How do you find the z-score for which 80% of the distribution's area lies between -z and z?

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May 19, 2015

Firstly we need to find $\alpha$ the area of which we don't want, which would be alpha = 1 - 80% = 1 - 0.8 = 0.2.
we also know that our Standard Normal Distribution is symmetric, so we would like to divide the area we don't want to be on either side of our area, so we solve for:

$\frac{\alpha}{2} = \frac{0.2}{2} = 0.1$

now it becomes easy to solve for $- z$ using our table.
Which gets us:

$- z = - 1.285$ and then $z = 1.285$

This is easier do do if you draw it out:

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