Respuesta :
Answer:
a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.
Step-by-step explanation:
This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.
The parameters for the normal distribution will be
[tex]\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353[/tex]
We can calculate the z values for x1=0.47 and x2=0.51:
[tex]z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28[/tex]
We can now calculate the probabilities:
[tex]P(0.47<p<0.51)=P(-0.85<z<0.28)=P(z<0.28)-P(z<-0.85)\\\\P(z<0.28)-P(z<-0.85)=0.61026-0.19766=0.41260[/tex]
If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b) If the sample size change, the standard deviation of the normal distribution changes:
[tex]\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05[/tex]
We can calculate the z values for x1=0.47 and x2=0.51:
[tex]z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2[/tex]
We can now calculate the probabilities:
[tex]P(0.47<p<0.51)=P(-0.6<z<0.2)=P(z<0.2)-P(z<-0.6)\\\\P(z<0.2)-P(z<-0.6)=0.57926-0.27425=0.30501[/tex]
If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.
