Use CLT to find approximation

Question

We are given $X_1,\dots,X_n$ i.i.d. Poisson($\lambda$) r.v.s. and assume $\lambda = 1$. We need to use CLT to find an approximation for:

$$P(X_1+X_2+\dots+X_{100} \leq 90)$$

What I have done:

• Let $Y = X_1+X_2 +\dots+ X_{100} $
• Therefore $Y\sim$ Poisson(100) (assuming $\lambda=1$)
• Also, if $\lambda$ is sufficiently large, $Y \sim N(100, 100)$

• Therefore:

  $ P(X_1+X_2+\dots+X_{100} \leq 90) = P\left(Z \leq \frac{90-100}{10}\right) = P(Z \leq -1) = P(Z \geq 1) = \Phi(1)$

I need to know if this is okay?

Answer 1

$\Phi(1)$ is wrong as it is greater than $\frac12$ and this probability is going to be less than $\frac12$. Instead you might consider $\Phi(-1)=1-\Phi(1)$. But you can do better.

It would probably be a better approximation if you use a continuity correction, as $X_1+X_2+\dots+X_{100} \leq 90$ is equivalent to $X_1+X_2+\dots+X_{100} \lt 91$ so you could try a normal approximation between these of $P\left(Z \leq \frac{90.5-100}{10}\right) = P(Z \leq -0.95) = 1-\Phi(0.95)$

The exact figure (up to rounding) can be found using R, and so we can compare the different approaches

• $\Phi(-1)$ with pnorm(-1) to give $0.1586553$
• $\Phi(-0.95)$ with pnorm(-0.95) to give $0.1710561$
• exact with ppois(90,100) to give $ 0.1713851$

and this is a demonstration of the improvement in the approximation from the continuity correction