# 问题及解答

## [Exer11-2] Exercise 29 of Book {Devore2017B} P.169

Posted by haifeng on 2020-04-29 17:22:10 last update 2020-04-29 18:23:46 | Edit | Answers (1)

Determine $z_{\alpha}$ for the following:

• (a) $\alpha=.0055$
• (b) $\alpha=.09$
• (c) $\alpha=.663$

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

Posted by haifeng on 2020-04-29 19:17:11

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, $z_{\alpha}=-0.4-0.02=-0.42$.