Estimate the probability of d i.i.d standard normal variables Let $Z_1,...,Z_d$ be i.i.d. standard normal variables where $d=1000$. Estimate the probability:
$$p = P\left(\left|\sum_{i=1}^{1000}(Z_i)\right|>10000\right)$$
I initially thought I could run a monte carlo simulation to estimate the probability, but realized I don't know the mean or the variance of the distribution to sample from it. Another approach I thought might be applicable was using convolutions because we know that each $Z$ has a common density (i.e. Gaussian) and we are looking at the sum of each of the $Z_i$s, but I'm not too sure how I can use that to estimate this probability.
Any ideas about how I can estimate this probability?
 A: Note that $\sum_{i=1}^{1000}(Z_i)\sim N(0,1000)$, or $\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)\sim N(0,1)$. Hence
$$\begin{align}p &= P\left(\left|\sum_{i=1}^{1000}(Z_i)\right|>10000\right)\\&=P\left(\left\{\sum_{i=1}^{1000}(Z_i)>10000\right\}\cup\left\{\sum_{i=1}^{1000}(Z_i)<-10000\right\}\right)\\&=P\left(\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)>100\sqrt{10}\right\}\cup\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)<-100\sqrt{10}\right\}\right)\\&=P\left(\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)>100\sqrt{10}\right\}\right)+P\left(\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)<-100\sqrt{10}\right\}\right)\\&=P\left(\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)<-100\sqrt{10}\right\}\right)+P\left(\left\{\frac1{\sqrt{1000}}\sum_{i=1}^{1000}(Z_i)<-100\sqrt{10}\right\}\right)\\&=2\Phi(-100\sqrt{10})\end{align}$$
where $\Phi(\cdot)$ is the cumulative distribution function of standard normal distribution. The second last equality is due to the symmetry of standard normal distribution, $P(Z>x)=P(Z<-x)$, where $Z\sim N(0,1)$ and $x\in\Bbb{R}$.
A: If $Z_1,\ldots,Z_d$ are independent and each distributed as $N(0,1)$ then $Z_1+\cdots+Z_d$ is normally distributed with expected value $0$ and variance $d.$
More generally, if $X_1,\ldots,X_d$ are independent and $X_i\sim N(\mu_i,\sigma_i^2)$ for $i=1,\ldots,d$ then $X_1+\cdots+X_d\sim N(\mu_1+\cdots+\mu_d,\sigma_1^2+\cdots+\sigma_d^2),$ i.e. the expected values and the variances just add up, and the distribution is still normal.
