2500 people have a car insurance, there are 1500 women, and 1000 men. Probability that a woman does a car accident is 1%. Probability that a man does a car accident is 1.5%.
exercise asks to estimate probability that in the next year there will be at least 25 car accidents.
here's what I've done:
I've used central limit theorem, and I have to find $P(X \geq25)$, which is equal to $$P(N(0, 1)\geq \frac{(365)-(25*365)}{\sqrt(365*25)})$$
now, I've rewritten it as $$1-\phi(-91.7839)$$
and I highly doubt this result is correct. So I've read the text again,
- 25 car accidents is the required data in the next year.
- of course, 365 is the total amount of days of the year.
in central limit theorem formula there's no space for probabilities (i.e P(woman), and P(man)), therefore these data are useless for the sake of central limit theorem.
I think I have to take into account the total number of people who have the car insurance. But I don't think it's correct, because the average and the variance of a normal distribution are: $$average = 365 * 25, \space variance = 365 * 25 $$
whenever I have an exercise like this, I have to compute the average and the variance in the same way. [because of theory. 365 is the number of events, and I have to multiply it by the average of the previous random variable]
I have no clue.
There are versions of CLT that does not require identically distributed random variables. See, e.g, wikipedia and http://personal.psu.edu/drh20/asymp/fall2002/lectures/ln04.pdf.
Let $W_i$, $i = 1,\dots,1500$ be a random variable such that $P(W_i = 0) = 0.01$ and $P(W_i = 0) = 0.99$. Similarly, let $M_j$, $j = 1,\dots,1000$ be such that $P(M_j = 1) = 0.015$ and $P(M_j = 0) = 0.985$. Then you need to calculate $$ P\left(\sum_i W_i + \sum_j M_i \ge 25\right). $$ Rewriting the event in the form $$ P\left(\frac{\sum_i (W_i - 0.01) + \sum_j (M_j - 0.015)}{\sqrt{1500 \cdot 0.01 \cdot 0.99 + 1000 \cdot 0.015 \cdot 0.985}} \ge \frac{-5}{\sqrt{29.625}} \right) = P\left(\mathcal{N}(0,1) \ge \frac{-5}{\sqrt{29.625}} \right) = 0.82098\dots, $$ which is more or less the same answer as this obtained by the method of @Matthew H.