Respuesta :
Answer:
a) [tex] \sigma_p = \sqrt{\frac{0.43 (1-0.43)}{200}}= 0.035[/tex]
b) [tex] P(p>0.33) = P(Z> \frac{0.33-0.43}{0.035}) = P(Z>-2.857) = 1-P(Z<-2.857) = 1-0.0021= 0.9979[/tex]
c) [tex] P(0.19 < p < 0.31) = P(\frac{0.19-0.43}{0.035} < Z<\frac{0.31-0.43}{0.035})[/tex]
[tex]P(0.19 < p < 0.31) =P(-6.857< Z< -3.429)= P(Z<-3.429)-P(Z<-6.857) =0.00030- 3.51x10^{-12}= 0.0003[/tex]
Step-by-step 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".
X represent the users who access the site on the way specified
[tex]\hat p=0.43[/tex] estimation for the sample proportion
n=200 sample size selected
We are interested in the distribution for the true proportion.
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And the reason is because the sample size is large enough and the estimated proportion is near to 0.5
a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users be?
For this case we know that the expected value would be:
[tex] \mu_p = \hat p =0.43[/tex]
And for the standard deviation we have:
[tex] \sigma_p = \sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
And if we replace we got:
[tex] \sigma_p = \sqrt{\frac{0.43 (1-0.43)}{200}}= 0.035[/tex]
What is the probability that the sample proportion of smart phone users is greater than 0.33?
For this case we can use the z score formula given by:
[tex] z = \frac{p -\mu_p}{\sigma_p}[/tex]
And we want this probability:
[tex] P(p>0.33)[/tex]
If we use the z score we got:
[tex] P(p>0.33) = P(Z> \frac{0.33-0.43}{0.035}) = P(Z>-2.857) = 1-P(Z<-2.857) = 1-0.0021= 0.9979[/tex]c) What is the probability that the sample proportion is between 0.19 and 0.31?
We can use the z score formula and we have:
[tex] P(0.19 < p < 0.31) = P(\frac{0.19-0.43}{0.035} < Z<\frac{0.31-0.43}{0.035})[/tex]
[tex] P(0.19 < p < 0.31) =P(-6.857< Z< -3.429)= P(Z<-3.429)-P(Z<-6.857) =0.00030- 3.51x10^{-12}= 0.0003[/tex]
