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Answer :
Answer:
The 90th percentile for the distribution of the total contributions is $6,342,525.
Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu*n[/tex] and the standard deviation is [tex]s = \sqrt{n}*\sigma[/tex]
In this question:
[tex]n = 2025, \mu = 3125*2025 = 6328125, \sigma = \sqrt{2025}*250 = 11250[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.28 = \frac{X - 6328125}{11250}[/tex]
[tex]X - 6328125 = 1.28*11250[/tex]
[tex]X = 6342525[/tex]
The 90th percentile for the distribution of the total contributions is $6,342,525.
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