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40 30 25 25 90 50 20
How many of the seven numbers listed are greater than the median but less than the average (arithmetic mean) of the seven numbers?
Let's break down what this question is asking us to find. We have seven numbers: 40, 30, 25, 25, 90, 50, 20.
We need to count how many of these numbers satisfy TWO conditions at the same time:
Think of it like finding numbers that fall in a specific "sweet spot" - bigger than the middle value but smaller than the average.
Process Skill: TRANSLATE - Converting the problem language into clear mathematical requirements
To find the median, we need to arrange our seven numbers from smallest to largest:
20, 25, 25, 30, 40, 50, 90
Since we have 7 numbers (an odd number), the median is simply the middle number - the 4th number in our ordered list.
Counting from the left: \(1\mathrm{st} = 20\), \(2\mathrm{nd} = 25\), \(3\mathrm{rd} = 25\), \(4\mathrm{th} = 30\)
So our \(\mathrm{median} = 30\)
The arithmetic mean is just the average of all seven numbers. We add them all up and divide by 7:
\(\mathrm{Sum} = 40 + 30 + 25 + 25 + 90 + 50 + 20 = 280\)
\(\mathrm{Arithmetic\ mean} = 280 ÷ 7 = 40\)
Now we need to find numbers that are:
So we're looking for numbers in the range: \(30 < \mathrm{number} < 40\)
Let's check each of our original numbers:
None of our numbers fall strictly between 30 and 40.
Process Skill: APPLY CONSTRAINTS - Systematically checking each number against both conditions
Zero numbers from our list are greater than the median (30) but less than the mean (40).
The answer is A. None
Students often miss that the question asks for numbers that satisfy BOTH conditions simultaneously (greater than median AND less than mean). They might count numbers that satisfy either condition instead of requiring both conditions to be true at the same time.
With 7 numbers, students might incorrectly think they need to average two middle values (as required for even-numbered datasets) rather than simply taking the single middle value that applies to odd-numbered datasets.
When adding the seven numbers \((40 + 30 + 25 + 25 + 90 + 50 + 20 = 280)\), students commonly make addition errors, especially with the repeated 25 or when handling the larger number 90. This leads to an incorrect mean calculation.
Students frequently include boundary values incorrectly. For example, they might count 30 as satisfying "greater than 30" or count 40 as satisfying "less than 40" when these strict inequalities exclude the boundary values themselves.
When arranging numbers from smallest to largest, students may incorrectly place numbers, particularly when there are repeated values like the two 25s in this dataset.
No likely faltering points - the final counting step is straightforward once the range is established and each number is properly evaluated against both conditions.