Question

The mean height of men in the United States ( ages 20-29 ) is 69.6 inches . A random sample of 58 men in this age group is selected . What is the probability that the mean height for the sample is greater than 71 inches ? Assume sigma = 3.05 inches .

Answer 1

z-score is calculated as follows:

$z=\frac{x-\mu}{\sigma}$

where x is the raw score, μ is the mean, and σ is the standard deviation.

Substituting with x = 71 in, μ = 69.6 in, and σ = 3.05 in, we get:

$z=\frac{71-69.6}{3.05}=\frac{1.4}{3.05}=0.46$

We want to know the probability of z greater than 0.46. We can do this with the help of the next table:

(Notice that the table shows the probability of z less than some z-score)

P(Z > 0.46) = 1 - P(Z < 0.46) = 1 - 0.6772 = 0.3228