Question

he labor force participation rate is the number of people in the labor force divided by the number of people in the country who are of working age and not institutionalized. The BLS reported in February 2012 that the labor force participation rate in the United States was 63.7% (Calculatedrisk). A marketing company asks 120 working-age people if they either have a job or are looking for a job, or, in other words, whether they are in the labor force. What is the probability that fewer than 60% of those surveyed are members of the labor force?

Answer 1

Answer:

$z= \frac{0.6-0.637}{0.0439} =-0.843$

So then we can find the probability like this:

$P(p<0.6) = P(Z<-0.843)$

And using the normal standard table or excel we got:

$P(p<0.6) = P(Z<-0.843)=0.1996$

Step-by-step explanation:

For this case we can check if we can use the normal approximation for the proportion and we have this:

$np = 120*0.637 =76.44 >10$

$n(1-p) = 120*(1-0.637) = 43.56>10$

Then we can conclude that we can use the normal approximation. And we have this:

$p\sim N (p, \sqrt{\frac{p(1-p)}{n}})$

So the mean is given by:

$\mu_p = 0.637$

And the deviation is given by:

$\sigma_p = \sqrt{\frac{0.637*(1-0.637)}{120}}= 0.0439$

And for this case we want to find this probability:

$P( p<0.6)$

And we can use the z score given by:

$z = \frac{p -\mu}{\sigma_p}$

And for this case the z score is:

$z= \frac{0.6-0.637}{0.0439} =-0.843$

So then we can find the probability like this:

$P(p<0.6) = P(Z<-0.843)$

And using the normal standard table or excel we got:

$P(p<0.6) = P(Z<-0.843)=0.1996$