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15 votes
7#Suppose that the heights of adult men in the United States are normally distributed with a mean of 71 inches and a standard deviation of 3.5 inches. What proportion of the adult men in the United States are at least 6 feet tall? (Hint: 6 feet =72 inches.) Round your answer to at least four decimal places.

User Vee
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1 Answer

12 votes
12 votes

ANSWER :

0.3859

EXPLANATION :

The z-score formula is :


z=\frac{x-mean}{standard\text{ }deviation}

The mean is 71 inches and the standard deviation is 3.5 inches

Calculate z when x = 6 feet or 72 inches


z=(72-71)/(3.5)=0.286\sim0.29

Using the normal distribution table :

0.6141 is the area under the curve or the percentage of adult men that are below 6 feet, since the area of the curve is to the left of z.

We need the area under the curve which is to the right of z.

Recall that the total area under the curve is 1.

Then we need to subtract 0.6141 from 1.

1 - 0.6141 = 0.3859

7#Suppose that the heights of adult men in the United States are normally distributed-example-1
User AJG
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