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Using the normal distribution, there is a 0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
The proportion estimate and the sample size are given as follows:
p = 0.45, n = 437.
Hence the mean and the standard error are:
- [tex]\mu = p = 0.45[/tex]
- [tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.45(0.55)}{437}} = 0.0238[/tex]
The probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3% is 2 multiplied by the p-value of Z when X = 0.45 - 0.03 = 0.42.
Hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (0.42 - 0.45)/0.0238
Z = -1.26
Z = -1.26 has a p-value of 0.1038.
2 x 0.1038 = 0.2076.
0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.
More can be learned about the normal distribution at https://brainly.com/question/28159597
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