Using the Central Limit Theorem and the Empirical Rule, it is found that the range of the sample proportion of adults over 65 in the sample according to the 95 part of the 68-95-99.7 rule would be of:
[tex]\left(p - 2\sqrt{\frac{p(1-p)}{689}}, p + 2\sqrt{\frac{p(1-p)}{689}}\right)[/tex]
The Central Limit Theorem states that for a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, there is a sample of [tex]n = 689[/tex], hence, the standard error is:
[tex]s = \sqrt{\frac{p(1-p)}{689}}[/tex]
The Empirical Rule(68-95-99.7 rule) states that 95% of the measures in a normal distribution are within 2 standard errors of the mean, hence, the range is:
[tex]\left(p - 2\sqrt{\frac{p(1-p)}{689}}, p + 2\sqrt{\frac{p(1-p)}{689}}\right)[/tex]
A similar problem is given at https://brainly.com/question/15581844
