Answer
(a) $\color{blue}{-0.44} $
(b) $\color{blue}{0.76}$
(c) $\color{blue}{0.41}$
(d) $\color{blue}{1.282\ \text{(interpolated)} \;\approx\; 1.28}$
(e) $\color{blue}{0.949\ \text{(interpolated)} \;\approx\; 0.95}$
Work Step by Step
(a)
$\begin{align*}
P(Z \le z) &= 0.33 \\
F_Z(z) &= 0.33 \\
F_Z(z) &= F_Z(-0.44) \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\
z &\color{blue}{= -0.44} \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ]
\end{align*}$
(b)
$\begin{align*}
P(Z \ge z) &= 0.2236 \\
P(Z \gt z) &= 0.2236 \\
P(Z\le z) &= 1-0.2236 \\
F_Z(z) &= 0.7764 \\
F_Z(z) &= F_Z(0.76) \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\
z &\color{blue}{= 0.76} \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ]
\end{align*}$
(c)
$\begin{align*}
P(-1.00 \le Z \le z) &= 0.5004 \\
P(-1.00 \lt Z \le z) &= 0.5004 \\
P(Z \le z) - P(Z\le -1.00) &= 0.5004 \\
F_Z(z) - F_Z(-1.00) &= 0.5004 \\
F_Z(z) - 0.1587 &= 0.5004 \\
F_Z(z) &= 0.6591 \\
F_Z(z) &= F_Z(0.41) \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\
z &\color{blue}{= 0.41} \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ]
\end{align*}$
(d)
$\begin{align*}
P(-z \le Z \le z) &= 0.80 \\
2P(0 \lt Z \le z) &=0.80 \qquad \text{[ since }\ f_z(z)\ \text{is symmetric about}\ x=0\ ] \\
P(0 \lt Z \le z) &= 0.40 \\
P(Z\le 0) + P(0\lt Z\le z) &= 0.50 + 0.40 \qquad \text{[ since }\ P(Z\le 0) = 0.50, Z\sim N(0,1)\ ] \\
P(Z\le z) &= 0.9000 \\
F_Z(z) &= 0.9000 \\
F_Z(z) &= F_Z(1.282) \qquad \text{[ see Appendix Table A.1, pp. 675-6; interpolate ]} \\
z &\color{blue}{= 1.282 \;\approx\; 1.28} \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ]
\end{align*}$
(e)
$\begin{align*}
P(z \le Z \le 2.03) &= 0.15 \\
P(z\lt Z \le 2.03) &= 0.15 \\
P(Z \le 2.03) - P(Z\le z) &= 0.15 \\
F_Z(2.03) - F_Z(z) &= 0.15 \\
0.9788 - F_Z(z) &= 0.15 \\
0.9788 - 0.15 &= F_Z(z) \\
0.8288 &= F_Z(z) \\
F_Z(0.949) &= F_Z(z) \qquad \text{[ see Appendix Table A.1, pp. 675-6; interpolate ]} \\
z &\color{blue}{= 0.949 \;\approx\; 0.95} \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ]
\end{align*}$
