Question

The average weekly work hours of full-time U.S. workers have approximately a normal distribution with mean 37 hours and standard deviation 10 hours. Suppose 200 U.S. citizens are randomly chosen. Find an approximate probability that less than 10 of them are working more than 50 hours a week on average (round off to third decimal place).

Answer 1

Ans:

z=(50-37)/10

z=1.3

P(z>1.3)=0.0968

Now use binomial distribution with n=200 and p=0.0968

P(x<10)=P(x<=9)=binomdist(9,200,0.0968,true)=0.005

or we can use normal approximation:

mean=200*0.0968=19.36

standard deviation=sqrt(200*0.0968*(1-0.0968))=4.182

P(x<10)=P(x<=9)

z(8.5)=(8.5-19.36)/4.182=-2.60

P(z<-2.60)=0.0047