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$Z_1, Z_2, .., Z_{100}$ 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?

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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.


Note: The statement about the mean and variance of the sum is always exactly true. The approximation is that the distribution of the sum isn't exactly normal, but the more RVs there are, the better the approximation is.

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  • $\begingroup$ What mean your letters? $\mu$ is expected value? $\sigma^2$ is variance. Is this right? But where is function? $\endgroup$ – roblind Oct 11 '17 at 13:48
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    $\begingroup$ 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 $k=100$). $\endgroup$ – MPW Oct 11 '17 at 13:56
  • $\begingroup$ Before I ask other question I ask is this correct formula? Because internet there is many different: $$\Phi(x) = \frac{1}{\sqrt{2\pi}} \cdot \int_{-\infty}^{x} e^{-\frac{t^2}{2}}dt$$? $\endgroup$ – roblind Oct 11 '17 at 14:43
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    $\begingroup$ 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. $\endgroup$ – MPW Oct 12 '17 at 16:43

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