Probability problem with Central limit theorem

Question

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.

Answer 1

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.

Answer 2

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 $$n->\infty$$