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?
z = ( x -mean)/sd
z = (72-68)/2
z = 4/2 = 2
To determine the percentage of men who are at least 72 inches tall, we can use the standard deviation and mean to calculate the z-score and then look up the corresponding probability in a standard normal distribution table.
The formula to calculate the z-score is:
z = (x - mean) / standard deviation
In this case, the value of x (height) is 72 inches. Substituting the given values into the formula, we get:
z = (72 - 68) / 2
z = 4 / 2
z = 2
Now, we need to find the probability associated with a z-score of 2 using a standard normal distribution table. The table will give us the proportion of the data that falls below a given z-score, so we want to find the area to the left of z = 2.
Looking up the z-score of 2 in the table, we find that the corresponding area is approximately 0.9772.
Since we want to determine the percentage of men who are at least 72 inches tall, we subtract the obtained area from 1 to get the area to the right of z = 2 (above 72 inches):
1 - 0.9772 = 0.0228
Finally, to express the probability as a percentage, we multiply the result by 100:
0.0228 * 100 = 2.28%
Therefore, approximately 2.28% of men have a height of at least 72 inches.