# Calculate probability for event $\sum_{i=1}^{100}Z_{i} \in \left(-10,10\right )$

$Z_1, Z_2, .., Z_{100}$ are independent identical distributed random variables with expected value $E(Z_i)=0$ and variance $Var(Z_i)=1$

Calculate the probability for the event $\sum_{i=1}^{100}Z_{i} \in \left(-10,10\right )$ approximatively.

Hint: We have that $\Phi(1) = 0.8413$ where $\Phi$ is the cumulative distribution function of a normally distributed random variable.

I don't know how solve this good.. But as other hint is given that $$P(|X_i| \geq 2) \leq \frac{1}{4}$$

And I think from this I need take inegral with limits $-10$ and $10$ then we have probability of event. Is this correct? But I need get function.. and no idea what to do with the cumulative distribution function because there is no function but just value from the function.. I need function to make the integral but where is it?

Hint: The sum $S$ of a large number $k$ of i.i.d. random variables $Z_i\sim Z$ is approximately normal. The mean of the sum is $\mu_S = k\mu_Z$ and the variance is $\sigma^2_S = k\sigma^2_Z$. That is, $S\sim N(k\mu_Z,k\sigma^2_Z)$ approximately.
• What mean your letters? $\mu$ is expected value? $\sigma^2$ is variance. Is this right? But where is function? – roblind Oct 11 '17 at 13:48
• Yes. You are using the CDF of the normal distribution that approximates the sum. It can be expressed in terms of the CDF $\Phi$ of the standard normal distribution $N(0,1)$. One generally must use a table of values for $\Phi$ since the integral cannot be computed exactly. Note you are being told $\mu_Z = 0$ and $\sigma^2_Z=1$, so that means $\mu_S=0$ and $\sigma^2_S=100$ (since $k=100$). – MPW Oct 11 '17 at 13:56
• Before I ask other question I ask is this correct formula? Because internet there is many different: $$\Phi(x) = \frac{1}{\sqrt{2\pi}} \cdot \int_{-\infty}^{x} e^{-\frac{t^2}{2}}dt$$? – roblind Oct 11 '17 at 14:43
• Yes, that's the CDF for the standard normal distribution (standard normal has $\mu = 0$ and $\sigma^2 = 1$). Note that for any RV $X$, the transformed variable $Y \equiv \frac{X-\mu_X}{\sigma_X}$ has $\mu_Y=0$ and $\sigma^2_Y=1$. The point is to use the fact that $$F_X(x)=\Pr(X\leq x) = \Pr(\frac{X-\mu_X}{\sigma_X}\leq\frac{x-\mu_X}{\sigma_X})=\Pr(Y\leq\frac{x-\mu_X}{\sigma_X})=F_Y(\frac{x-\mu_X}{\sigma_X})$$so you really only have to consider standardized variables. – MPW Oct 12 '17 at 16:43