The middle 80 percent of the heights for adult men in the given population ranges from approximately 64.88 inches to 75.12 inches, calculated based on the z-scores corresponding to the 10th and 90th percentiles in a normal distribution with a mean of 70 inches and a standard deviation of 4 inches.
To find the middle 80 percent of the heights for adult men with a normal distribution, we look for the heights that correspond to the 10th percentile and the 90th percentile, because these percentiles divide the lowest and highest 10 percent from the middle 80 percent. We use the standard normal distribution and transform it using the mean and standard deviation of the population.
Steps to Find the Middle 80 Percent of Heights
Identify the z-scores that correspond to the 10th and 90th percentiles. Since the distribution is symmetric, the z-scores will be the same distance from the mean but with opposite signs.
Consult a z-table, statistical software, or a graphing calculator to find the z-scores that correspond to the cumulative probabilities of 0.1 and 0.9, respectively.
Use the formula X = μ + (z • σ) to convert the z-scores to the normal distribution with mean (μ) of 70 inches and standard deviation (σ) of 4 inches.
Calculate the heights for the 10th and 90th percentiles.
Assuming a normal distribution, the z-score for the 10th percentile is approximately -1.28 and for the 90th percentile is approximately 1.28. Using the formula, the heights corresponding to these z-scores are:
10th percentile: X = 70 + (-1.28 • 4) = 64.88 inches
90th percentile: X = 70 + (1.28 • 4) = 75.12 inches
Therefore, the middle 80 percent of the heights range from approximately 64.88 inches to 75.12 inches.