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Given that for a standard normal variable $Z$,$p(0<z<0.8) =0.2881$
The value of $p($$\vert$$z$$\vert$ $\geq$$0.8)=?$

I already know how to find $p(z$$\geq$$0.8)$ which is equal to $0.21186$.
But I dont know how to find that of the above question.

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

2 Answers 2

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

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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:

$1 = p(z<-0.8) + p(-0.8<z<0)+p(0>z>0.8)+p(z>0.8)$.

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

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

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

$p(-0.8<z<0)=p(0>z>0.8)$.

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 $a = 0.8$.

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