Question

Suppose 42% of the population has a retirement account.

If a random sample of size 711 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 4%? Round your answer to four decimal places.

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.420
sample size       =n=	711
std error of proportion=σp=√(p*(1-p)/n)=	0.0185

probability that the proportion of persons with a retirement account will differ from the population proportion by less than 4%:

probability =

P(0.38<X<0.46)

=

P(-2.16<Z<2.16)=

0.9846-0.0154=

0.9692