according to an article from huffingtonpost.com, some experts believe that 20% of all freshwater fish in the united states have such high levels of mercury that they are dangerous to eat. suppose a fish market has 250 fish we consider randomly sampled from the population of edible freshwater fish. use the central limit theorem (and the empirical rule) to find the approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.20. you can use the central limit theorem because the fish were randomly sampled; the population is more than 10 times 250; and n times p is 50, and n times (1 minus p) is 200, and both are more than 10.

The probability that the market will contain a percentage of fish with dangerously high mercury levels that is more than two standard errors over 0.20 is approximately 95% by Central Limit Theorem.

What is Central Limit Theorem?

The Central Limit Theorem states that if large samples with n > 30 are taken from an unknown population and the sample percentage for each sample is calculated, the distribution of the sample proportion obtained from the samples will conform to a Normal distribution.

Now,

The sample proportion's sampling distribution's mean is as follows:

[tex]\mu_p[/tex] = 0.20

This sampling distribution's sample proportion's standard deviation is:

[tex]\sigma_p=\sqrt{\frac{p(1-p)}{n}}[/tex]

The central limit theorem can be used to approximate the sampling distribution of the sample proportion by the normal distribution because the sample size is big, i.e., n = 250 > 30.

Do the mean and standard deviation calculations as follows:

[tex]\sigma_p=\sqrt{\frac{0.20(1-0.20)}{250}} = \sqrt{000.64}=0.02529[/tex]

The empirical rule, also known as the 68-95-99.7 rule, is a shortcut in statistics that is used to remember that 68%, 95%, and 99.7% of the Normal distribution fall within one, two, and three standard deviations of the mean, respectively.

Then,

P (µ-σ < X < µ+σ) ≈ 0.68
P (µ-2σ <X < µ+2σ) ≈ 0.95
P (µ-3σ <X < µ+3σ) = 0.997

Therefore, there is a roughly 0.95 percent chance that the market will contain a percentage of fish with dangerously high mercury levels that is more than two standard errors over 0.20.

That is:

[tex]\mu_p-2\sigma_p < x < \mu_p+2\sigma_p[/tex]

=> 0.20 - 0.0508 < x < 0.20 + 0.0508

=> 0.14 < x < 0.2508.

Hence, The probability that the market will contain a percentage of fish with dangerously high mercury levels that is more than two standard errors over 0.20 is approximately 95% by Central Limit Theorem.

To learn more about Central Limit Theorem, Refer to the link: https://brainly.com/question/18403552

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