Poisson approximation to Binomial A life insurance company found that the probability  that a randomly selected application contains an error is $3 \%$. The application are mutually independent. An auditor randomly select $100$ applications. Calculate the probability that $95 \%$ or less of the selected applications are error-free
try
We want to find the probability that fewer than 95 applications are error-free, which is the same as 5 error or more. Let $X$ be the number of errors in an application. We could use binomial with $n=100$ and $p=0.03$, but since $p$ is small and $n$ large we could use Poisson with $\lambda = pn = 0.03 \times 100 = 3 $. Thus,
$$ P(X \geq 5) = 1 - P(X<5) = 1 - \sum_{i=0}^4 \frac{ 3^i e^{-3}}{i!} $$
is this correct?
 A: Note: I am carrying out @lulu's suggestions because I think there are
important insights to be gained from doing so.
Binomial: The number $X$ of errors is has the distribution $\mathsf{Binom}(n=100, p=.03).$ You seek $P(X \ge 5) = 1 - P(X \le 4) = 0.1821.$ For an exact
evaluation with the binomial PDF (or PMF) you would need to sum five terms.
Using software (statistical calculator or computer program) may be easier.
With R statistical software:
1 - pbinom(4, 100, .03)
## 0.1821452

Poisson: The number of errors is approximately $Y \sim \mathsf{Pois}(\lambda = 3),$ where $\lambda = np = 3,$ which is the same as the binomial mean.
You seek $P(Y \ge 5) = 1 - P(Y \le 4) = 0.1847.$ For an exact
evaluation with the Pois PDF (or PMF) one would need to sum five terms, as you have shown in your Question.
With R statistical software:
1 - ppois(4, 3)
## 0.1847368

In this case, the Poisson approximation to binomial gives two decimal place
accuracy. For practical purposes, that may be good enough. It would take
hundreds of such audits to distinguish between the two probabilities.
Normal: The number of errors is approximately $Z \sim \mathsf{Norm}(\mu = 3, \sigma = 1.706),$ where $\mu = np = 3, \sigma = \sqrt{np(1-p)} = 1.705872.$
You seek $$P(Z \ge 5) = 1 - P(Z \le 4) = 1 - P(Z \le 4.5) = 0.1896.$$
The use of 4.5, instead of 4 or 5, is known as the 'continuity correction' (which you can google or search on this site). In R the result is as follows:
1 - pnorm(4.5, 3, 1.706)
## 0.1896329

You can also find the normal approximation by standardizing and using
printed normal tables. Because of the rounding required to use normal
tables, values obtained that way may differ slightly from software values.
Different textbooks give various 'rules of thumb' for values of $n$ and $p$
that lead to a useful normal approximation (often not better than two decimal
place accuracy). Because $n$ is large here, this situation satisfies some
of those rules; but because $p$ is far from $1/2$ it does not satisfy others.
Generally, normal approximations work better when $0.3 < p < 0.7,$ whatever
guideline you are using. 
The figure below shows binomial probabilities (solid blue bars), Poisson
probabilities (dotted orange), and the approximating normal density function
(black curve). The exact binomial probability is the sum of the heights of
the blue bars to the right of the heavy purple vertical line. (Probabilities for more than about ten errors are negligible.)

