问题及解答

Let $Z$ be a standard normal random variable.

In each case, determine the value of the constant $c$ that makes the probability statement correct.

• (a) $\Phi(c)=.9838$
• (b) $P(0\leqslant Z\leqslant c)=.291$
• (c) $P(c\leqslant Z)=.121$
• (d) $P(-c\leqslant Z\leqslant c)=.668$
• (e) $P(c\leqslant|Z|)=.016$
   

(a)

From the Table A.3, we find that $\Phi(2.1+0.04)=0.9838$, hence $c=2.1+0.04=2.14$

(b)

\[
.291=P(0\leqslant Z\leqslant c)=\Phi(c)-\Phi(0)
\]

By the Table A.3, $\Phi(0)=0.5$. Thus, $\Phi(c)=\Phi(0)+0.291=0.5+0.291=0.791$. Again, from Table A.3, we have

$\Phi(0.8+0.01)=0.7910$. Hence, $c=0.8+0.01=0.81$.

(c)

\[
.121=P(c\leqslant Z)=1-P(Z < c)=1-P(Z\leqslant c)=1-\Phi(c)
\]

It infers that $\Phi(c)=1-0.121=0.879$. From Table A.3, we have $\Phi(1.1+0.07)=0.8790$. Hence, $c=1.1+0.07=1.17$.

(d)

\[
.668=P(-c\leqslant Z\leqslant c)=2\cdot P(0\leqslant Z\leqslant c)=2\cdot(\Phi(c)-\Phi(0))=2\cdot(\Phi(c)-0.5)
\]

It infers that $\Phi(c)=0.668/2+0.5=0.834$. From Table A.3, we have $\Phi(0.9+0.07)=0.834$. Hence, $z=0.9+0.07=0.97$.

(e)

\[
\begin{split}
.016=P(c\leqslant |Z|)&=1-P(|Z| < c)=1-P(|Z|\leqslant c)=1-2\cdot P(0\leqslant Z\leqslant c)\\
&=1-2\cdot(\Phi(c)-\Phi(0))=1-2\cdot(\Phi(c)-0.5)=2-2\cdot\Phi(c)
\end{split}
\]

It infers that $\Phi(c)=(2-0.016)/2=0.992$. From Table A.3, we find $\Phi(2.4+0.01)=0.9920$. Therefore, $c=2.4+0.01=2.41$.