Final answer:
Approximately 99.89% of college-aged males fall between 58.7 inches and 77.9 inches.
The standardized z-score for a college-aged male of 75 inches is approximately 2.09375.
Step-by-step explanation:
To calculate the percentage of college-aged males that fall between 58.7 inches and 77.9 inches, you can use the empirical rule.
This rule states that approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
So, to find the percentage between 58.7 inches and 77.9 inches, you need to calculate how many standard deviations each value is from the mean.
To do this, subtract the mean from each value and divide it by the standard deviation. The z-scores for 58.7 inches and 77.9 inches are -3.2813 and 3.2813, respectively.
Then, you can use a z-table to find the percentage of data that falls between these two z-scores.
From the z-table, you can see that the area to the left of -3.2813 is 0.00055 and the area to the left of 3.2813 is 0.99945.
To find the percentage between these two z-scores, subtract the smaller area from the larger area:
0.99945 - 0.00055 = 0.9989.
Therefore, approximately 99.89% of college-aged males fall between 58.7 inches and 77.9 inches.
To find the standardized z-score for a college-aged male of 75 inches, you can use the formula:
z = (x - mean) / standard deviation.
Plug in the values: x = 75 inches, mean = 68.3 inches, and standard deviation = 3.2 inches.
Then, calculate: z = (75 - 68.3) / 3.2
= 2.09375.
Therefore, the standardized z-score for a college-aged male of 75 inches is approximately 2.09375.