Question

A global research study found that the majority of​ today's working women would prefer a better​ work-life balance to an increased salary. One of the most important contributors to​ work-life balance identified by the survey was​ "flexibility," with 41​% of women saying that having a flexible work schedule is either very important or extremely important to their career success. Suppose you select a sample of 100 working women. Answer parts​ (a) through​ (c).

If a sample of 400 is​ taken, how does this change your answers to​ (a) through​ (c)?

A) The probability that in the sample fewer than 47​% say that having a flexible work schedule is either very important or extremely important to their career success is?

B) The probability that in the sample between 36% and 47​% say that having a flexible work schedule is either very important or extremely important to their career success is?

C) The probability that in the sample more than 37​% say that having a flexible work schedule is either very important or extremely important to their career success is?

Answer 1

a)

population proportion ,p=   0.41      
n=   400      
          
std error , SE = √( p(1-p)/n ) =    0.024592      
          
sample proportion , p̂ =   0.47      
Z=( p̂ - p )/SE=   2.440      
P ( p̂ <    0.47   ) =P(Z<( p̂ - p )/SE) =  
          
=P(Z <    2.440   ) =    0.9927

The probability that in the sample fewer than 47​% say that having a flexible work schedule is either very important or extremely important to their career success is 0.9927

B)

                       population proportion ,p=   0.41          
                       n=   400          
                                      
                       std error , SE = √( p(1-p)/n ) =    0.0246          
                                      
                       we need to compute probability for               
                       0.36   < p̂ <   0.47      
                                      
                       Z1 =( p̂1 - p )/SE=   -2.033          
                       Z2 =( p̂2 - p )/SE=   2.440          
P(   0.36   < p̂ <   0.47   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -2.033   < Z <   2.440   )
                                      
= P ( Z <   2.440   ) - P (    -2.033   ) =    0.9927   -   0.021   =   0.9716  
The probability that in the sample between 36% and 47​% say that having a flexible work schedule is either very important or extremely important to their career success is = 0.9716

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c)

population proportion ,p=   0.41      
n=   400      
          
std error , SE = √( p(1-p)/n ) =    0.025      
          
sample proportion , p̂ =   0.37      
Z=( p̂ - p )/SE=   -1.627      
P ( p̂ >    0.37   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   -1.627   ) =    0.9481

The probability that in the sample more than 37​% say that having a flexible work schedule is either very important or extremely important to their career success is 0.9481