Sale-price observations are $290,000, $310,000, $325,000, $340,000, $385,000. Compute the arithmetic mean. Report to the nearest dollar.
Correct Answer
A) $330,000
Sum=1650000; n=5; mean=1650000/5=330000. Common errors: computing the median instead of the mean gives $325,000; reporting the lowest observation gives $290,000; reporting the highest observation gives $385,000.
Why This Is the Correct Answer
Option A is correct because $290,000 + $310,000 + $325,000 + $340,000 + $385,000 = $1,650,000, and dividing by 5 observations gives $330,000. The arithmetic mean uses the size of each value.
Why the Other Options Are Wrong
Option B: $325,000
Option B, $325,000, is the median, the middle of the five ordered prices. The mean is $1,650,000 / 5 = $330,000.
Option C: $290,000
Option C, $290,000, is the lowest observation. The mean is $1,650,000 / 5 = $330,000.
Option D: $385,000
Option D, $385,000, is the highest observation. The mean is $1,650,000 / 5 = $330,000.
Mean: add, then share
The mean is like splitting a total equally: add all the prices, then share the total evenly across the count.
How to use: Use this to compute the mean of any list of values.
Exam Tip
Separate the mean from the median, minimum and maximum, which are the usual distractors.
Common Mistakes to Avoid
- -Reporting the median instead of the mean
- -Dividing by the wrong number of observations
- -Selecting the highest or lowest value
Concept Deep Dive
Analysis
This question tests calculating the arithmetic mean. Add all five prices and divide by 5. The sum is $1,650,000, and $1,650,000 / 5 = $330,000.
Background Knowledge
The arithmetic mean is the sum of the values divided by the number of values. It reflects the size of each observation, so a single high or low value can shift it. Here the $385,000 sale lifts the mean slightly above the median of $325,000.
Real-World Application
An appraiser summarizing five sales of similar ranch homes reports a mean of $330,000 and a median of $325,000. The close figures suggest the sales are fairly evenly spread without strong outliers.
More Statistics Questions
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