<s>[INST] (Dice), population mean and variance of the odd numbers. 
A fair die is rolled 100 times. Find the population mean and variance of the sum of the odd numbers that appear. 

I am aware that the expected value for discrete multinomial distributions is $\Sigma \Sigma ... g(x_1 , x_2,...)f(x_1 ,x_2...)$. In this case, to find the expected value of the sum of the odd numbers that were rolled, should I not simply find the sum of the expected values for each of the individual odd numbers?
Couldn't something similar be done for variance? Or must I compute it using something like $Y= X_1 + 3X_3 +5X_5$?
 [/INST] For the mean, we find:
$$E[100X] = 100 E[X] = 100 \bigg(\frac{1}{6} \cdot 1 + \frac{1}{6} \cdot 3 + \frac{1}{6} \cdot 5 + \frac{3}{6} \cdot 0\bigg) = 100 \cdot \frac{3}{2} = 150$$
For the variance, we find:
$$Var[100X] = 100 Var[X] = 100 \cdot \frac{\big(1-\frac{3}{2}\big)^2 + \big(3-\frac{3}{2}\big)^2 + \big(5-\frac{3}{2}\big)^2 + 3 \big(0-\frac{3}{2}\big)^2}{6}$$
$$= 100 \cdot \frac{0.5^2 + 1.5^2 + 3.5^2 + 3 \cdot 1.5^2}{6} = 100 \cdot  \frac{43}{12} \approx 358$$
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