Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To determine whether the electron wave function is normalized, we calculate the area under the $|\psi(x)|^2$ curve. If it equals 1, the wave function is normalized; if not, it is not.
$$A=2\left[\frac{1}{2}(1\;{\rm cm})(1\;{\rm cm^{-1}})\right]=1$$
Therefore, yes, it is normalized.
$$\color{blue}{\bf [b]}$$
We know that the graph of $\psi(x)$ is the square root of $|\psi(x)|^2$.
See the graph below.
$$\color{blue}{\bf [c]}$$
$\bullet$ We know that the probability of detecting a particle in a region of $\delta x$ is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x=0)=|\psi(0)|^2(0.001)=(0)(0.001)=\color{red}{\bf 0}\%$$
where at $x=0$, $|\psi(0)|^2=0$ as shown in the given graph.
$\bullet$ At $x=0.5$ cm, we can see that $|\psi(0)|^2=0.5\;\rm cm^{-1}$
$$\text{Prob}( \text{in } \delta x \text{ at } x=0.5{\;\rm cm}) =(0.5)(0.001)=\color{red}{\bf 5\times 10^{-4}}\%$$
$\bullet$ At $x=0.999$ cm, we can see that $|\psi(0)|^2=0.5\;\rm cm^{-1}$
$$\text{Prob}( \text{in } \delta x \text{ at } x=0.999{\;\rm cm}) =(0.999)(0.001)=\color{red}{\bf 9.99\times 10^{-4}}\%$$
$$\color{blue}{\bf [c]}$$
We know that
$$\text{Prob}( \text{in } \delta x \text{ at } x=0) =\text{Area under the $|\psi(x)|^2$ curve}=\dfrac{N}{N_{tot}}$$
where $N$ is the number of detected electrons while $N_{tot}$ is the total number of electron.
Thus,
$$N=N_{tot}\cdot \text{Prob}( \text{in } \delta x\;\text{ at }-0.3\;{\rm cm}\leq x \leq0.3{\;\rm cm})$$
Finding the area under the curve for the chosen region.
$$N=N_{tot}A=N_{tot}\left[ 2\left(\frac{1}{2}bh\right)\right]$$
where $b$ is the width in the $x$-direction and $h$ is the height in the $y$-direction.
Plug the known
$$N=N_{tot}A=(10^4)\left[ 2\left(\frac{1}{2}(0.3)(0.3)\right)\right]$$
$$N=\color{red}{\bf 900}\;\rm electron$$
