Answer:
For this case we know this:
![n=37 , p=0.2](https://tex.z-dn.net/?f=%20n%3D37%20%2C%20%20p%3D0.2)
We can find the standard error like this:
![SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658](https://tex.z-dn.net/?f=%20SE%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Chat%20p%20%281-%5Chat%20p%29%7D%7Bn%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B0.2%2A0.8%7D%7B37%7D%7D%3D%200.0658)
So then our random variable can be described as:
![p \sim N(0.2, 0.0658)](https://tex.z-dn.net/?f=%20p%20%5Csim%20N%280.2%2C%200.0658%29)
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
![P(p>0.4)](https://tex.z-dn.net/?f=%20P%28p%3E0.4%29)
We can use the z score given by:
![z = \frac{p -\mu_p}{SE_p}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7Bp%20-%5Cmu_p%7D%7BSE_p%7D)
And using this we got this:
![P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z](https://tex.z-dn.net/?f=%20P%28p%3E0.4%29%20%3D%201-P%28z%3C%20%5Cfrac%7B0.4-0.2%7D%7B0.0658%7D%29%20%3D%201-P%28z%3C3.04%29%20%3D%200.0012)
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:
![np = 37 *0.2 = 7.4 \geq 5](https://tex.z-dn.net/?f=%20np%20%3D%2037%20%2A0.2%20%3D%207.4%20%5Cgeq%205)
![n(1-p) = 37*0.8 = 29.6\geq 5](https://tex.z-dn.net/?f=%20n%281-p%29%20%3D%2037%2A0.8%20%3D%2029.6%5Cgeq%205)
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
:
For this case we know this:
![n=37 , p=0.2](https://tex.z-dn.net/?f=%20n%3D37%20%2C%20%20p%3D0.2)
We can find the standard error like this:
![SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658](https://tex.z-dn.net/?f=%20SE%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Chat%20p%20%281-%5Chat%20p%29%7D%7Bn%7D%7D%3D%20%5Csqrt%7B%5Cfrac%7B0.2%2A0.8%7D%7B37%7D%7D%3D%200.0658)
So then our random variable can be described as:
![p \sim N(0.2, 0.0658)](https://tex.z-dn.net/?f=%20p%20%5Csim%20N%280.2%2C%200.0658%29)
Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:
![P(p>0.4)](https://tex.z-dn.net/?f=%20P%28p%3E0.4%29)
We can use the z score given by:
![z = \frac{p -\mu_p}{SE_p}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7Bp%20-%5Cmu_p%7D%7BSE_p%7D)
And using this we got this:
![P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z](https://tex.z-dn.net/?f=%20P%28p%3E0.4%29%20%3D%201-P%28z%3C%20%5Cfrac%7B0.4-0.2%7D%7B0.0658%7D%29%20%3D%201-P%28z%3C3.04%29%20%3D%200.0012)
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)