I have a problem:
Let $X$ be a random variable from a dice, defined by $X:=(k-3)^2$ for $k\in\{1,\ldots,6\}$. The exercise is to find the mean and the variance of $X$.
- The probabilities are $$P(X=0)=\frac{1}{6},\ P(X=1)=\frac{1}{3},\ P(X=4)=\frac{1}{3},\ P(X=9)=\frac{1}{6}.$$ Is it true that $$\mathbb{E}(X) = \sum_{k\in\{1,\ldots,6\}} P(X=(k-3)^2)\cdot (k-3)^2?$$ If the answer is yes, the mean of $X$ will be $$\mathbb{E}(X)=P(X=4)\cdot4+P(X=1)\cdot 1+0+P(X=1)\cdot 1+P(X=4)\cdot 4+P(X=9)\cdot 9\tag{1}$$ or $$\mathbb{E}(X)=2\cdot P(X=1)\cdot 1+2\cdot P(X=4)\cdot 4+ P(X=9)\cdot 9?\tag{2}$$ I'm confused with the definition of mean.
Since you have already found the probability space, it is easier (especially for the variance) to calculate it as $$\mathbb{E}(X) = \sum_{n\in\{0,1,4,9\}} P(X=n)\cdot n=\frac{19}{6}$$