Suppose $X$ is Poisson with parameter $\lambda$, and $Y$ is normal with mean and variance $\lambda$. It seems to me that the appropriate comparison is between $\Pr(X = n)$ and $\Pr(Y \in [n-\frac12,n+\frac12])$. Here for simplicity I write $n = \lambda + \alpha \sqrt\lambda$, that is, we are interested when $n$ corresponds to $\alpha$ standard deviations from the mean.
So I cheated. I used Mathematica. So both $\Pr(X = n)$ and $\Pr(Y \in [n-\frac12,n+\frac12])$ are asymptotic to
$$ \frac 1{\sqrt{2\pi \lambda}} e^{-\alpha^2/2} $$
as $\lambda \to \infty$. But their difference is asymptotic to
$$ \frac{\alpha \left(\alpha ^2-3\right) e^{-\alpha ^2/{2}}}{6 \sqrt{2
\pi } \lambda } $$
If you plot this as a function of $\alpha$, you will get the same curve as is shown in the second to last figure in http://www.johndcook.com/blog/normal_approx_to_poisson/.
Here are the commands I used:
n = lambda + alpha Sqrt[lambda];
p1 = Exp[-lambda] lambda^n/n!;
p2 = Integrate[1/Sqrt[2 Pi]/Sqrt[lambda] Exp[-(x-lambda)^2/2/lambda], {x, n-1/2, n+1/2}];
Series[p1, {lambda, Infinity, 1}]
Series[p2, {lambda, Infinity, 1}]
Also, with a bit of experimentation, it seems to me that a better asymptotic approximation to $\Pr(X = n)$ is $\Pr(Y \in [n-\alpha^2/6,n+1-\alpha^2/6])$. Then the error is
$$ -\frac{\left(5 \alpha ^4-9 \alpha ^2-6\right) e^{-{\alpha ^2}/{2}}
}{72 \sqrt{2 \pi } \lambda ^{3/2} } $$
which is about $\sqrt\lambda$ times smaller.
If $X \sim N(0,1)$, then $Y = \sigma X + \mu \sim N(\mu,\sigma^2)$. So,
leave the RNG and Box-Muller alone, and when you get the generated
variate, just multiply by $\sigma$ and add $\mu$. You could stick these
inside the Box-Muller code if you really want just simple code by
modifying
let y1 sqrt (2 - ln x1) * cos (2 * pi * x2)
to
let y1 sigma * sqrt (-2 * ln x1) * cos (2 * pi * x2) + mu
and similarly for y2.
Best Answer
I don't know what exactly is wanted here but with $\Phi(\cdot)$ denoting the cumulative probability distribution function of the standard normal random variable, we can say the following.
If $a, b > 0$, then $\Phi(a) + \Phi(b) \leq 2\Phi\left(\frac{a+b}{2}\right)$ with equality holding only when $a=b$.
If $a, b < 0$, then $\Phi(a) + \Phi(b) \geq 2\Phi\left(\frac{a+b}{2}\right)$ with equality holding only when $a=b$.
If $a < 0, b > 0$, then $\Phi(a) + \Phi(b)$ is greater than $1$ or less than $1$ according as $|a| < b$ or $|a| > b$ with equality only when $a = -b$.