Respuesta :
Using the normal distribution and the central limit theorem, it is found that there is a 0.0166 = 1.66% probability of a sample proportion of 0.59 or less.
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]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sampling proportions of a proportion p in a sample of size n has mean [tex]\mu = p[/tex] and standard error [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
In this problem:
- 1,190 adults were asked, hence [tex]n = 1190[/tex]
- In fact 62% of all adults favor balancing the budget over cutting taxes, hence [tex]p = 0.62[/tex].
The mean and the standard error are given by:
[tex]\mu = p = 0.62[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.62(0.38)}{1190}} = 0.0141[/tex]
The probability of a sample proportion of 0.59 or less is the p-value of Z when X = 0.59, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.59 - 0.62}{0.0141}[/tex]
[tex]Z = -2.13[/tex]
[tex]Z = -2.13[/tex] has a p-value of 0.0166.
0.0166 = 1.66% probability of a sample proportion of 0.59 or less.
You can learn more about the normal distribution and the central limit theorem at https://brainly.com/question/24663213
