Question

The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard deviation 4 inches. Which of the following represents the middle 80 percent of the heights ? A. 2.5% B. 5% C. 16% D. 1%

Answer 1

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have

$0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)$

Thus

$\phi(y) = 1-0.1 = 0.9$

by looking at the table, we find that y = 1.28, therefore

$\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12$

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

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