Approximating Poisson Probabilities I have a random variable X with Poisson distribution with mean 38.
I have to find the value that give the approximate value of the probability  that is obtained using the central limit theorem with a continuity correction for: 
$$
P(35 \leq X < 39)
$$
How I would like to solve it:
$$
= P(35 \leq X \leq 38)
$$
$$
= P(39.5) - P(34.5)
$$
$$
= \Phi(0.5157436) - \Phi(0.4633072)
$$
From that calculation I get 0.0524364. Is that right? 
UPDATE:
Hi @Stefan, as you proposed. I do standarized them as the following:
$$
 P(X \le 35) = (35.5 - 38)/\sqrt(38)
$$
$$
 P(X \le 38) = (38.5 - 38)/\sqrt(38)
$$
$$
\Phi(P(X\le38)-P(X\le35))
$$
is that right?
 A: See for example this. It states that a Poisson distribution with parameter $\lambda>10$ can be approximated by a normal distribution with mean $\mu=\lambda$ and variance $\sigma^2=\lambda$ if a continuity correction is used, i.e.
$$
P(X\leq x)=P(Z\leq x+0.5),
$$
where $Z\sim\mathcal{N}(\lambda,\lambda)$. In terms of $\Phi$ this is
$$
P(X\leq x)=P(Z\leq x+0.5)=P\left(\frac{Z-\lambda}{\sqrt{\lambda}}\leq\frac{x+0.5-\lambda}{\sqrt{\lambda}}\right)=\Phi\left(\frac{x+0.5-\lambda}{\sqrt{\lambda}}\right),
$$
because $\frac{Z-\lambda}{\sqrt{\lambda}}$ follows a standard normal distribution. Now, you should be able to approximate $P(35\leq X\leq 38)$.

Let us calculate $P(X\leq 38)$. By the above this can be approximated by 
$$
P(X\leq 38)=\Phi\left(\frac{38+0.5-38}{\sqrt{38}}\right)\approx \Phi\left(0.08111\right)\approx 0.5323
$$
and similarly $P(X\leq 34)\approx \Phi(-0.56778)\approx 0.28509$, and so (note that I subtract $P(X\leq 34)$ and not $P(X\leq 35)$ as you)
$$
P(35\leq X\leq 38)=P(X\leq 38)-P(X\leq 34)\approx 0.2472.
$$
