Probability problem with Central limit theorem An insurance company for sport planes has 100 planes insured with each planes insurance fee being 10.000 (for a year). In case of a plane crash the company pays 1 million to the owner. The probability for a plane crash is independent and equal to $P(crash) = 0.008$. What is the probability that the company will have to pay more than than they earn in a year.
My thoughts:
Let $X_i$ be a s.v. that shows if the $X_i$ plane has crashed or not.
We also know the company will have a loss if:
$Y = 10^6(1-X)$
So we are looking for the probability that:
$P(Y<0)$ or $P(X>1)$ or $1-P(X<=1)$
I thought I could try to use the normal approximation to solve it but I don't understand why it doesn't work.
 A: I agree with @ERR that you want $P(X > 2),$ where $X \sim Binom(100, .008).$
$Exact\; binomial.$ The easiest computation is to find $1 - P(X \leq 1).$
But $P(X \leq 1) = P(X = 0) + P(X = 1),$ where both terms are easy
to compute on a calculator. The answer 0.1909 to your problem
is correct to four places.
$Poisson\; approximation.$ If you want to use an approximation, then the Poisson approximation
would work reasonably well. Suppose $Y \sim Pois(\lambda),$ where
$\lambda = 100(.008) = 0.8.$ Thus $X$ and $Y$ have the same 
expected value: $E(X) = E(Y) = 0.8.$ Then you could
approximate the answer to your problem as
$1 - P(Y \le 1) = 1 - e^{-.8} - .8e^{-.8} = 0.1912.$
If binomial $n$ is large and $p$ is small, then $Pois(np)$ is
often a good approximation to $Binom(n, p).$
$Attempted\; normal\; approximation.$ The "best-fitting" normal distribution would have $\mu = .8$
and $\sigma = \sqrt{np(1-p)} = .8908.$ But it does not
fit well according to the rule of thumb quoted by @copper.hat. Standardizing and
using the continuity correction, you would get approximately
probability 0.2160. Perhaps not a horrible approximation, but not a very
good one either. (And the plot below shows that getting this close may be an accident of the particular values for which you need to find probabilities.) Furthermore, the exact binomial answer
and the well-approximating Poisson answers are easier to
compute---and they do not require tables.
The graph below shows exact binomial probabilities as vertical bars
and approximate Poisson probabilities as small circles. (Within the accuracy of the plot, roughly two places, they are hardly distinguishable from the tops of the bars.) The
attempted normal approximation would include the area to the right of
the dotted vertical line. All three distributions have negligible
probability to the right of 5.

A: Ok soo, we have:
An insurance company for sport planes has 100 planes insured with each plane's insurance fee being $10,000
Revenue = 100*10,000 = 1,000,000
The question asks: What is the probability that the company will have to pay more than than they earn in a year. Ie what is the probability that the claims exceed $1,000,000
We're given that "In case of a plane crash the company pays 1 million to the owner".
So you would need 2 plane crashes to exceed $1,000,000.
As calculus points out in the comments above, your random variable $$X∼Bin(100;0.008)$$
$$ P(X\geq2) =\sum _{n=2}^{100}(^{100}_n)*.008^n(1-.008)^{100-n}$$
by the definition of the binomial distribution. 
Putting this into Wolfram Alpha I get $$P(X\geq2) = .1909 \:approximately$$
P.S. You don't need/can't use the CLT, because you only have one random variable X. The CLT applies when you're taking the sum of iid random variables, for instance, if you had $$X_1, X_2, ....., X_n$$ binomial distributed random variables you could then use the CLT and say that the sum & or average of these R.V.s becomes approximately normal (or "approaches the normal in distribution") as $$n->\infty$$
