The distribution of heights of adult American men is approximately normal with a mean of 68 inches and a standard deviation of 2 inches. What percent of mean are at least 72 inches tall?
Thanks!!
z = (x -mean)/ sd
z = (72-68)/2
z= 4/2= 2
1-.9772 = .0228
2.28%
To find the percentage of men who are at least 72 inches tall, we need to calculate the area under the normal distribution curve to the right of 72 inches. This area represents the percentage of men who are taller than 72 inches.
To do this, we can use the Z-score formula to standardize the value 72. The Z-score formula is given by:
Z = (X - μ) / σ,
where X is the value we are interested in (72 inches in this case), μ is the mean of the distribution (68 inches in this case), and σ is the standard deviation of the distribution (2 inches in this case).
Calculating the Z-score:
Z = (72 - 68) / 2 = 4 / 2 = 2.
Once we have the Z-score, we can look up the corresponding area to the right of this Z-score in a standard normal distribution table or use statistical software. The area under the curve to the right of a Z-score represents the percentage of values greater than that Z-score.
Using a standard normal distribution table, we find that the area to the right of a Z-score of 2 is approximately 0.0228. This means that approximately 0.0228 (or 2.28%) of men are at least 72 inches tall.
Therefore, the percentage of men who are at least 72 inches tall is about 2.28%.