<s>[INST] Standard normal distribution - Calculating the probability of an electron Figure $1$  shows $Bohr$'s model for the hydrogen atom. We let $(X, Y)$ denote the position of the electron in relation to the proton-core. In quantum mechanics one can never determine $(X, Y)$ exactly, but we are simplifying the model by using probability distribution. We assume $X∼N(0,1)$ and $Y∼N(0,1)$, i.e. standard normal distribution with expectation $E(X)=0$ and variance $V(X)=1$. $X$ and $Y$ are independent.


Figure 1: Bohrs model with an electron in orbit around the proton-core. The black circle has radius r = 2,15.

(A)
What is the probability that $-2,15 <= x <= 2,15$? What is the probability that the electron is inside the black square? Hint: We assume $X$and $Y$ are independent random variables.
My answer:
To determine $P(a <= x <= b)$ when $X$ is a norml random variable with parameters $\mu$ and $\sigma$, we must evaluate
\begin{align}\\
& \int_{a}^b\frac{1}{\sqrt{2\pi}\sigma}e^{\left( \frac{-(x-\mu)^2}{2\sigma ^2} \right)} dx\\ 
\end{align}
So we have:
\begin{align}\\
& \Rightarrow P(-2,15\le X\le 2,15)=\int_{-2,15}^{2,15}\frac{1}{\sqrt{2\pi }} * {e^{\frac{-x^2}{2}}} dx \\ 
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\left( \frac{x}{2}\right) ^2} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^u du \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left[ e^u + C \right]_{-2,15}^{2,15} \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( e^{\left( \frac{2,15}{2}\right) ^2} - e^{\left( \frac{-2,15}{2}\right) ^2} \right) \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( 0 \right) \\
\end{align}
I feel like I have done something wrong with the calculations but I donw know what. Hpw do I calculate the probability for if the electron is within the black square?
 [/INST] You cannot integrate the normal density by substitution over a finite interval. You need tables or a computer.
Your answer should have been $\Phi(2.15)-\Phi(-2.15) \approx 0.968$
Among other errors, you have the false steps $$\int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \to \int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx$$ $$\int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx \to \int_{-2,15}^{2,15} e^{u} du$$ neither of which can be justified
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