Overview
A mean describes a total distributed across a count; a median describes a central position after sorting. Changing a data set can affect these statistics differently. We will use totals to find a new observation and combine groups, use ordered positions to update a median, and then combine order and total constraints to find the largest possible median. That final task needs more than one plausible data set: it needs a proof of a bound and a way to reach it.
Recover a missing observation from two totals
Six measurements have mean 18, so their total is 6·18 = 108. Adding a seventh raises the mean to 20, requiring a new total of 7·20 = 140. The new measurement is the difference, 140 − 108 = 32.
The answer twenty is the new mean, not the added value. Because the added observation must lift the average of all seven above the previous eighteen, it must exceed the new mean. That qualitative check supports the result. Working with totals avoids the misleading idea that an increase of two in the mean means an added measurement of two. The change is spread across every observation in the new count, so both the old and new counts belong in the calculation.
Weight group means by their group sizes
One group of twelve students has mean score 75, while a group of eight has mean 90. Their score totals are 12·75 = 900 and 8·90 = 720. The combined twenty students have total 1,620 and mean 1,620/20 = 81.
The answer 165/2, or 82.5, averages the two group means equally. That would give each group the same weight even though one contains more students. Reconstructing the totals makes the correct weights unavoidable. The combined mean should lie between 75 and 90 and closer to 75 because the larger group has that mean. This is a useful reasonableness check, but only the weighted calculation determines the exact result.
Recount the middle positions after adding data
The ordered data set 2, 4, 4, 7, 9 has five entries. Adding 10 gives six ordered entries: 2, 4, 4, 7, 9, 10. With an even count, the median is the average of the third and fourth values, so it is (4 + 7)/2 = 11/2.
The answer four keeps the old median without updating the count; seven selects only the upper middle value. Sorting and indexing are the defining operations for a median, whereas adding all values is relevant to a mean. Repeated values still occupy separate positions in the ordered list. When a new observation is inserted, determine its position and the new count before deciding which entries are central. A value added at the end can change the middle positions even though it does not lie near the center itself.
Prove the greatest possible median and attain it
Five real numbers have mean 8, minimum 2 and maximum 14. Write them in order as 2, a, m, b, 14. Their total is forty, so the three interior values satisfy a + m + b = 24. Since a ≥ 2 and b ≥ m, this sum is at least 2 + 2m. Therefore 24 ≥ 2 + 2m and m ≤ 11.
The list 2, 2, 11, 11, 14 satisfies all conditions and has median eleven, so the bound is attainable. Together the inequality and witness prove that eleven is the greatest possible value. Showing that eight or ten can occur does not prove a maximum. For transfer, use a total equation for the mean and ordering inequalities for the median. In optimization questions, always give both parts: a reason no larger value can work and an explicit configuration reaching the proposed answer. That combination turns a plausible guess into a complete argument.
The extremal list is allowed to repeat its minimum and median because the prompt asks for real numbers, not distinct real numbers. If distinctness were imposed, the witness would need to change and attaining the same bound could become impossible. In an optimization problem, check not only a numerical inequality but also whether the equality conditions are compatible with every stated restriction. The witness is where that compatibility becomes concrete.
Apply the method to a new case
Four observations have mean 8 and are listed in increasing order as 2, x, 10 and 12. Find x and the median. The mean fixes the total at 4·8 = 32. The known observations total 24, so x = 8. The two middle values are 8 and 10, giving median 9. Check the ordering, 2 ≤ 8 ≤ 10 ≤ 12, before using x as the second observation. Reporting 8 for the median confuses the given mean with the center of the ordered list. Replacing x by a value inferred from the median would reverse the supplied condition. Verify the total and the two middle positions independently; an average of all observations and an average of the middle pair describe different features and need not agree. A solved value that violated the stipulated order would reveal inconsistent givens.