We throw Fake Cube(Containing $1,2,3,4,5,6$) $~300$ times .
Probability of geting $6$ is $\frac{1}{2}$
Probability of geting $1,2,3,4,5~$ is $~0.1$
What is the probability that that the Sum : $~~S \leq 1400$.
What I tried :
Let $x_i$ be the number of times we get $i$ in the $300$ throws. ($1 \leq i \leq 6$)
we know that $\sum_1^6 x_i ~~$~ Binomal distribuation
we Shall approximate to Normal Disturbution.
so $x^{Normal}_i = \frac{x_i - E(x_i)}{\sigma(x_i)}$ , ie $x_i^{Normal}$~$N~(E(x_i),\sigma^2(x_i))$
OverAll I get that $S$~$~N(300,210)$ but moving to the standard normal Z is large $Z=75.94$
any ideas why ? or even a soultion to this simple problem.
Instead of having a variable for each outcome let us denote $S$ as the sum of all the outcomes. That is $S=\sum_{i=1}^{300} X_i$, where $X_i$ is the outcome of one roll.
Since $$E[X_i]= 0.1\cdot(1+2+3+4+5) + 0.5\cdot 6 = 4.5$$ and $$E[X_i^2] = 0.1\cdot (1+4+9+16+25) + 0.5\cdot 36 = 23.5$$ we have $$Var(X_i) = 23.5 - 4.5^2 = 3.25$$ Which means that approximately $S \sim N(4.5 \cdot 300 , 3.25 \cdot 300)= N(1350, 975)$, and therefore $P(S\leq 1400)\approx 0.945..$
EDIT: Now suppose that we want to use your approach (which is a slightly more complicated route). The first problem that needs to be addressed is that, if we let $X_i$ denote the amount of times $i$ is observed, then it is true that $X_i \sim Binom(300, p_i)$ where $p_i \in \{ 0.1, 0.5\}$, but it is not true that $X_1,...,X_6$ are independent. We can however use that $(X_1,...,X_6)$ follows a multinomial distribution https://en.wikipedia.org/wiki/Multinomial_distribution . Now the sum of $X_i$'s is always 300, but to get $S$ we need to multiply $X_i$ with $i$ yielding that $$S=X_1 + 2X_2 + 3X_3 + 4X_4 + 5X_5 + 6X_6.$$
And therefore \begin{align*}E[S] = \sum_{i=1}^6 iE[X_i] = \sum_{i=1}^6 i300\cdot p_i = 1350 \end{align*} Calculating the variance becomes slightly more complicated, since we no longer have independence: \begin{align*} Var(S) &= Cov(\sum_{i=1}^6 i X_i , \sum_{j=1}^6 j X_j) \\ &= \sum_{i=1}^6 \sum_{j=1}^6 i j Cov(X_i, X_j) \end{align*} The above sum can be calculated using that $Var(X_i)=300p_i(1-p_i)$ and $Cov(X_i,X_j)=-300p_ip_j$ for $i\neq j$.