Question

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%

Answer 1

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,$\hat{p}\sim N(\mu,\sigma^2)$

$\mu_{\hat{p}}=p=0.58$

$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$\sigma_{\hat{p}}=\sqrt{\frac{0.58\times 0.42}{570}}$

$\sigma_{\hat{p}}=0.02067$

Now,

$P(\hat{p}<0.57)=P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}<\frac{0.57-0.58}{0.02067})$

$P(\hat{p}<0.57)=P(Z<-0.483)$

$P(\hat{p}<0.57)=0.31561\times 100$

$P(\hat{p}<0.57)$=31.561%

Hence,  the probability that the proportion of persons with a retirement account will be less than 57%=31.561%