Answer
a) $\color{blue}{0.5782}$
b) $\color{blue}{0.8264}$
c) $\color{blue}{0.9306}$
d) $\color{blue}{0.0000}$
Work Step by Step
a)
$\begin{align*}
\int_{-0.44}^{1.33} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= P(Z\le 1.33) - P(Z\le -0.44),\quad Z\sim N(0,1) \\
&= F_Z(1.33) - F_Z(-0.44) \\
&= 0.9082 - 0.3300 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\
\int_{-0.44}^{1.33} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= \color{blue}{0.5782}
\end{align*}$
b)
$\begin{align*}
\int_{-\infty}^{0.94} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= P(Z\le 0.94) - P(Z\le -\infty),\quad Z\sim N(0,1) \\
&= F_Z(0.94) - F_Z(-\infty) \\
&= 0.8264 - 0 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]}\\
\int_{-\infty}^{0.94} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= \color{blue}{0.8264}
\end{align*}$
c)
$\begin{align*}
\int_{-1.48}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= P(Z\le \infty) - P(Z\le -1.48),\quad Z\sim N(0,1) \\
&= F_Z(\infty) - F_Z(-1.48) \\
&= 1 - 0.0694 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]}\\
\int_{-1.48}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= \color{blue}{0.9306}
\end{align*}$
d)
$\begin{align*}
\int_{-\infty}^{-4.32} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= P(Z\le -4.32) - P(Z\le -\infty),\quad Z\sim N(0,1) \\
&= F_Z(-4.32) - F_Z(-\infty) \\
&= 0.000 0- 0.0000 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]}\\
\int_{-\infty}^{-4.32} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\ dz &= \color{blue}{0.0000}
\end{align*}$
