Question

a) Let Z be the standard normal distribution and a a real number in (0,1). Calculate the following probability b) Find the probability P(Z < in + zi-o) c) Find α given that 4-a)--0.8

Answer 1

a) $P(-z_{\frac{\alpha }{2}} \leq Z \leq z_{\frac{\alpha }{2}}) = P(Z \leq z_{\frac{\alpha }{2}})-P(Z \leq -z_{\frac{\alpha }{2}})$

$\Rightarrow P(-z_{\frac{\alpha }{2}} \leq Z \leq z_{\frac{\alpha }{2}}) = \Phi(z_{\frac{\alpha }{2}})-\Phi(-z_{\frac{\alpha }{2}})$

$\Rightarrow P(-z_{\frac{\alpha }{2}} \leq Z \leq z_{\frac{\alpha }{2}}) =(1 - \frac{\alpha }{2})-\frac{\alpha }{2}$

$\Rightarrow P(-z_{\frac{\alpha }{2}} \leq Z \leq z_{\frac{\alpha }{2}}) =1 - \alpha$

since by definition $P(Z>z_{\frac{\alpha}{2} }) = \frac{\alpha}{2}$

b) $P(Z<z_{\alpha } + z_{1 - \alpha }) = P(Z<z_{\alpha } - z_{\alpha }) = P(Z<0) = \Phi(0) = 0.5$

since $z_{1 - \alpha } = -z_{\alpha }$

c) $\Phi(-z_{\alpha }) = 0.8$

$\Rightarrow \Phi(z_{1 - \alpha }) = 0.8$

$\Rightarrow z_{1 - \alpha } = \Phi^{-1}(0.8) = 0.841621$

$\Rightarrow z_{1 - \alpha } = 0.841621$

$\Rightarrow z_{1 - \alpha } = 0.841621$

$\Rightarrow \Phi^{-1}(\alpha ) = 0.841621$

$\Rightarrow \alpha = \Phi(0.841621) = 0.8$ (ans)

It is known that by definition

$P(Z>z_{1 - \alpha }) = 1 -\alpha$

$\Rightarrow 1 - P(Z\leq z_{1 - \alpha }) =1 - \alpha$

$\Rightarrow P(Z\leq z_{1 -\alpha }) = \alpha$

$\Rightarrow \Phi(z_{1 -\alpha }) = \alpha$

$\Rightarrow z_{1 -\alpha } = \Phi^{-1}(\alpha)$