I was given the task to calculate the probability that the sum of $50$ independent random variables are greater than $54$. That is
$\mathbb{S}_{50} = \Sigma_{i=1}^{50} \mathbb{X}_i$
$P(\mathbb{S}_{50} > 54)$
For each of the random variables we have the following:
$P(\mathbb{X}_i = 0) = 0.1$
$P(\mathbb{X}_i = 1) = 0.8$
$P(\mathbb{X}_i = 2) = 0.1$
I am supposed to do this using the central limit theorem. So, I take what I know and insert this into the theorem:
$P({{\mathbb{S}_n-nm}\over{\sigma\sqrt{n}}}<a) \to \Phi(a) = P({{\mathbb{S}_{50}-50m}\over{\sigma\sqrt{50}}}<54) \to \Phi(54)$
Since I'm looking for $P(\mathbb{S}_{50}>54)$, not $P(\mathbb{S}_{50}<54)$, I end up with
$1-P({{\mathbb{S}_{50}-50m}\over{\sigma\sqrt{50}}}<54) \to 1-\Phi(54)$
This seems correct enough, according to the book, if you take into account that if $\mathbb{X} \in N(m,\sigma)$ gives that $\mathbb{Y} = {{(\mathbb{X}-m)}\over{\sqrt{\sigma}}}$
The thing is that the book also states that for each $\mathbb{X}_i$ we have $E(\mathbb{X}_i)=1$ and $\sigma = 0.2$, which results in the equation
$1-\Phi({{\mathbb{S}_{50}-50}\over{\sqrt{10}}}) = 1-\Phi(0.63)$
I see how this is true, but I don't see how I am supposed to come up with the expected value and standard deviation of the variables $\mathbb{X}_i$. I'd be very greatful if someone could explain this.
For a discrete random variable $X$ whose support is $\mathcal S$, we have $${\rm E}[X] = \sum_{x \in \mathcal S} x \Pr[X = x];$$ that is, for each possible outcome of the random variable $X$, we multiply its value by the probability of observing it, then add it all up. Similarly, the variance is defined as ${\rm Var}[X] = {\rm E}[(X - {\rm E}[X])^2] = {\rm E}[X^2] - {\rm E}[X]^2$, or explicitly, $${\rm Var}[X] = \sum_{x \in \mathcal S} x^2 \Pr[X = x] - \left( \sum_{x \in \mathcal S} x \Pr[X = x] \right)^{\!2}.$$ These should be easy for you to calculate.
Next, the Central Limit Theorem basically says that we can approximate the sum of $n$ IID random variables $S = X_1 + X_2 + \cdots + X_n$ as a normal distribution with mean $\mu = n{\rm E}[X_1]$, and variance $\sigma^2 = n{\rm Var}[X_1]$. So the random variable $$Z = \frac{S - \mu}{\sigma}$$ is approximately standard normally distributed (mean 0, variance 1).