If a random sample of size 774 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3%

Suppose 43% of the population has a retirement account. If a random sample of size 774 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3%?

the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is 0.9082

Step-by-step explanation:

Given that:

sample size n = 774

Let P be the population proportion for having a retirement account  = 0.43

Also

Let consider be the sample proportion of having a retirement account.

However; as n is > 30 ; we can say:

 ;  

⇒  

⇒

So; we need  P( the sample proportion will differ from 'p' by less than 3% i.e 0.03)

From Normal Cumulative Distribution Function Table

= 1.9082 - 1

= 0.9082

Thus; the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is 0.9082

30 feet

step-by-step explanation:

that would be 2 1/2 * 12

= 24 + 6

= 30 feet   (answer)