Exercise :
A car insurance company has $10.000$ customers. The average annual compensation per customer is $240$ with Standard Deviation $800$. Calculate by approximation the probability of the total annual compensations to be bigger than $2.500.000$.
Attempt:
I think the correct approach is by the Central Limit Theorem (CLT).
$$n=10.000$$
$$μ=240$$
$$σ=800$$
$$\frac{X - μ}{σ/\sqrt{n}}=\frac{\sum x_i - nμ}{σ\sqrt{n}}$$
$$\Rightarrow$$
$$\frac{X - μ}{σ/\sqrt{n}} = \frac{2.500.000 - 10.000*240}{800*\sqrt{10.000}}$$
So :
$$P(X > 2.500.000)= P\bigg(Z > \frac{X - μ}{σ/\sqrt{n}} = \frac{2.500.000 - 10.000*240}{800*\sqrt{10.000}}\bigg)= 1- P(Z \leq 1.25)$$
Is this approach correct ? If not I would really appreciate a thorough solution and explanation.