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When dealing with normal distributions, the Z-score helps us standardize individual data points. A Z-score tells us how many standard deviations an element is from the mean, using the formula: \( Z = \frac{X - \mu}{\sigma} \). Here, \(X\) represents the value we're examining, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For example, if a man's height is 64 inches, with \(\mu = 68.6\) inches and \(\sigma = 2.8\) inches, we find the Z-score: \( Z = \frac{64 - 68.6}{2.8} \approx -1.64 \).

This Z-score of -1.64 tells us 64 inches is 1.64 standard deviations below the mean height.

Cumulative probabilities let us determine the likelihood of a Z-score being below a certain value. These probabilities are found using the Z-table, which provides the area under the curve of a standard normal distribution up to a given Z-score.

For instance, if we look up a Z-score of -1.64 on the Z-table, we get approximately 0.0505. This number tells us that only about 5.05% of men are shorter than 64 inches.

To find the proportion of men within a certain height range, subtract the cumulative probability at the lower limit from the upper limit. For heights between 64 inches and 77 inches, the cumulative probabilities are approximately 0.0505 and 0.9987 respectively, giving us a final probability of 0.9482, or 94.82%.

Percentile ranks represent the percentage of scores below a specific point in a dataset. In a normal distribution, certain Z-scores correspond to specific percentiles.

For instance, the 2.5th percentile corresponds to a Z-score of -1.96, and the 97.5th percentile corresponds to a Z-score of 1.96. These values help determine thresholds that exclude the top and bottom 2.5% of a distribution.

Using these percentiles ensures that the height requirements exclude only the shortest and tallest men, offering a more focused range.

The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. All normal distributions can be transformed to this standard form by converting individual elements into Z-scores.

With a Z-score table, you can easily find the probabilities associated with any value. This forms the foundation for solving problems involving probabilities and percentiles in normally distributed data.

In our example, the U.S. Air Force has height requirements for pilots. Originally, these were between 64 inches and 77 inches.

By converting these limits to Z-scores, we find \(Z_{lower} = -1.64\) and \(Z_{upper} = 3.0\).

Changing these requirements to exclude only the shortest 2.5% and the tallest 2.5% corresponds to new Z-scores. With Z-scores of -1.96 and 1.96, we can calculate the adjusted height limits: \(X_{lower} = (-1.96 \times 2.8) + 68.6 \approx 63.1\) inches and \(X_{upper} = (1.96 \times 2.8) + 68.6 \approx 74.1\) inches.

These new limits ensure a well-balanced selection of pilot height requirements that exclude outliers at both ends.