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If n is large enough then a variance should be around 0 but not 1?

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    $\begingroup$ Haven't you missed spotting the $1/\sqrt{n}$ in the denominator...? $\endgroup$
    – Xi'an
    Oct 9, 2020 at 8:02
  • $\begingroup$ @Xi'an you are right. Xn ~ Normal(true_mean, true_variance/n)? $\endgroup$
    – Oleg Dats
    Oct 9, 2020 at 11:37

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