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I'm quite confused about how to find the probability when the variance , sample mean and a different sample size is given.

I have found the variance an the sample mean through the MLE (maximum likelihood estimator). And now I have to find the probability of some occurrence of the RV(in a normal distribution) takes when a new sample size is given. Please can someone help with this issue.

Thanks in advance!

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  • $\begingroup$ You can only solve this problem if you make some assumption about the underlying distribution (e.g., it is Gaussian, or Poissonian, or...). $\endgroup$ – David G. Stork Aug 13 '18 at 1:52
  • $\begingroup$ Yes it is a Gaussian distribution, is there any possible assumption that I could make $\endgroup$ – user9815591 Aug 13 '18 at 1:57
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If you know (or assume) your distribution is Gaussian with estimated mean $\mu$ and estimated variance $\sigma^2$, then the probability a single draw occurs between $x_l$ and $x_u$ is:

$$P[x_l < x < x_u] = \int\limits_{x=x_l}^{x_u} {1 \over \sqrt{2 \pi} \sigma} e^{-(x-\mu)^2/(2 \sigma^2)}\ dx$$,

which can also be expressed using error functions.

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