Poisson distribution with exact value of x and mean is a larger number $X $ follows $ Po(80)$. 
I used a normal approximation to get $P(55\leq X\leq 75)$ which I got correct. 
I need to find $P(X=80)$. I tried the Poisson distribution directly; however, my calculator shows an error when calculating $80^{80}$.
The correct answer is 0.0446.
Do we need to use normal distribution? 
Thanks.
 A: Here I discuss several approaches. If I wasn't near my computer (on which I'd just do dpois(80,80) in R), I'd probably use the first one below.
1) You can help your calculator out a bit by doing some algebra first, and trying Stirling's approximation for $80!$
You can try taking the series for the Stirling approximation out several terms but it won't be necessary in this case, just use the "ordinary" approximation ($n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$), which hugely simplifies the calculation, and has about 3 significant figure accuracy in this region (which will be sufficient). [Taking one more term in the series - $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n(1+\frac{1}{12n})$ - gives about six figure accuracy.]
So take the formula for the probability:
$$P(X=80) = \frac{\exp(-80) 80^{80}}{80!}$$
Then just replace 80! by the above approximation, and cancel out all those large terms, leaving a very simple calculation.
(I bet you can do it from there.)
2) The "right" way to do that is to use a function (or a series, if you are working on a calculator) that returns a log-factorial (or log-gamma-function), and work on the log-scale until the final step ... which gives the "exact" answer of about 0.04455667. While quite accurate, this may involve a bit of work unless you're using a computer.
3) If you use the normal approximation, the usual way would be to use the continuity correction. Let $X$ be the original Poisson, and let $X^*$ be an approximating normal ($N(80,(\sqrt{80})^2$), and let $Z$ be a standard normal.
$$P(X=80)\approx P(79.5<X^*<80.5)$$ 
$$= P(\frac{79.5-80}{\sqrt{80}}<\frac{X^*-80}{\sqrt{80}}<\frac{80.5-80}{\sqrt{80}})$$
$$= P(\frac{-0.5}{\sqrt{80}}<Z<\frac{0.5}{\sqrt{80}})$$
$$= P(-0.055902<Z<0.055902)$$
$$= 2\,P(0<Z<0.055902)$$
$$= 2\times  0.02229$$
$$= 0.04458$$
... which is also sufficiently accurate.
4) Another way to do the approximation would be to approximate it by $f_{X^*}(80)\times 1\approx 0.0446$, but you have to be careful to justify that properly (refer to the diagrams and discussion at the link on continuity corrections I gave above).
A: You could get a better calculator or use logarithms for large powers: 
$$\log_{10}\left(80^{80}\right)=80\times \log_{10}\left(80\right)\approx 152.247199$$
$$80^{80} \approx 10^{0.247199} \times 10^{152} \approx 1.76685\times 10^{152}$$
If you used a normal approximation, you have two choices:


*

*the density at $x=80$ of a normal distribution with mean $80$ and standard deviation $\sqrt{80}$, which is $\tfrac{1}{\sqrt{160 \pi}}$ or about  $0.044603$ 

*the probability of a value between $79.5$ and $80.5$ from a normal distribution with mean $80$ and standard deviation $\sqrt{80}$, which is about $0.044580$


The actual value from a Poisson distribution  is closer to $0.044557$ so these approximations are not bad.
A: $\ln(Poisson_{80}(80))=\exp(\ln(Poisson_{80}(80)))=\exp(80 \ln 80-80-\ln(80!))$
For $\ln(80!)$ use Ramanujan's approximation.
