Question

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

Answer 1

The 90th percentile for the distribution of the total contributions is $6,342,525.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n, the mean is
`\mu*n` and the standard deviation is
`s = √(n)*\sigma`

In this question:

`n = 2025, \mu = 3125*2025 = 6328125, \sigma = √(2025)*250 = 11250`

The 90th percentile for the distribution of the total contributions

This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`1.28 = (X - 6328125)/(11250)`

`X - 6328125 = 1.28*11250`

`X = 6342525`

The 90th percentile for the distribution of the total contributions is $6,342,525.