91Ó°ÊÓ

Over roughly the past 100 years, the mean monthly April precipitation in Williamstown, Massachusetts, equaled 3.6 inches with a standard deviation of 1.6 inches. (Source: http://web.williams.edu/ weather/) a. In April \(1983,\) the wettest April on record, the precipitation equaled 8.4 inches. Find its z-score. If the distribution of precipitation were roughly normal, would this be unusually high? Explain. b. Assuming a normal distribution, an April precipitation of 4.5 inches corresponds to what percentile? c. Of the 119 measurements of April precipitation on record (reaching as far back as 1892 ), \(66.5 \%\) fell within one, \(97.5 \%\) within two, and \(99.1 \%\) within three standard deviations of the mean. Do you think that the distribution of April precipitation is approximately normal? Why or why not?

Step by step solution

01

Calculate the Z-score for April 1983

To find the Z-score for April 1983, we use the formula for the Z-score, which is \( Z = \frac{X - \mu}{\sigma} \). Here, \( X = 8.4 \) inches (the precipitation in April 1983), \( \mu = 3.6 \) inches (mean precipitation), and \( \sigma = 1.6 \) inches (standard deviation). Thus, \( Z = \frac{8.4 - 3.6}{1.6} = \frac{4.8}{1.6} = 3.0 \).

02

Interpret the Z-score

A Z-score of 3.0 indicates that the April 1983 precipitation was three standard deviations above the mean. In a normal distribution, a Z-score of 3 or greater is considered unusually high, as it suggests the event is in the extreme upper tail of the distribution.

03

Calculate the Z-score for April precipitation of 4.5 inches

We repeat the Z-score calculation for \( X = 4.5 \) inches. Using the same formula \( Z = \frac{X - \mu}{\sigma} \), we find \( Z = \frac{4.5 - 3.6}{1.6} = \frac{0.9}{1.6} \approx 0.5625 \).

04

Convert Z-score to percentile

We use a standard normal distribution table to find the percentile corresponding to a Z-score of approximately 0.5625. This Z-score converts to about the 71st percentile, meaning 71% of April precipitation records are below 4.5 inches.

05

Analyze the distribution for normality

To determine if the precipitation distribution is normal, consider the empirical rule, which states that for a normal distribution: about 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3. Given the provided percentages for Williamstown (66.5%, 97.5%, 99.1%), the data roughly follows these rules, suggesting the distribution is approximately normal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score

The Z-score is an important concept in statistics when dealing with normal distribution. It tells you how many standard deviations an element is from the mean. Knowing how to calculate a Z-score can help you understand if a data point is typical or unusual.

• Formula for Z-score: \[Z = \frac{X - \mu}{\sigma}\]- Here, \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
• A Z-score of 0 means the data point is exactly average.
• Positive Z-score: Data point is above the mean.
• Negative Z-score: Data point is below the mean.

In the case of Williamstown, the April 1983 precipitation had a Z-score of 3.0. This tells us it was three standard deviations above the mean of 3.6 inches, signaling an unusually high event in terms of precipitation.

Standard Deviation

Standard deviation is a key measure of how spread out the numbers in a data set are. It quantifies the amount of variation or dispersion. It's crucial when determining how different data sets are distributed.

• Low standard deviation: Data points are close to the mean.
• High standard deviation: Data points are spread out over a wider range.

For example, in Williamstown's April precipitation data, the standard deviation was 1.6 inches. This means that most of the month's precipitation figures fall within 1.6 inches of the average over the years. A smaller standard deviation means less variation, while a larger one indicates more variability.

Percentile

Percentiles are used to understand and interpret the relative standing of a data point within a data set. In the context of a normal distribution, they help identify the position of a score in relation to the rest of the data.

• A percentile indicates the percentage of scores that fall below a particular value.
• For example, the 50th percentile is the median.
• Finding a percentile involves looking at the cumulative distribution function (CDF).

In our example, a Z-score of approximately 0.5625 converted to the 71st percentile. This means that 71% of the recorded April precipitation totals were below 4.5 inches, which helps users understand how common or rare a certain occurrence is in the broader context.

Empirical Rule

The empirical rule is a rough guideline for understanding the distribution of data in a normal distribution. Sometimes called the 68-95-99.7 rule:

• 68% of the data falls within one standard deviation of the mean.
• 95% within two standard deviations.
• 99.7% within three standard deviations.

Applying these percentages to the data from Williamstown’s April precipitation, 66.5% fell within one standard deviation, 97.5% within two, and 99.1% within three standard deviations. The close match to the empirical rule suggests the data is approximately normal, confirming the assumption used in such statistical analyses.