Respuesta :
Answer:
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
For this case we know this:
[tex] n=37 , p=0.2[/tex]
We can find the standard error like this:
[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]
So then our random variable can be described as:
[tex] p \sim N(0.2, 0.0658)[/tex]
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
[tex] P(p>0.4)[/tex]
We can use the z score given by:
[tex] z = \frac{p -\mu_p}{SE_p}[/tex]
And using this we got this:
[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)
Step-by-step explanation:
We need to check if we can use the normal approximation:
[tex] np = 37 *0.2 = 7.4 \geq 5[/tex]
[tex] n(1-p) = 37*0.8 = 29.6\geq 5[/tex]
We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
For this case we know this:
[tex] n=37 , p=0.2[/tex]
We can find the standard error like this:
[tex] SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658[/tex]
So then our random variable can be described as:
[tex] p \sim N(0.2, 0.0658)[/tex]
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
[tex] P(p>0.4)[/tex]
We can use the z score given by:
[tex] z = \frac{p -\mu_p}{SE_p}[/tex]
And using this we got this:
[tex] P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z<3.04) = 0.0012[/tex]
And we can find this probability using the Ti 84 on this way:
2nd> VARS> DISTR > normalcdf
And the code that we need to use for this case would be:
1-normalcdf(-1000, 3.04; 0;1)
Or equivalently we can use:
1-normalcdf(-1000, 0.4; 0.2;0.0658)
