Question

Suppose 42% of the population has a retirement account. If a random sample of size 711...

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.

Homework Answers

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
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