely dry year is significantly below the mean (greater than two standard deviations in the negative direction).
By calculating these values, you can classify each year accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff56f1b7-36da-490f-a87a-f784147acbf5%2Fb79d860f-7334-4ad0-abf4-9d2e4a572144%2Feh2in5j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Seattle, Washington, averages \( \mu = 34 \) inches of annual precipitation. Assuming that the distribution of precipitation amounts is approximately normal with a standard deviation of \( \sigma = 6.5 \) 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
**Analysis:**
This exercise requires understanding of a normal distribution where:
- The mean (\( \mu \)) is 34 inches.
- The standard deviation (\( \sigma \)) is 6.5 inches.
To classify the years, we can calculate the z-scores for each amount of precipitation and determine their position relative to the mean.
1. **Z-score Calculation:**
\[
Z = \frac{X - \mu}{\sigma}
\]
- For 41.8 inches:
\[
Z = \frac{41.8 - 34}{6.5}
\]
- For 49.6 inches:
\[
Z = \frac{49.6 - 34}{6.5}
\]
- For 28.0 inches:
\[
Z = \frac{28.0 - 34}{6.5}
\]
2. **Interpretation:**
- A typical year falls within one standard deviation from the mean.
- An extremely wet year is significantly above the mean (greater than two standard deviations).
- An extremely dry year is significantly below the mean (greater than two standard deviations in the negative direction).
By calculating these values, you can classify each year accordingly.
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