Respuesta :
Answer:
a) [tex] p \sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]
With the following parameters:
[tex]\mu_p = 0.25[/tex]
[tex]\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559[/tex]
b) [tex]\mu_p = 0.25[/tex]
[tex]\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559[/tex]
c) [tex]P(\frac{0.13-0.25}{0.0559}< Z< \frac{0.47-0.25}{0.0559}) = P(-2.147< Z< 3.936)[/tex]
And for this case we can use the following difference and the normal standard distribution table or excel and we got:
[tex] P(-2.147< Z< 3.936) = P(z<3.936)-P(Z<-2.147) = 0.999959-0.01590=0.9841[/tex]
Step-by-step explanation:
For this case we have the following info:
n = 60 represent the sample size
p = 0.25 represent the proportion of success
For this case we can check the conditions to use the normal distribution:
1) np= 60*0.25=15>10
2) n(1-p)= 60(1-0.25) = 45>10
So then we can use the normal distribution as an approximate distribution for p
Part a
[tex] p \sim N (p, \sqrt{\frac{p(1-p)}{n}})[/tex]
Part b
With the following parameters:
[tex]\mu_p = 0.25[/tex]
[tex]\sigma_p = \sqrt{\frac{0.25*(1-0.25)}{60}}= 0.0559[/tex]
Part c
For this case we want this probability:
[tex] P(0.13< p<0.47)[/tex]
And for this case we can use the z score given by:
[tex] z = \frac{p -\mu_p}{\sigma_p}[/tex]
And using this formula we got:
[tex]P(\frac{0.13-0.25}{0.0559}< Z< \frac{0.47-0.25}{0.0559}) = P(-2.147< Z< 3.936)[/tex]
And for this case we can use the following difference and the normal standard distribution table or excel and we got:
[tex] P(-2.147< Z< 3.936) = P(z<3.936)-P(Z<-2.147) = 0.999959-0.01590=0.9841[/tex]
