Respuesta :
Answer:
[tex] P(p<0.68)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{p - \mu_p}{\sigma_p}[/tex]
For this case we have:
[tex] \mu_p = 0.69[/tex]
[tex]\sigma_p = \sqrt{\frac{0.69*(1-0.69)}{162}}= 0.036[/tex]
And using the z score formula we got:
[tex] P(p<0.68) = P(z<\frac{0.68-0.69}{0.036}) = P(z<-0.275)[/tex]
And using the normal standard table or excel we got:
[tex] P(p<0.68)=P(z<-0.275) = 0.392[/tex]
Step-by-step explanation:
For this case we have the following data given:
[tex] n = 162[/tex] represent the random sample selected
[tex] p = 0.69[/tex] represent the proportion os students in the United States enroll in college directly after high school graduation
We can check if we can use the normal approximation with the following conditions:
1) [tex] np = 162*0.69 = 111.78>10[/tex]
2) [tex] n(1-p) = 162*(1-0.69) = 50.22>10[/tex]
3) Independence between the events
For this case the 3 conditions are satisfied so then we can approximate the distribution for the population proportion is given by:
[tex] p \sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
For this case we want this probability:
[tex] P(p<0.68)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{p - \mu_p}{\sigma_p}[/tex]
For this case we have:
[tex] \mu_p = 0.69[/tex]
[tex]\sigma_p = \sqrt{\frac{0.69*(1-0.69)}{162}}= 0.036[/tex]
And using the z score formula we got:
[tex] P(p<0.68) = P(z<\frac{0.68-0.69}{0.036}) = P(z<-0.275)[/tex]
And using the normal standard table or excel we got:
[tex] P(p<0.68)= P(z<-0.275) = 0.392[/tex]
