An insurance company issues 1250 vision care insurance policies. The number of claims filed by a policyholder under a vision care insurance policy during one year is a Poisson random variable with mean 2. Assume the numbers of claims filed by different policyholders are mutually independent.

Calculate the approximate probability that there is a total of between 2450 and 2600 claims during a one-year period.

• 0.68
• 0.82
• 0.87
• 0.95
• 1.00

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Analyze Individual Policy Claims

The number of claims filed by a single policyholder is given as a Poisson random variable with a mean of 2.

• Poisson Distribution Properties:

For a Poisson random variable [math]X[/math] with mean [math]\lambda[/math]:

• Expected Value (Mean): [math]E[X] = \lambda[/math]
• Variance: [math]Var[X] = \lambda[/math]

In this case, for a single policyholder, [math]\lambda = 2[/math]. Therefore, for one policy:

• Mean of claims = 2
• Variance of claims = 2

Step 2: Determine the Total Number of Claims Distribution

We have 1250 policyholders, and their claim numbers are mutually independent. Let [math]X_i[/math] represent the number of claims for policyholder [math]i[/math]. The total number of claims, [math]S[/math], is the sum of claims from all policyholders:

[[math]]S = \sum_{i=1}^{1250} X_i[[/math]]

Since each [math]X_i[/math] is an independent Poisson random variable with mean [math]\lambda = 2[/math], the sum [math]S[/math] is also a Poisson random variable with a mean equal to the sum of the individual means.

• Mean of Total Claims:

[[math]]E[S] = \sum_{i=1}^{1250} E[X_i] = 1250 \times 2 = 2500[[/math]]

• Variance of Total Claims:

Due to the independence of claims, the variance of the sum is the sum of the individual variances.

[[math]]Var[S] = \sum_{i=1}^{1250} Var[X_i] = 1250 \times 2 = 2500[[/math]]

Step 3: Apply Normal Approximation

For a Poisson distribution with a large mean (in this case, [math]E[S] = 2500[/math]), it can be accurately approximated by a Normal distribution.

Parameters for the Normal Approximation:

• Mean ([math]\mu[/math]): [math]\mu = E[S] = 2500[/math]
• Variance ([math]\sigma^2[/math]): [math]\sigma^2 = Var[S] = 2500[/math]
• Standard Deviation ([math]\sigma[/math]): [math]\sigma = \sqrt{2500} = 50[/math]

We need to calculate the approximate probability that the total number of claims is between 2450 and 2600, which can be written as [math]P(2450 \lt S \lt 2600)[/math].

Step 4: Standardize the Range using Z-scores

To use the standard normal distribution table, we convert the range values to Z-scores using the formula:

[[math]]Z = \frac{X - \mu}{\sigma}[[/math]]

• Lower Bound Z-score:

For [math]X = 2450[/math]:

[[math]]Z_1 = \frac{2450 - 2500}{50} = \frac{-50}{50} = -1[[/math]]

• Upper Bound Z-score:

For [math]X = 2600[/math]:

[[math]]Z_2 = \frac{2600 - 2500}{50} = \frac{100}{50} = 2[[/math]]

So, we are looking for the probability [math]P(-1 \lt Z \lt 2)[/math].

Step 5: Calculate the Probability

Using the standard normal distribution tables, we find the cumulative probabilities for the Z-scores:

• [math]P(Z \lt 2) = 0.9772[/math]
• [math]P(Z \lt -1) = 1 - P(Z \lt 1)[/math]
• [math]P(Z \lt 1) = 0.8413[/math]
• So, [math]P(Z \lt -1) = 1 - 0.8413 = 0.1587[/math]

The probability [math]P(-1 \lt Z \lt 2)[/math] is given by:

[[math]]P(-1 \lt Z \lt 2) = P(Z \lt 2) - P(Z \lt -1)[[/math]]

[[math]]= 0.9772 - 0.1587 = 0.8185[[/math]]

The approximate probability that there is a total of between 2450 and 2600 claims during a one-year period is 0.8185.

Key Insights

• The sum of independent Poisson random variables is itself a Poisson random variable. Its mean and variance are the sum of the individual means and variances, respectively.
• For a Poisson distribution, the mean and variance are equal ([math]\lambda[/math]). This property is maintained for the sum of independent Poisson variables.
• A Poisson distribution with a sufficiently large mean (typically [math]\lambda \gt 10[/math]) can be accurately approximated by a Normal distribution. The mean and variance of the approximating Normal distribution are set equal to the Poisson distribution's parameters.
• To calculate probabilities for a continuous random variable using a normal approximation, it is essential to standardize the values using Z-scores: [math]Z = \frac{X - \mu}{\sigma}[/math].
• Probabilities for a range [math](a, b)[/math] in a standard normal distribution are calculated as [math]P(a \lt Z \lt b) = P(Z \lt b) - P(Z \lt a)[/math].

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: B

A single policy has a mean and variance of 2 claims. For 1250 polices the mean and variance of the total are both 2500. The standard deviation is the square root, or 50. The approximate probability of being between 2450 and 2600 is the same as a standard normal random variable being between (2450 – 2500)/50 = –1 and (2600 – 2500)/50 = 2. From the tables, the probability is 0.9772 – (1 – 0.8413) = 0.8185.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.