Question

The Bureau of Labor Statistics reported that in 2016, the median weekly earnings for people employed full time

in the United States was $837. A sample of 150 full time employees is chosen. What is the probability that

more than 55% of them earned more than $837?

Answer 1

0.1112

Explanation:

Given data:

sample size ( N ) = 150

proportion of population (p) = 50% = 0.500

Determine :

P( x > 0.55 ) = P ( Z >
`(0.55 - 0.5)/(0.0408)` ) -------- (1)

бp =
`\sqrt{(p(1-p))/(n) }` =
`\sqrt{(0.5(1-0.5))/(150) }` = 0.0408 ( standard error of proportion )

Back to equation 1

P( x > 0.55 ) = P ( Z > 1.22 )

= 1 - P ( Z < 1.22 )

Hence

P( x < 0.55 ) = 0.1112