A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mean 3125 and standard deviation 250.

Calculate the approximate 90th percentile for the distribution of the total contributions received.

• 6,328,000
• 6,338,000
• 6,343,000
• 6,784,000
• 6,977,000

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

Step 1: Identify Given Information

We are given information about a series of mutually independent and identically distributed contributions.

Contribution Parameters

Parameter	Value
Number of contributions ([math]n[/math])	2025
Mean of a single contribution ([math]\mu[/math])	3125
Standard deviation of a single contribution ([math]\sigma[/math])	250

Step 2: Calculate the Mean and Standard Deviation of the Total Contributions

Let [math]S[/math] be the total contributions received. Since the contributions are mutually independent and identically distributed, the mean of the total contributions is the sum of the individual means, and the variance of the total contributions is the sum of the individual variances.

• Mean of Total Contributions ([math]E[S][/math]):

The mean of the sum of [math]n[/math] independent and identically distributed random variables is [math]n[/math] times the mean of a single variable.

[[math]]E[S] = n \times \mu = 2025 \times 3125 = 6,328,125[[/math]]

• Standard Deviation of Total Contributions ([math]SD[S][/math]):

The variance of the sum of [math]n[/math] independent and identically distributed random variables is [math]n[/math] times the variance of a single variable. The standard deviation is the square root of the variance.

[[math]]Var[S] = n \times \sigma^2[[/math]]

[[math]]SD[S] = \sqrt{n \times \sigma^2} = \sqrt{n} \times \sigma = \sqrt{2025} \times 250 = 45 \times 250 = 11,250[[/math]]

Step 3: Apply the Central Limit Theorem (CLT)

According to the Central Limit Theorem, for a large number of independent and identically distributed random variables, the distribution of their sum (or average) approaches a normal distribution, regardless of the shape of the individual distributions. Given that there are 2025 contributions, which is a large number, the total contributions [math]S[/math] can be approximated by a normal distribution with the mean and standard deviation calculated in Step 2.

Step 4: Determine the Z-score for the 90th Percentile

To find the 90th percentile of a normally distributed variable, we need to find the Z-score that corresponds to a cumulative probability of 0.90. From standard normal distribution tables or calculators, the Z-score for the 90th percentile ([math]Z_{0.90}[/math]) is approximately 1.282.

Step 5: Calculate the 90th Percentile of Total Contributions

The formula for calculating a specific percentile of a normal distribution is:

[[math]]P_k = \text{Mean} + Z_k \times \text{Standard Deviation}[[/math]]

Using the values derived:

• Mean of total contributions ([math]E[S][/math]): 6,328,125
• Standard deviation of total contributions ([math]SD[S][/math]): 11,250
• Z-score for 90th percentile ([math]Z_{0.90}[/math]): 1.282

[[math]]P_{90} = 6,328,125 + 1.282 \times 11,250[[/math]]

[[math]]P_{90} = 6,328,125 + 14,422.5[[/math]]

[[math]]P_{90} = 6,342,547.5[[/math]]

Rounding to the nearest whole number, the approximate 90th percentile for the distribution of the total contributions received is 6,342,548. This corresponds to option C when considering the given choices.

Key Insights

• The Central Limit Theorem (CLT) is crucial for approximating the distribution of a sum (or average) of many independent and identically distributed random variables with a normal distribution.
• For a sum of [math]n[/math] independent and identically distributed random variables, if each has mean [math]\mu[/math] and standard deviation [math]\sigma[/math], the sum will have a mean of [math]n\mu[/math] and a standard deviation of [math]\sqrt{n}\sigma[/math].
• Percentiles of a normal distribution are calculated using the formula [math]\text{Percentile} = \text{Mean} + Z \times \text{Standard Deviation}[/math], where [math]Z[/math] is the Z-score corresponding to the desired percentile.

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

Solution: C

The mean and standard deviation for the 2025 contributions are 2025(3125) = 6,328,125 and 45(250) = 11,250. By the central limit theorem, the total contributions are approximately normally distributed. The 90th percentile is the mean plus 1.282 standard deviations or 6,328,125 + 1.282(11,250) = 6,342,548.

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