Respuesta :
Answer:
[tex] P(p>0.38) = P(z>-2.79) [/tex]
And using the complement rule we got:
[tex] P(p>0.38) = P(z>-2.79)=1-P(Z<-2.79)= 1-0.00814=0.9919[/tex]
Explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case wwe know that p = 0.52 and n = 99 and we can check if we can use the normal approximation for the proportion distribution.
[tex] np= 99*0.52= 51.48>10 [/tex]
[tex]n(1-p) =99*(1-0.52) = 47.52>10[/tex]
So then we can use the normal approximation.
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The mean is given by:
[tex]\mu_{p} = 0.52[/tex]
And the standard error is given by:
[tex]\sigma_{p} = \sqrt{\frac{0.52*(1-0.52)}{99}}= 0.0502[/tex]
We want to calculate this probability:
[tex] P(p >0.38)[/tex]
And for this case we can calculate the z score given by:
[tex] z = \frac{p \mu_p}{\sigma_p}[/tex]
And replacing we got:
[tex]z=\frac{0.38-0.52}{0.0502} = -2.79[/tex]
And using this formula:
[tex] P(p>0.38) = P(z>-2.79) [/tex]
And using the complement rule we got:
[tex] P(p>0.38) = P(z>-2.79)=1-P(Z<-2.79)= 1-0.00814=0.9919[/tex]
