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Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?
Let's break down what we know in simple terms:
We need to figure out which of the three statements MUST be true - meaning they're true in every possible scenario that fits our constraints.
Think of it this way: imagine you're arranging 15 home prices from lowest to highest on a table. We know the middle price (8th position) is \(\$\mathrm{130,000}\), and when we add all 15 prices and divide by 15, we get \(\$\mathrm{150,000}\).
Process Skill: TRANSLATE - Converting the statistical terms into a concrete arrangement we can visualize
Since we have 15 homes (an odd number), the median is the middle value when all prices are arranged from lowest to highest. The middle position is the 8th home.
This tells us something very specific:
Wait - I need to be more careful here. The median being \(\$\mathrm{130,000}\) means:
Process Skill: INFER - Drawing the correct conclusions about what median position means for the distribution
Now let's use the average price information. If the mean is \(\$\mathrm{150,000}\) for 15 homes, then:
Total value of all homes = \(\mathrm{15} \times \$\mathrm{150,000} = \$\mathrm{2,250,000}\)
This is our key constraint. No matter how the individual prices are distributed, they must add up to exactly \(\$\mathrm{2,250,000}\).
Here's the crucial insight: Since the median (\(\$\mathrm{130,000}\)) is less than the mean (\(\$\mathrm{150,000}\)), the distribution is skewed. Some homes must be pulling the average up above the median.
For each statement, I'll try to create a scenario where it's false while keeping our mean = \(\$\mathrm{150,000}\) and median = \(\$\mathrm{130,000}\). If I can't create such a scenario, the statement must be true.
Let me try to disprove this by creating a scenario where NO home sold for more than \(\$\mathrm{165,000}\).
Extreme case: Make the prices as equal as possible while maintaining median = \(\$\mathrm{130,000}\)
Total = \(\mathrm{8} \times \$\mathrm{130,000} + \mathrm{7} \times \$\mathrm{165,000} = \$\mathrm{1,040,000} + \$\mathrm{1,155,000} = \$\mathrm{2,195,000}\)
But we need \(\$\mathrm{2,250,000}\)! We're \(\$\mathrm{55,000}\) short.
Since we can't reach the required total without exceeding \(\$\mathrm{165,000}\), Statement I MUST be true.
Process Skill: CONSIDER ALL CASES - Testing the extreme boundary condition to see if it's possible
Let me try to disprove this by avoiding the range \(\$\mathrm{130,000} < \text{price} < \$\mathrm{150,000}\).
Scenario: Use only prices \(\leq \$\mathrm{130,000}\) or \(\geq \$\mathrm{150,000}\)
Total = \(\mathrm{8} \times \$\mathrm{130,000} + \mathrm{7} \times \$\mathrm{150,000} = \$\mathrm{1,040,000} + \$\mathrm{1,050,000} = \$\mathrm{2,090,000}\)
We need \(\$\mathrm{2,250,000}\), so we need \(\$\mathrm{160,000}\) more. I can achieve this by raising some of the higher-priced homes above \(\$\mathrm{150,000}\). This works! Statement II is NOT necessarily true.
Let me try to disprove this by having NO homes sell for less than \(\$\mathrm{130,000}\).
Scenario: All homes sell for \(\$\mathrm{130,000}\) or more, with median still \(\$\mathrm{130,000}\)
Check: \(\mathrm{8} \times \$\mathrm{130,000} + \mathrm{7} \times \$\mathrm{160,000} = \$\mathrm{1,040,000} + \$\mathrm{1,120,000} = \$\mathrm{2,160,000}\)
We still need \(\$\mathrm{90,000}\) more, so homes 9-15 need to average about \(\$\mathrm{172,857}\) each.
This scenario works: median = \(\$\mathrm{130,000}\), mean = \(\$\mathrm{150,000}\), and no home sells below \(\$\mathrm{130,000}\). Statement III is NOT necessarily true.
Process Skill: APPLY CONSTRAINTS - Systematically checking each constraint while testing scenarios
Only Statement I must be true in every scenario that satisfies our constraints.
The answer is A. I only
1. Misunderstanding what "must be true" means: Students often confuse "must be true" (true in ALL possible scenarios) with "could be true" (true in SOME scenarios). They may select statements that work in one example they create, without checking if those statements hold in every possible arrangement that satisfies the given constraints.
2. Incorrectly interpreting median constraints: Students frequently assume that with median = \(\$\mathrm{130,000}\), exactly 7 homes are below \(\$\mathrm{130,000}\), exactly 1 home is at \(\$\mathrm{130,000}\), and exactly 7 homes are above \(\$\mathrm{130,000}\). They miss that multiple homes can be sold at the median price, meaning "at most 7" homes are below the median, not "exactly 7."
3. Failing to use both constraints simultaneously: Students often analyze the mean and median separately rather than understanding they must work together. They don't realize that the relationship between mean (\(\$\mathrm{150,000}\)) being higher than median (\(\$\mathrm{130,000}\)) creates specific limitations on how prices can be distributed.
1. Arithmetic errors in constraint checking: When testing scenarios, students make calculation mistakes when computing totals (\(\mathrm{15} \times \$\mathrm{150,000} = \$\mathrm{2,250,000}\)) or when checking if their proposed price distributions actually sum to the required total.
2. Creating invalid test scenarios: Students construct scenarios that violate the median constraint. For example, they might place the median price in the wrong position or create arrangements where the 8th value (when sorted) is not \(\$\mathrm{130,000}\).
3. Incomplete testing of boundary conditions: When trying to disprove statements, students often don't push their test scenarios to the true extremes. For Statement I, they might test with some high values but not maximize the constraint to see if they can avoid exceeding \(\$\mathrm{165,000}\) entirely.
1. Confusing proved vs. disproved statements: After testing, students sometimes select statements they successfully disproved (showed could be false) instead of statements they couldn't disprove (must be true). They reverse their logic when interpreting their test results.
2. Selecting based on partial analysis: Students may correctly identify that Statement I must be true but then assume other statements are also necessarily true without properly testing them. They select "I and II" or "I and III" based on incomplete verification.