Question

Weekly paycheck: The Bureau of Labor Statistics reported that in a recent year, the median weekly earnings for people employed full time in the United States was $755.

a) What proportion of full-time employees had weekly earnings of more than $755?

b) A sample of 150 full-time employees is chosen. What is the probability that more than 55% of them earned more than $755 per week?

c) What is the probability that less than 60% of the sample of 150 employees earned more than $755 per week?

d) What is the probability that between 45% and 55% of the sample of 150 employees earned more than $755 per week?

e) Would it be unusual if less than 45% of the sample of 150 employees earned more than $755 per week?

Answer 1

a)proportion of full-time employees had weekly earnings of more than $755 =0.5

b)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.500
sample size       =n=	150
std error of proportion=σp=√(p*(1-p)/n)=	0.0408

  probability that more than 55% of them earned more than $755 per week:

probability =

P(X>0.55)

=

P(Z>1.22)=

1-P(Z<1.22)=

1-0.8888=

0.1112

c)

probability that less than 60% of the sample of 150 employees earned more than $755 per week

probability =

P(X<0.6)

=

P(Z<2.45)=

0.9929

d)

probability =

P(0.45<X<0.55)

=

P(-1.22<Z<1.22)=

0.8888-0.1112=

0.7776

e)

probability of having less than 45% =0.1112 ; as this is not less than 0.05 therefore it is not unusual