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?

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 > )   -------- (1)

бp = =  = 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

i think it might be a but you might want a second opinion

$3720

step-by-step explanation:

taxable income = $31000

as per the 2018-2019 us tax bracket, the tax rate is 12% for an individual whose taxable income is between $9,526 to $38,700.

so, the tax to be paid by him = 12% of 31000

= 0.12 * 31000     [12% converted to decimal as 0.12]

= $3720