Approximating Poisson Probabilities

Question

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?

Answer 1

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 $P(35\leq X\leq 38)$.

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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) $$ P(35\leq X\leq 38)=P(X\leq 38)-P(X\leq 34)\approx 0.2472. $$