Let X be the average of a sample of 16 independent normal random variables with mean 0 and variance 1. Determine c such that P(|X| < c) = .5 Let $\overline{X}$  be the average of a sample of $16$ independent normal random variables with mean $0$ and variance $1$. Determine c such that
$P(| \overline{X} | < c) = .5$
I am having a lot of trouble with this question. I know it is related to chi-square but I don't know how to even start. 
 A: HINT- The sum of independent normal variables also follows the normal distribution with
$$N(\sum_{i=1}^n {\mu_i},\sum_{i=1}^n {\sigma_i^2})$$
Also, if a random variable $X$ has mean $\mu$ and variance $\sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $k\mu$ and variance $k^2\sigma^2$
A: If $X$ is distributed as $X\sim \mathcal N(0,1)$, then $\frac1n \sum\limits_{i=1}^{16} X_i=\overline X$ is distributed as $\overline X\sim \mathcal N\left( 0, \frac1{16}\right)$
Now we have to evaluate how $|\overline X|$ is distributed in terms of $\overline X$
$P(|\overline X| \leq c)=P(-c \leq \overline X \leq c)$
$=P(\overline X \leq c)-P(\overline X \leq -c)$
$=P(\overline X \leq c)-\left[ 1-P(\overline X \leq c) \right]$
$=2 \cdot P(\overline X \leq c)-1=2\cdot F_{\overline X}(c)-1$
At the next step we standardize $\overline X$ to be able to use the cdf of the standard normal distribution.
$2\cdot P(\overline X \leq c)-1=2\cdot \Phi\left(\frac{c-0}{\sqrt{\frac1{16}}} \right)-1=2\cdot \Phi\left(4c \right)-1=0.5$
$2\cdot \Phi\left(4c \right)=1.5$
$\Phi\left(4c \right)=0.75$
$4c=\Phi^{-1}\left(0.75\right)$
$\Phi^{-1}\left(p\right)$ is the inverse function of $\Phi\left(z\right)$
$4c=0.674\Rightarrow c=0.1685$
