Question

An institute reported that

68% of its members indicate that lack of ethical culture within financial firms has contributed most to the lack of trust in the financial industry. Suppose that you select a sample of 100 institute members. Complete parts ​(a) through ​(d) below.

a. What is the probability that the sample percentage indicating that lack of ethical culture within financial firms has contributed the most to the lack of trust in the financial industry will be between

66

and 72​%?

0.4704

​(Type an integer or decimal rounded to four decimal places as​ needed.)

b. The probability is 70

that the sample percentage will be contained within what symmetrical limits of the population​ percentage?The probability is 70% that the sample percentage will be contained above 63​%

and below 73%

​(Type integers or decimals rounded to one decimal place as​ needed.)

c. The probability is

97% that the sample percentage will be contained within what symmetrical limits of the population​ percentage? The probability is 97% that the sample percentage will be contained above nothing​% and below nothing​%.

What's the answer to C

Answer 1

a)

population proportion ,p=   0.68                              
n=   100                              
                                  
std error , SE = √( p(1-p)/n ) =    0.0120                              
                                  
we need to compute probability for                                   
0.66   < p̂ <   0.72                          
                                  
Z1 =( p̂1 - p )/SE= (   (0.66-0.68)/0.012)=       -1.667                      
Z2 =( p̂2 - p )/SE= (   (0.72-0.68)/0.012)=       3.333                      
P(   0.66   < p̂ <   0.72   ) = P(   -1.667   < Z <   3.333   )  
= P ( Z <   3.333   ) - P (    -1.667   ) =    0.9996   -   0.0478   =   0.9518

b)

population proportion ,p=   0.59                      
n=   300                      
                          
std error , SE = √( p(1-p)/n ) =    0.0284                      
we need to compute probability for                           
P(p̂1<p̂<p̂2)=   0.7                      
proportion left    0.30   is equally distributed both left and right side of normal curve                   
P(p̂ < p̂ 1) =   0.15   and P(p̂ > p̂2)=   0.85              
Z value at    0.15   =   -1.036   (excel formula =NORMSINV(   0.30   / 2 ) )  
Z value at    0.85   =   1.036   (excel formula =NORMSINV(   0.30   / 2 ) )  
                          
Z =( p̂ - p )/SE=                          
so, p̂ = Z*SE+p                          
                          
p̂ 1 = Z*SE+p =   -1.036   *   0.0284   +   0.59   =   0.5606
p̂ 2 = Z*SE+p =   1.036   *   0.0284   +   0.59   =   0.6194

answer: 56.1% , 61.9%

c)

we need to compute probability for                           
P(p̂1<p̂<p̂2)=   0.97                      
proportion left    0.03   is equally distributed both left and right side of normal curve                   
P(p̂ < p̂ 1) =   0.015   and P(p̂ > p̂2)=   0.985              
Z value at    0.015   =   -2.170   (excel formula =NORMSINV(   0.03   / 2 ) )  
Z value at    0.985   =   2.170   (excel formula =NORMSINV(   0.03   / 2 ) )  
                          
Z =( p̂ - p )/SE=                          
so, p̂ = Z*SE+p                          
                          
p̂ 1 = Z*SE+p =   -2.170   *   0.0284   +   0.59   =   0.528
p̂ 2 = Z*SE+p =   2.170   *   0.0284   +   0.59   =   0.652

answer: (52.8% , 65.2%)