Answer:
The probability that the proportion of persons with a retirement account will be less than 57%=31.561%
Step-by-step explanation:
We are given that
n=570
p=58%=0.58
We have to find the probability that the proportion of persons with a retirement account will be less than 57%.
q=1-p=1-0.58=0.42
By takin normal approximation to binomial then sampling distribution of sample proportion follow normal distribution.
Therefore,[tex]\hat{p}\sim N(\mu,\sigma^2)[/tex]
[tex]\mu_{\hat{p}}=p=0.58[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.58\times 0.42}{570}}[/tex]
[tex]\sigma_{\hat{p}}=0.02067[/tex]
Now,
[tex]P(\hat{p}<0.57)=P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}<\frac{0.57-0.58}{0.02067})[/tex]
[tex]P(\hat{p}<0.57)=P(Z<-0.483)[/tex]
[tex]P(\hat{p}<0.57)=0.31561\times 100[/tex]
[tex]P(\hat{p}<0.57)[/tex]=31.561%
Hence, the probability that the proportion of persons with a retirement account will be less than 57%=31.561%
