# How to find value of $\vert$$z$$\vert$ in normal distribution

Given that for a standard normal variable $$Z$$,$$p(0
The value of $$p(\vertz\vert$$ $$\geq0.8)=?$$

I already know how to find $$p(z\geq0.8)$$ which is equal to $$0.21186$$.
But I dont know how to find that of the above question.

• $$|Z|\geqslant 0.8\iff Z\geqslant 0.8 \text{ or }Z\leqslant -0.8$$ Commented Oct 10, 2018 at 20:45
• Draw a picture of the region $|Z|\geq 0.8$. It should then be obvious what to do. Commented Oct 11, 2018 at 9:10

Normal is symmetric about $$z=0$$, so $$p(z \geq 0.8) = p(z \leq -0.8)$$. Also, because its symmetric, $$p(|z| \geq 0.8) = p(z \geq 0.8) + p(z \leq -0.8)$$.

I would start by thinking about the fact that the sum of probability over the whole domain is 1. Given the way you have written the problem, I write the sum of probability as:

$$1 = p(z<-0.8) + p(-0.8z>0.8)+p(z>0.8)$$.

The quantity of interest is

$$p(\left | z \right |>0.8)$$

which I will rewrite as

$$p(\left | z \right |>0.8) = p(z<-0.8) + p(z>0.8)$$.

So going back to the sum of probability, I can make the following substitution

$$1 = p(-0.8z>0.8)+p(\left | z \right |>0.8)$$.

Because the distribution is defined to be a standard normal distribution and thus symmetric about zero, the following is true,

$$p(-0.8z>0.8)$$.

Using this information, I again rewrite the sum of probabilities,

$$1 = 2\,p(0>z>0.8) + p(\left | z \right |>0.8)$$

and solve for the quantity of interest

$$p(\left | z \right |>0.8) = 1 - 2\,p(0>z>0.8)$$.

Hopefully that makes sense, I tried to do this in terms of the quantities on hand. However, integrating over the probability density function (pdf) is the simplest path to the answer:

$$p(\left | z \right |> a) = 1 - \int_{-a}^a$$pdf$$(x)\,dx$$

In this case, the pdf is a a standard normal distribution, and $$a = 0.8$$.