Determine $z_{\alpha}$ for the following:
Here $z_{\alpha}$ denote the value on the measurement axis for which $\alpha$ of the area under the $z$ curve lies to the right of $z_{\alpha}$.
1
By definition of $z_{\alpha}$, $P(Z\geqslant z_{\alpha})=\alpha$.
(a)
For $\alpha=0.0055$, we have $P(Z\geqslant z_{\alpha})=0.0055$. That is,
\[
0.0055=P(Z\geqslant z_{\alpha})=1-P(Z < z_{\alpha})=1-P(Z\leqslant z_{\alpha})=1-\Phi(z_{\alpha})
\]
It infers that $\Phi(z_{\alpha})=1-0.0055=0.9945$. By Table A.3, we have $\Phi(2.5+0.04)=0.9945$. Hence, $z_{\alpha}=2.5+0.04=2.54$.
(b)
For $\alpha=0.09$, we have $P(Z\geqslant z_{\alpha})=0.09$. That is,
\[
0.09=P(Z\geqslant z_{\alpha})=1-P(Z < z_{\alpha})=1-P(Z\leqslant z_{\alpha})=1-\Phi(z_{\alpha})
\]
It infers that $\Phi(z_{\alpha})=1-0.09=0.91$. By Table A.3, we have $\Phi(1.3+0.04)=0.9099$, which is close to $0.9100$ . Hence, $z_{\alpha}=1.3+0.04=1.34$.
(c)
For $\alpha=0.663$, we have $P(Z\geqslant z_{\alpha})=0.663$. That is,
\[
0.663=P(Z\geqslant z_{\alpha})=1-P(Z < z_{\alpha})=1-P(Z\leqslant z_{\alpha})=1-\Phi(z_{\alpha})
\]
It infers that $\Phi(z_{\alpha})=1-0.663=0.337$. By Table A.3, we have $\Phi(-0.4-0.02)=0.3372$, which is close to $0.3370$ . Hence, $z_{\alpha}=-0.4-0.02=-0.42$.