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

Answer:

0.1112

Step-by-step explanation:

Given data:

sample size ( N ) = 150

proportion of population (p) = 50% = 0.500

Determine :

P( x > 0.55 ) = P ( Z > $\frac{0.55 - 0.5}{0.0408}$ )   -------- (1)

бp = $\sqrt{\frac{p(1-p)}{n} }$ = $\sqrt{\frac{0.5(1-0.5)}{150} }$  = 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