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 42​% 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​ (d).

a. What is the probability that in the sample fewer than 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success?

ANSWER= .7283

​(Round to four decimal places as​ needed.)

b. What is the probability that in the sample between 34​% and 45​% say that having a flexible work schedule is either very important or extremely important to their career​ success?

​(Round to four decimal places as​ needed.)

c. What is the probability that in the sample more than 39​% say that having a flexible work schedule is either very important or extremely important to their career​ success?

​(Round to four decimal places as​ needed.)

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

The probability that in the sample fewer than 45​% say that having a flexible work schedule is either very important or extremely important to their career success is ___ .

The probability that in the sample between 34% and 45​% say that having a flexible work schedule is either very important or extremely important to their career success is ___ .

The probability that in the sample more than 39​% 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=   0.42      
n=   100      
          
std error , SE = √( p(1-p)/n ) =    0.049356      
          
sample proportion , p̂ =   0.45      
Z=( p̂ - p )/SE=   0.608      
P ( p̂ <    0.45   ) =P(Z<( p̂ - p )/SE) =  
          
=P(Z <    0.608   ) =    0.7284(answer)

b)

                       population proportion ,p=   0.42          
                       n=   100          
                                      
                       std error , SE = √( p(1-p)/n ) =    0.0494          
                                      
                       we need to compute probability for               
                       0.34   < p̂ <   0.45      
                                      
                       Z1 =( p̂1 - p )/SE=   -1.621          
                       Z2 =( p̂2 - p )/SE=   0.608          
P(   0.34   < p̂ <   0.45   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -1.621   < Z <   0.608   )
                                      
= P ( Z <   0.608   ) - P (    -1.621   ) =    0.7284   -   0.053   =   0.6758 (answer)

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c)

population proportion ,p=   0.42      
n=   100      
          
std error , SE = √( p(1-p)/n ) =    0.049      
          
sample proportion , p̂ =   0.39      
Z=( p̂ - p )/SE=   -0.608      
P ( p̂ >    0.39   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   -0.608   ) =    0.7284(answer)

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d)

population proportion ,p=   0.42      
n=   400      
          
std error , SE = √( p(1-p)/n ) =    0.024678      
          
sample proportion , p̂ =   0.45      
Z=( p̂ - p )/SE=   1.216      
P ( p̂ <    0.45   ) =P(Z<( p̂ - p )/SE) =  
          
=P(Z <    1.216   ) =    0.8879
The probability that in the sample fewer than 45​% say that having a flexible work schedule is either very important or extremely important to their career success is 0.8879

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                       population proportion ,p=   0.42          
                       n=   400          
                                      
                       std error , SE = √( p(1-p)/n ) =    0.0247          
                                      
                       we need to compute probability for               
                       0.34   < p̂ <   0.45      
                                      
                       Z1 =( p̂1 - p )/SE=   -3.242          
                       Z2 =( p̂2 - p )/SE=   1.216          
P(   0.34   < p̂ <   0.45   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -3.242   < Z <   1.216   )
                                      
= P ( Z <   1.216   ) - P (    -3.242   ) =    0.8879   -   0.001   =   0.8873  
The probability that in the sample between 34% and 45​% say that having a flexible work schedule is either very important or extremely important to their career success is 0.8873

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population proportion ,p=   0.42      
n=   400      
          
std error , SE = √( p(1-p)/n ) =    0.025      
          
sample proportion , p̂ =   0.39      
Z=( p̂ - p )/SE=   -1.216      
P ( p̂ >    0.39   ) =P(Z > ( p̂ - p )/SE) =  
          
=P(Z >   -1.216   ) =    0.8879
The probability that in the sample more than 39​% say that having a flexible work schedule is either very important or extremely important to their career success is 0.8879