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Suppose I have a population whose distribution is definitely not normal but both the population and sample size will be large. Is there any way I can prove/ show that the mean of a large enough sample on the population will confidently approximate the mean of the entire population? Is there also a way to calculate the percentage of the confidence interval since I obviously would want the approximation to be of high confidence?

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    $\begingroup$ Please research the Weak Law of Large Numbers, the Strong Law of Large Numbers, and the Central Limit Theorem. And could you explain what you mean by the "percentage of the confidence interval"? $\endgroup$
    – whuber
    Jun 27, 2022 at 18:52

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Supposing your samples are IID, and supposing the population variance $\sigma^2$ is finite, you can use the central limit theorem. Under these conditions, if the sample size $n$ is sufficiently large, then the sample mean $\bar{X}_n$ is approximately $N(\mu, \sigma^2/n)$ with $\mu$ the population mean. You can use this normal to create confidence intervals and show convergence of $\bar{X}_n$ to $\mu$ as $n\to\infty$.

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    $\begingroup$ The CLT does not hold this generally. $\endgroup$
    – whuber
    Jul 13, 2023 at 16:53
  • $\begingroup$ @whuber, thanks! I edited answer to include needed hypotheses. Please correct if this is not correct. $\endgroup$ Jul 13, 2023 at 17:44
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    $\begingroup$ Very good. But there's no need to make that special assumption if instead you invoke a law of large numbers. $\endgroup$
    – whuber
    Jul 13, 2023 at 17:54
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    $\begingroup$ I see, you can get this result with less knowledge/assumptions. $\endgroup$ Jul 13, 2023 at 18:04

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