Let $ 0 \lt \alpha \lt 1$. $z_a$ is a solution to $\Phi(z_a)=\alpha $.

Question

Let $ 0 \lt \alpha \lt 1 $. $z_a$ is a solution to $\Phi(z_a)=\alpha $.

1.) What is the relation between $z_a$ and $z_{(1-a)}$

2.) Find $z_a$ (with an error that does not exceed 0.01) for the values:

$\alpha = 0.02,0.1,0.6,0.95,0.995.$

I have no idea how to properly grasp what the first question is asking.

The second question, we know $\Phi(x)$= $0.5\left[1+erf\left(\frac{x}{\sqrt2} \right) \right] $

Now is it as simple as plugging in values of alpha and solve for $z_a$?

Answer 1

For every $x\in\mathbb R$:$$\Phi\left(-x\right)=1-\Phi\left(x\right)$$

Substituting $x=z_a$ leads to$$\Phi\left(-z_{a}\right)=1-\Phi\left(z_a\right)=1-a=\Phi\left(z_{1-a}\right)$$ Conclusion: $$-z_a=z_{1-a}$$

Or equivalently (but a bit nicer): $$z_a+z_{1-a}=0$$