Question

Suppose that the helghts of aduit men in the United States are normally distributed with mean of 69 inches and a standard deviation of 3.5 inches. What proportion of the adult men in the United 5 tates are less than 6 feet tall? (Hint: 6 feet =72 inches.) Round your answer to at least faur becimal places

Answer 1

Approximately 80.47% of adult men in the United States are less than 6 feet tall.

Step-by-step explanation:

To find the proportion of adult men in the United States that are less than 6 feet tall, which is equal to 72 inches, we need to use the properties of the normal distribution.

Given that the heights of adult men are normally distributed with a mean of 69 inches and a standard deviation of 3.5 inches, we can use the z-score formula to convert the height of 72 inches into a z-score.

The formula for calculating the z-score is: z = (x - mean) / standard deviation. Plugging in the values, we get z = (72 - 69) / 3.5 = 0.857.

Now, we can use a standard normal distribution table or a calculator to find the proportion of values that are less than a given z-score. From the table, we find that the proportion to the left of 0.857 is 0.8047. Therefore, approximately 80.47% of adult men in the United States are less than 6 feet tall.