enter image description hereif we are given the sample variance value as 129.77, how can we calculate the sample mean.

Moreover, the following is known:

$\sum x_i^2 = 123\,117.538\,151$

  • $\begingroup$ What other information do you have? $\endgroup$ – user2974951 Feb 12 at 13:01
  • $\begingroup$ If this is scaled version of your sample variance, then your sample mean is $0$. $\endgroup$ – gunes Feb 12 at 13:22
  • $\begingroup$ @user2974951, Other than the ∑x2i, I only have the sample variance value of 129 $\endgroup$ – Malick Feb 12 at 13:24
  • $\begingroup$ What is $\sum x2i$? $\endgroup$ – user2974951 Feb 12 at 13:25
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    $\begingroup$ If you have the sample variance, please edit the question. It seems as if you call $\sum x_i^2$ as sample variance. $\endgroup$ – gunes Feb 12 at 13:54

You can't - you can have different samples with different means, yet having the same variance.

Variance is not a function of the mean.

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    $\begingroup$ Note: it can be done if we make assumptions about the population, such as exponential. Our favorite distribution assumption, normal, does not wrk for this, however. $\endgroup$ – Dave Feb 12 at 13:09
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    $\begingroup$ @Dave That's incorrect, because the sample will not enjoy the same mathematical relationships among its moments as its underlying generative distribution. $\endgroup$ – whuber Feb 12 at 15:17
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    $\begingroup$ whuber brings up a subtle but important point that I missed. Take Poisson$(\lambda)$ as an example, where the population mean and population variance are equal and equal to the $\lambda$ parameter. Our sample, even if it is drawn from a distribution that is perfectly Poisson, (probably) is not perfectly Poisson$(\lambda)$, so the relationship that $\mu = \sigma^2 = \lambda$ does not hold in the sample, despite it holding in the population, so we are not able to say exactly what the mean of the sample would have been if we had calculated it while we still had access to the data. $\endgroup$ – Dave Feb 12 at 15:34

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