Question

If the random variable z is the standard normal score and a > 0, is it true that P(z > -a) = P(z < a)? Why or why not?

The answer someone gave me was False as there is a distribution in negative side even. why would it be even?

Answer 1

No it should be true. The standard normal distribution is symmetric, so the distance between quantile \(a\) and the mean, is the same as the distance of quantile \(-a\) and the mean.
Let me illustrate the probabilities given below:

If you need more convincing, you can check the probability values for \(P(Z>-a)\) and \(P(Z<a)\) by using a few different values of \(a\).

Like You can try \(a=1\) and see \(P(Z>-1)=P(Z<1)\)
or even \(a=2.5\), and see that \(P(Z>-2.5)=P(Z<2.5)\).

You could also verify that in the same way above by finding the appropriate integral with, say using 2.5 again (using software/Wolfram Alpha):
\[ \large P(Z>-2.5)=\int_{-2.5}^{\infty}\frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx=0.99379\] http://www.wolframalpha.com/input/?i=int+1%2F%28sqrt%7B2*pi%7D%29*e%5E%28-x%5E2%2F2%29+from+x%3D-2.5..oo

\[ \large P(Z<2.5)=\int_{-\infty}^{2.5}\frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx=0.99379\] http://www.wolframalpha.com/input/?i=int+1%2F%28sqrt%7B2*pi%7D%29*e%5E%28-x%5E2%2F2%29+from+x%3D-oo..2.5