Using the normal distribution and the central limit theorem, it is found that there is a 0.937 = 93.7% probability of getting a sample statistics like the telephone polls or higher.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem, we have that p = 0.22, n = 1000, hence the mean and the standard error are given by:
[tex]\mu = p = 0.22[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{1000}} = 0.0131[/tex]
The probability of getting a sample statistics like the telephone polls or higher is one subtracted by the p-value of Z when X = 0.2, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.22}{0.0131}[/tex]
Z = -1.53
Z = -1.53 has a p-value of 0.063.
1 - 0.063 = 0.937.
0.937 = 93.7% probability of getting a sample statistics like the telephone polls or higher.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213