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If repeated simple random samples of 689 residents are taken, what would be 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?

Answer :

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

Final answer:

The range of the sample proportion of adults over 65, given repeated simple random samples of 689 residents, can be estimated using the 95 part of the 68-95-99.7 rule in statistics. The range is (p - 2*sqrt[p(1-p)/689], p + 2*sqrt[p(1-p)/689]), with p being the proportion of adults over 65 in the whole population.

Explanation:

The 95 part of the 68-95-99.7 rule refers to the fact that in a standard normal distribution, 95% of the values lie within 2 standard deviations of the mean. This rule is often used in statistics to estimate the proportion of values within a particular range, given a large sample size.

Assuming the sample proportion of adults over 65 in the whole population is p, and the sample size is 689, we can calculate the standard deviation of the sample proportion as sqrt[p(1-p)/689]. According to the 95 rule, 95% of the sample proportions would lie within 2 standard deviations from the population proportion, which gives the range (p - 2*sqrt[p(1-p)/689], p + 2*sqrt[p(1-p)/689]).

If the exact value of p is not known, previous studies or a pilot survey can be conducted to estimate this population proportion.

Learn more about Sample Proportion Range here:

https://brainly.com/question/32965609

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