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Let's say I want to generate some random data that follows the normal distribution, with a mean of 5. Will setting the standard deviation to 3 or 5 affect the distribution, that is, will it still follow the normal distribution?

I am confused on this question because the data's mean and variance are based on the data, and there are specific way of calculating them for different distribution.

*edit for clarification: I want to generate TWO data sets: both of them follow the normal distribution with a mean of 5. The 1st data set has $\mu =5$ AND $\sigma= 5 $. If I generate the 2nd data set by changing the $\sigma$ from 5 $\rightarrow$3, will the data still follow the normal distribution.

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  • $\begingroup$ please clarify: you want to generate part of dataset with normal distribution with variance 3 and then rest with variance 5 and question is if resulting dataset follows normal distribution? $\endgroup$ – quester Oct 12 '19 at 19:24
  • $\begingroup$ @quester edit ! $\endgroup$ – GarlicSTAT Oct 12 '19 at 19:56
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If you take one data sample $\{X_i\}_{i=1}^{100} \sim \mathcal N(5, 5^2)$, that is, $X_1$ through $X_{100}$ each following a normal distribution with mean and standard deviation $5$ – and then a different data sample $\{Y_i\}_{i=1}^{100} \sim \mathcal N(5, 3^3)$, then:

$X$ and $Y$ each still follow a normal distribution. They don't affect each other in any way.

If you combine the two datasets together, they don't follow a normal distribution anymore; instead, they follow a mixture of Gaussians. You can find the mean and find the variance of this mixture pretty easily with a very small amount of math. This is what quester's answer does empirically.

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for example if I will generate 6 observations from $N(5, 3)$ and 4 from $N(5, 5)$ all 10 observations should be considered to follow distribution $N(5, 3^2*\frac{6}{10} + 5^2*\frac{4}{10}) = N(5,15.6)$

so in general data as a whole will follow mixture of normal distributions but with variance being weighted sum of variances of original distributions

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  • $\begingroup$ So no matter what mean or variance I select, it will still follows a normal distribution? It will not suddenly follows an exponential distribution? $\endgroup$ – GarlicSTAT Oct 12 '19 at 20:20
  • $\begingroup$ see edit :) I wanted to be sure $\endgroup$ – quester Oct 12 '19 at 20:49
  • $\begingroup$ I think you have misunderstood the question. $\endgroup$ – Glen_b Oct 13 '19 at 0:32

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