Let's say you want to compute the probability [tex]\mathbb P(a\le X\le b)[/tex] where [tex]X[/tex] converges in distribution to [tex]Y[/tex], and [tex]Y[/tex] follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing [tex]Y[/tex] such that its mean and variance are the same as those for [tex]X[/tex].
Example: If [tex]X[/tex] is binomially distributed with [tex]n=100[/tex] and [tex]p=0.1[/tex], then [tex]X[/tex] has mean [tex]np=10[/tex] and variance [tex]np(1-p)=9[/tex]. So you can approximate a probability in terms of [tex]X[/tex] with a probability in terms of [tex]Y[/tex]:
[tex]\mathbb P(a\le X\le b)\approx\mathbb P(a\le Y\le b)=\mathbb P\left(\dfrac{a-10}3\le\dfrac{Y-10}3\le\dfrac{b-10}3\right)=\mathbb P(a^*\le Z\le b^*)[/tex]
where [tex]Z[/tex] follows the standard normal distribution.
