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.
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.420 |
sample size =n= | 711 |
std error of proportion=σ_{p}=√(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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