statistics question please help

Question

An article indicates that 20% of male employees state that they have to pick up the slack for moms working flextime schedules. A random sample of 100 male employees working for companies offering flextime,

1)What is the probability that 22% or fewer male employees will indicate that they have to pick up the slack for moms working flextime? otherwise P(p<0.22)

2)What's the probability that 15% or fewer will indicate that they have to pick up the slack for moms working flextime? otherwise P(p < 0.15)

Answer 1

First, is the distribution of the sample approximately normal? Yes, n=100 makes this a good approximation even if the population distribution is only slightly bell shaped.

 

The standard error of a proportion of a sample is given by √(p*q/n) = √[(.2)(.8)/100] = .04.

 

A z score is calculated using the standard formula z = (x-µ)/σ where σ is the standard error because the question is in regard to the distribution of the sample mean.

 

1) z = (.22 - .2)/.04 = .02/.04 = 0.5

 

(Use probability table or calculator for the standard normal distribution)

 

P(p<.22) = P(Z<0.5) = 0.69146

 

2) z = (.15 - .2)/.04 = .05/.04 = -1.25

 

P(p<.15) = P(Z<-1.25) = 0.10565