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].

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