Suppose I have a group of $40\%$ girls, $60\%$ boys. The girls have Normal dist of $\mu=100$ and $\sigma=3$ and the boys have Normal dist $\mu=130$ and $\sigma=5$.
I would like to combine these two normals, which I think is normally easily done by just averaging the normals together, however I'm not sure how the weight of the group comes into play here.
For the mean of the group, should it be $(.4)\cdot 100 +(.6)\cdot130$ so $118$?
Would the standard deviation follow the same? $(.4)\cdot3 +(.6)\cdot5$ so $4.2$? Or do I have to square the probabilities? $(.4)^2\cdot3 +(.6)^2\cdot5$?
Let $X$ be the measurement from the randomly chosen person in this population of $40\%$ girls and $60\%$ boys. Then \begin{align} \Pr(a<X<b) = {} & 0.4\Pr(a<X<b\mid \text{girl} ) + 0.6 \Pr(a<X<b\mid \text{boy}) \\[12pt] \operatorname E(X) = {} & 0.4\operatorname E(X\mid \text{girl}) + 0.6\operatorname E(X\mid \text{boy}) \\[12pt] \operatorname{var}(X) = {} & 0.4\operatorname{var}(X\mid\text{girl}) + 0.6\operatorname{var}(X\mid \text{boy}) \\[2pt] & {} + (0.4\times0.6)\cdot(130-100)^2. \end{align} In that last term, $0.4\times0.6$ is the variance of a random variable that is $0$ with probability $0.4$ and $1$ with probability $0.6,$ or that is $0$ with probability $0.6$ and $1$ with probability $0.4.$