A survey was conducted with a population size of 500 resulted in 1/3 of all taxpayers having adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of itemized deduction was $16,642 from a sample size of 75 and a standard deviation of $2,400. What is the probability the itemized deductions were within $200 of the sample mean?

52.84% probability the itemized deductions were within $200 of the sample mean

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.

In this question, we have that:

What is the probability the itemized deductions were within $200 of the sample mean?

This is the pvalue of Z when X = 16642 + 200 = 16842 subtracted by the pvalue of Z when X = 16642 - 200 = 16442. So

X = 16842

By the Central Limit Theorem

has a pvalue of 0.7642

X = 16442

has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability the itemized deductions were within $200 of the sample mean