Question

The average income tax refund for the 2009 tax year was $3045. Assume the refund per person follows the nomal probabily distribution with a standard deviation of S989. Complete parts a d below. a. What is the probabilfy that a randomly selected tax return retund will be more than $1900 ? (Round to four decimat places as needed.)

Answer 1

To find the probability of a tax return refund being more than $1900, calculate the Z-score using the mean of $3045 and standard deviation of $989, then find the corresponding probability in a normal distribution table or using a calculator.

Step-by-step explanation:

The question involves finding the probability that a randomly selected tax return refund from the 2009 tax year will be more than $1900, given that the average (mean) refund was $3045 and followed a normal probability distribution with a standard deviation of $989.

To find this probability, we would use the Z-score formula, which is Z = (X - μ) / σ, where X is the value in question ($1900), μ is the mean ($3045), and σ is the standard deviation ($989). After calculating the Z-score, we would look up the corresponding probability in a standard normal distribution table or use a Z-score calculator to find the area to the right of the Z-score, which represents the probability of a refund being more than $1900.

The process is a common procedure in statistics for dealing with normal distributions and requires understanding of Z-scores and their relation to probabilities.