How do you find the z-score for which 95% of the distribution's area lies between -z and z?

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Answer 1 · 2015-05-19T19:32:22

for this question, we would first need to find $α$ $alpha$ the area underneath the cure which does not lay between $- z$ $-z$ and $z$ $z$

$α = 100 % - 95 % = 1 - 0.95 = 0.05$ $alpha = 100% - 95% = 1 - 0.95 = 0.05$

now we also know that our Standard Normal Distribution is symmetrical, so we divide $α$ $alpha$ to equally be on either side of our wanted area.

so we get:

$\frac{α}{2} = \frac{0.05}{2} = 0.025$ $alpha/2 = 0.05/2 = 0.025$

which we can use our table to find $- z$ $-z$ where $Phi(-z) = 0.025$
thus $-z = -1.96$ and $z = 1.96$

this is easier understood with images.

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