Methods for calculating the mean and variance of a distribution created from the addition of two normally distributed quantities I'm trying to understand how to interpret the following which refers to determination of the mean and variance of a distribution that's the result of adding two normally distributed random variables.
$\sum_{i=1}^{n}X_i \sim N(\sum_{i=1}^{n}\mu_i, \sum_{i=1}^{n}\sigma^{2}_i)$
Let's imagine I've a bag of 100 lotto balls each with some number printed on the side, and the values of these numbers are distributed normally about the mean 1 with a variance of 1. 
(OK, so normal lotto balls are integers, but let's assume a continuous notion of 'value' here) 
Now, if I add another bag of 100 lotto balls, which are also normally distributed about the mean, this time of 100, with a variance of 1. 
What is my expected mean and variance of the new population?
If I start from scratch and calculate the new mean, it's clear that the mean is now $101/2 = 50.5$, but from the addition of distribution formula above, I'd expect the new mean to be $\sum_{i=1}^n\mu_{i}$ which would, in this case be 101. 
Similarly, the variance of such a distribution would not be 2 (i.e. $1^2 + 1^2$ ) but something much larger. 
So my question here is, how am I misinterpreting this formula?
Is there a more appropriate function that will provide me the new mean and variation of a random variable that's the result of adding or mixing two known populations, calculable from totals, means, variances or other descriptive data, without resorting to recalculation from the elemental data?
And part two of my question is, under what circumstances would you use the distribution addition formula above? I can see how by adding heights of (say) paired elements I might find the results described, but how does this work if the distributions describe populations of different sizes?
 A: Your question (and to an extent also the Answer by @ZoranLoncarevic) fails to distinguish explicitly between the ideas of 'adding' random variables and 'mixing' their distributions. If I add $X \sim Norm(\mu=0, \sigma^2 = 4)$ and $Y \sim Norm(\mu=8, \sigma^2 = 4)$, then I get the random variable $T = X + Y \sim Norm(\mu = 8, \sigma^2 = 8)$. If I make a 50-50 mixture of these two distributions, I get a bimodal distribution with mean 4. In the
latter case the PDF of the mixture is an average of the PDFs of
$X$ and $Y$. (The mixture is bimodal if the means are far enough
apart; here more than twice the common SD $\sigma = 2.$)
The two histograms below illustrate the distinction. At the
end of the program, approximate means, variances, and SDs of
the two simulated distributions are shown.
 x = rnorm(10^5, 0, 2);  y = rnorm(10^5, 8, 2)  # SDs are 2
 t = x + y;  # sum
 h = rbinom(10^5, 1, 1/2); v = h*x + (1-h)*y    # 50-50 mixture
 par(mfrow=c(1,2))
 hist(t, br=20, prob=T, col="wheat", main="Sum of Two Normals")
   curve(dnorm(x, 8, sqrt(8)), lwd=2, col="blue", add=T)
 hist(v, prob=T, col="wheat", main="Mixture of Two Normals")
   curve(.5*dnorm(x, 0, 2) + .5*dnorm(x, 8, 2), lwd=2, col="darkgreen", add=T)
 par(mfrow=c(1,1))

 mean(t);  var(t);  sd(t)
 ## 7.991537  # Approx E(T) = 8
 ## 8.026132  # Approx Var(T) = 8
 ## 2.833043  # Approx SD(T)
 mean(v);  var(v);  sd(v)
 ## 3.988405  # Approx E(V) = 4
 ## 19.99138  # Approx Var(V)
 ## 4.471172  # Approx SD(V)


