(Dice), population mean and variance of the odd numbers.

Question

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 $Y= X_1 + 3X_3 +5X_5$?

Answer 1

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:

$$Var[100X] = 100 Var[X] = 100 \cdot \frac{\big(1-\frac{3}{2}\big)^2 + \big(3-\frac{3}{2}\big)^2 + \big(5-\frac{3}{2}\big)^2 + 3 \big(0-\frac{3}{2}\big)^2}{6}$$ $$= 100 \cdot \frac{0.5^2 + 1.5^2 + 3.5^2 + 3 \cdot 1.5^2}{6} = 100 \cdot \frac{43}{12} \approx 358$$

Answer 2

Another way to think about the variance:

Because we know the rolls are independent,

$$Var(\sum\limits_{i=1}^{100}x_i)=\sum\limits_{i=1}^{100}Var(x_i)$$

We can then look at just $Var(x_i)$

$$Var(x_i) = E(x_i^2) - E(x_i)^2$$

$$E(x_i^2) = \frac{1}{6}*1^2 + \frac{1}{6}*3^2 + \frac{1}{6}*5^2 = \frac{35}{6}$$

$$Var(x_i)=\frac{35}{6}-(\frac{3}{2})^2=\frac{43}{12}$$

and then just plug back into the intitial equation

$$Var(\sum\limits_{i=1}^{100}x_i)=\sum\limits_{i=1}^{100}\frac{43}{12}=100*\frac{43}{12}$$