Question

Course Note Packet Chapter 7 Page 20-22 (Proportion Probabilities Template) 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 contbutors to work life balance identified by the survey was exibility with 49% 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 57% 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) b. What is the probability that in the sample between 44% and 57% 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 56% 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 57% 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 44% and 57% 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 56% say that having a flexible work schedule is either very important or extremely important to their career success is Enter your answer in the edit fields and then click Check Answer Check Answer Clear All 6:40 PM All parts showing A 3/4/2020 0 e 9

Answer 1

a)

population proportion ,p=   0.49                          
n=   100                          
                              
std error , SE = √( p(1-p)/n ) =    0.0500                          
                              
sample proportion , p̂ =   0.57                          
Z=( p̂ - p )/SE= (   0.570   -   0.49   ) /    0.050   =   1.600  
P ( p̂ <    0.570   ) =P(Z<( p̂ - p )/SE) =                      
                              
=P(Z <    1.600   ) =    0.9452                   (answer)

b)

                       population proportion ,p=   0.49                      
                       n=   100                      
                                                  
                       std error , SE = √( p(1-p)/n ) =    0.0120                      
                                                  
                       we need to compute probability for                           
                       0.44   < p̂ <   0.57                  
                                                  
                       Z1 =( p̂1 - p )/SE= (   0.44   -   0.49   ) /    0.0120   =   -4.167
                       Z2 =( p̂2 - p )/SE= (   0.57   -   0.49   ) /    0.0120   =   6.667
P(   0.44   < p̂ <   0.57   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -4.167   < Z <   6.667   )          
                                                  
= P ( Z <   6.667   ) - P (    -4.167   ) =    1.0000   -   0.0000   =   0.999985              

c)

population proportion ,p=   0.49                      
n=   100                      
                          
std error , SE = √( p(1-p)/n ) =    0.0500                      
                          
sample proportion , p̂ =   0.56                      
Z=( p̂ - p )/SE= (   0.56   -   0.49   ) /    0.0500   =   1.400
P ( p̂ >    0.56   ) =P(Z > ( p̂ - p )/SE) =                  
                          
=P(Z >   1.400   ) =    0.0807              

d)

i) 0.9993

ii) 0.999985
iii) 0.0026

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