Whether you're a student or a professional, IDNLearn.com has answers for everyone. Join our knowledgeable community to find the answers you need for any topic or issue.

Suppose 58% of the population has a retirement account. If a random sample of size 570 is selected, what is the probability that the proportion of persons with a retirement account will be less than 57%

Sagot :

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%

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.