Final answer:
Using the Central Limit Theorem, the total mean and variance of the claims filed can be calculated as 12500. The standard deviation is ~111.8, allowing us to use a normal approximation to compute the required probability.
Step-by-step explanation:
The central question involves determining if the Central Limit Theorem (CLT) can be used to approximate the probability that the total number of claims filed is between 2450 and 2600 inclusive, when each policyholder files claims according to a Poisson distribution with a mean of 5. Since the number of insurance policies is 2500, and the mean number of claims per policyholder is 5, we have a large number of independent Poisson trials, which is suitable for approximation using the CLT.
To use the CLT, we find the mean (μ) and the variance (σ2) of the total number of claims. For each policyholder, μ = 5, and since the variance of a Poisson distribution equals its mean, we also have σ2 = 5. For n = 2500 policyholders, the mean of the total number of claims is nμ = 2500 × 5 = 12500, and the variance of the total number of claims is nσ2 = 2500 × 5 = 12500.
The standard deviation (σ) of the total number of claims is therefore the square root of the variance, σ = √12500, which is approximately 111.8. We can then use the normal approximation to find the probability that the total number of claims is between 2450 and 2600. Because the interval is symmetric about the mean, we can use standard z-scores to determine this probability.
If we denote the total number of claims by C, the z-score for C = 2450 is (2450 - 12500) / 111.8 and the z-score for C = 2600 is (2600 - 12500) / 111.8. We can look up these z-scores in the standard normal distribution table to find the probabilities and calculate the difference to get the final probability.