Choosing the variance of a random normal variable to fulfill some criteria Suppose you create a random normal variable $X$ with mean zero and variance $\sigma^2$. You wish to choose $\sigma^2$ such that 80% of the time (or $a$% of the time, to make it more general), $X$ is between -1 and 1 (or between $-b$ and $b$ to make it more general). How to calculate $\sigma^2$?
First, note that

$$\mathbb{P}(-1 \leq X \leq 1) = \mathbb{P}(X \leq 1) - \mathbb{P}(X \leq -1). $$

Furthermore, we have that $\mathbb{P}(X \leq -1) = \mathbb{P}(X \geq 1) = 1 - \mathbb{P}(X \leq 1)$ by symmetry of the density of the normal cdf. Hence, we obtain
$$\mathbb{P}(-1 \leq X \leq 1) = 2 \mathbb{P}(X \leq 1) - 1 = 0.8 \Leftrightarrow  \mathbb{P}(X \leq 1) = 0.9$$
Standardizing yields
$$\mathbb{P}(X \leq 1) = \mathbb{P}\left(Z \leq \frac{1}{\sigma}\right) = 0.9 ,$$
where $Z \sim N(0,1)$. Looking in the table for the $z$-scores of the normal distribution, we find that
$$\frac{1}{\sigma} = 1.282 \Rightarrow \sigma \approx 0.78$$
Hint: 

Find $\sigma^2$ such that:
$$\mathbb{P}(-1 \leq X \leq 1) = 0.8,$$
knowing that $X \sim N(0,\sigma^2)$.