# Will changing the standard deviation affect the distribution?

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.

• 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? Commented Oct 12, 2019 at 19:24
• @quester edit ! Commented Oct 12, 2019 at 19:56

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.

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

• So no matter what mean or variance I select, it will still follows a normal distribution? It will not suddenly follows an exponential distribution? Commented Oct 12, 2019 at 20:20
• see edit :) I wanted to be sure Commented Oct 12, 2019 at 20:49
• I think you have misunderstood the question. Commented Oct 13, 2019 at 0:32