Question

According to the National Health Survey, the heights of adult males in the United States are normally distributed with mean 69.0 inches and standard deviation 2.8 inches. (a) What is the probability that an adult male chosen at random is between 66 and 72 inches tall? (Round your answer to three decimal places.)

Answer 1

Answer:

P(66 < x < 72) = 0.715

Step-by-step explanation:

you must find the probability P(66 < x < 72)

so you can find both, P(x<66) and P(x<72), then substract them.

therefore, using the normal distribution:

let´s start with P(x<66)

finding z:

Z = (x-μ)/σ/($\sqrt{n}$) [1]

where:

x is the raw measurement

μ is the mean

σ is the standard deviation

n is the sample size

our information is:

μ = 69.0 inches, σ = 2.8 inches, n = 1, x = 66 and x = 72

so replacing:

$Z = \frac{(66-69)}{\frac{2.8}{\sqrt{1} }}\\\\Z = -\frac{(3)}{2.8}\\\\Z = -1.07$

now you look at the normal distribution table and find Z score:

an this is 0.1423, you can check in the attached table.

so P(x<66) = 0.1423.

now finding P(x<72), doing the same above but this time x = 72:

$Z = \frac{(72-69)}{\frac{2.8}{\sqrt{1} }}\\\\Z = \frac{(3)}{2.8}\\\\Z = 1.07$

looking in the table the Z score:

this is 0.8577, check the table.

now:

P(66 < x < 72) = P(x<72) - P(x < 66)

P(66 < x < 72) = 0.8577 - 0.1423

P(66 < x < 72) = 0.7154 or 71.54%

rounding to three decimal places

P(66 < x < 72) = 0.715