Respuesta :
Answer:
[tex] z= \frac{0.21- 0.39}{0.0373}= -4.829[/tex]
[tex] z= \frac{0.32- 0.39}{0.0373}=-1.877[/tex]
And we can find the probability with this difference:
[tex] P(-4.829<z< -1.877) = P(Z<-1.877) -P(Z<-4.829)=0.0303- 6.86x10^{-7}=0.0303[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n = 171[/tex] represent the sample size
[tex]p =0.39[/tex] the proportion of interest
We want to find the following probability:
[tex] P( 0.21 < \hat p < 0.32)[/tex]
We can use the normal approximation for this case since np >10 and n (1-p) >10
For this case we know that the distribution for the sample proportion is given by:
[tex]\hat p \sim N( p , \sqrt{\frac{p (1-p)}{n}} )[/tex]
And we can use the following parameters:
[tex] \mu_{\hat p}= 0.39[/tex]
[tex] \sigma_{\hat p} =\sqrt{\frac{0.39*(1-0.39)}{171}}= 0.0373[/tex]
And we can apply the z score formula given by:
[tex] z = \frac{p \\mu_{\hat p}}{\sigma_{\hat p}}[/tex]
And using this formula we got:
[tex] z= \frac{0.21- 0.39}{0.0373}= -4.829[/tex]
[tex] z= \frac{0.32- 0.39}{0.0373}=-1.877[/tex]
And we can find the probability with this difference:
[tex] P(-4.829<z< -1.877) = P(Z<-1.877) -P(Z<-4.829)=0.0303- 6.86x10^{-7}=0.0303[/tex]
