Sale prices are $390,000, $410,000, $430,000, $450,000. Find the arithmetic mean of their absolute deviations from the sample mean. Report to the nearest dollar.
Correct Answer
B) $20,000
The center is $420,000. Absolute deviations are $30,000, $10,000, $10,000, $30,000. Their mean is $20,000. Averaging absolute deviations and taking the median of those deviations are different operations.
Why This Is the Correct Answer
Option B is correct because the mean is ($390,000 + $410,000 + $430,000 + $450,000) / 4 = $420,000. The absolute deviations are $30,000, $10,000, $10,000 and $30,000, which sum to $80,000, and $80,000 / 4 = $20,000.
Why the Other Options Are Wrong
Option A: $420,000
Option A, $420,000, is the sample mean itself, the first step of the calculation. The question asks for the average distance from that mean, which is $80,000 / 4 = $20,000.
Option C: $60,000
Option C, $60,000, is the range, $450,000 − $390,000, not an average of deviations. The mean absolute deviation is ($30,000 + $10,000 + $10,000 + $30,000) / 4 = $20,000.
Option D: $80,000
Option D, $80,000, is the sum of the absolute deviations, $30,000 + $10,000 + $10,000 + $30,000, with the final averaging step left out. Dividing that sum by the four prices gives $80,000 / 4 = $20,000.
Center, distance, average
Three steps: find the center (mean), measure each distance from it ignoring signs, then average the distances.
How to use: Use this to work any mean absolute deviation problem step by step.
Exam Tip
Watch for answer choices that stop at an intermediate step, such as the mean or the total of the deviations, or that substitute the range. Confirm your final figure is an average of distances.
Common Mistakes to Avoid
- -Reporting the mean instead of the deviation
- -Adding the absolute deviations but forgetting to divide by the count
- -Reporting the range as if it were an average deviation
Concept Deep Dive
Analysis
This question tests the average absolute deviation about the mean. First find the mean, then measure how far each price is from it without regard to sign, and finally average those distances. Keeping the steps in order prevents confusing the result with the mean itself or the range.
Background Knowledge
The average absolute deviation measures typical distance from a center, using absolute values so that positive and negative deviations do not cancel out. It differs from the range, which uses only the two extreme values, and from the median absolute deviation, which takes the median of the distances instead of their mean.
Real-World Application
An appraiser reviewing four comparable sales wants a simple measure of how tightly they cluster. A mean absolute deviation of $20,000 around a $420,000 average tells the reader the typical sale is within about $20,000 of the mean.
More Statistics Questions
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Compared with analyzing one matched sale pair for one feature, what can a suitably specified multiple regression do?
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Assessment-to-sale ratios are 0.94, 0.96, 0.98, 1, 1.01. Compute the arithmetic mean. Report as a decimal ratio to three places.
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