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I have a discrete distribution $X$ with mean $56.87500$ and standard deviation $70.725$. I also have that the support of $X$ is contained in $[0, 8750]$, the maximum value of $X$ is $8750$ and $p(X=0)=(0.99)^{75}$. Why a normal distribution would not give a good approximation to X?

Thanks in advance for any help.

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    $\begingroup$ Please add more information on the characteristics of the data and on what you want to analyse. It really depends on the type of data you are studying. If you can have negative and positive values of your variable, the Normal approximation might end up being a good approximation. On the other hand, if you have a high skewness or kurtosis, or if the values belong to a set with a small number of elements (e.g., if the possible values are only in $\{0;100\}$), then it is likely that the normality should not be assumed. $\endgroup$ – Ertxiem Mar 22 at 9:51
  • $\begingroup$ Thank you for your comment. I have edited my question. $\endgroup$ – Mainnet Mar 22 at 10:29
  • $\begingroup$ Is it $(0.99)^{75}$? $\endgroup$ – Jimmy R. Mar 22 at 10:29
  • $\begingroup$ Yes, it was a typo $\endgroup$ – Mainnet Mar 22 at 10:30
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    $\begingroup$ Here $P(X=0)\approx0.47$, so a lot of probability is concentrated on the edges of the interval. This is not well approximated by a normal distribution, which has mass concentration around its mean and not on the "boundary". $\endgroup$ – Jimmy R. Mar 22 at 10:35
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One major issue with a normal approximation is that a $N(57, 70)$ distribution gives a significant weigh---about 20%---to $(-\infty,0)$, which is outside your support. If your data is necessarily non-negative (maybe it counts something), then approximating with a distribution that gives 20% weight to negative numbers doesn't seem like a good idea.

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