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Answer :
Final answer:
The probability that a randomly selected young American man aged 18 to 24 will be greater than 6 feet tall is 5.48%, after calculating the z-score and subtracting the cumulative probability from 1.
Explanation:
To find the likelihood that a randomly selected young man between the age of 18 and 24 is greater than 6 feet tall, we would first need to convert 6 feet into inches, as the average height is given in inches. Since 1 foot equals 12 inches, 6 feet is 72 inches. We can use the standard normal distribution to find this probability. The z-score is calculated using the formula z = (X - μ) / σ, where X is the height of interest (72 inches), μ is the mean (68 inches), and σ is the standard deviation (2.5 inches).
So, the z-score for 72 inches is:
z = (72 - 68) / 2.5 = 1.6
Using a standard normal distribution table or calculator, we find the probability for a z-score of 1.6. The table gives us the area to the left of the z-score, so to find the probability of a young man being taller than 6 feet, we subtract this value from 1.
If we denote the cumulative probability of a z-score of 1.6 as P(z < 1.6), then the probability that a young man is taller than 6 feet is 1 - P(z < 1.6). Suppose P(z < 1.6) = 0.9452, then the probability of being taller than 6 feet is 1 - 0.9452 = 0.0548 or 5.48%.
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