A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea
1 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 and standard deviation
, the zscore of a measure X is given by:
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 and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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 and the standard deviation is
In this question:
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
By the Central Limit Theorem
The 90th percentile for the distribution of the total contributions is $6,342,525.
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Answer:
(a) <em> </em><em>n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b) The correct option is (b).
Step-by-step explanation:
Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.
The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.
Assuming that the random variable <em>X </em>follows a normal distribution.
(a)
Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:
- For a sample size of <em>n</em> = 20
- For a sample size of <em>n</em> = 50
- For a sample size of <em>n</em> = 100
- For a sample size of <em>n</em> = 500
<em> n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b)
The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample () approaches the whole population mean (
).
Consider the probabilities computed in part (a).
As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.
So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
Thus, the correct option is (b).