Suppose 52% of the population has a college degree. If a random sample of size 563 is selected, what is the probability that the proportion of persons with a college degree will differ from the population proportion by more than 5%? Round your answer to four decimal places.
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.520 |
sample size =n= | 563 |
std error of proportion=σ_{p}=√(p*(1-p)/n)= | 0.0211 |
proportion of persons with a college degree will differ from the population proportion by more than 5%
=P(P>0.57)+P(P<0.47)=1-P(0.47<X<0.57)=1-P(-0.05/0.0211<Z<0.05/0.0211)=1-P(-2.37<Z<2.37)=1-(0.9912-0.0088)
=0.0176 (please try 0.0178 if this comes wrong due to rounding error and revert)
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