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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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