Why is the median of 7, 2, 9, 4, 5 equal to 5 and not 9?

Because the median is the middle value of the sorted list, not the middle of the list as given. Sorting 7, 2, 9, 4, 5 gives 2, 4, 5, 7, 9. The middle value (position 3 of 5) is 5. The value 9 sits in position 3 of the original unsorted list — but the order you write the values down in is irrelevant; only the sorted order matters.

How do you find a missing value when the mean is given?

Use total = mean × n to find the total of all values, then subtract the values you know. For example: four numbers have a mean of 10, and three of them are 5, 8, and 9. Total = 10 × 4 = 40. Known sum = 5 + 8 + 9 = 22. Fourth number = 40 − 22 = 18. The trap is to average the three known values (22 ÷ 3 ≈ 7.33) — that gives the mean of the knowns, not the missing value.

Why do you divide by Σf and not by the number of rows when finding the mean from a frequency table?

Because n in the mean formula is the total number of data points, not the number of categories. In a frequency table, Σf gives the total number of data points. For example, with rows x = 0, 1, 2, 3 and frequencies 2, 5, 8, 5, Σf = 20 students (not 4 rows). The mean is Σ(f × x) ÷ Σf = 36 ÷ 20 = 1.8. Dividing by 4 gives 9, which is meaningless.

GCSE Maths Foundation · AQA · Averages & spread

Computing mean and median: why you must sort first, use total = mean × n, and divide by Σf

Averages questions trip students who apply the recipe to the numbers as they appear on the page. For the median, they pick the physically middle entry of the list as written — without sorting. For a reverse-mean question, they average the values they can see — because the mean recipe “only runs forwards.” For a frequency table, they divide by the number of rows rather than by the total number of data points.

The thirty-second fix: the median requires a sorted list; the mean always links to a total via total = mean × n; and in a frequency table, n is Σf — the sum of the frequencies, never the number of rows.

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How to spot it in your own work

You gave $9$ as the median of 7, 2, 9, 4, 5 — because 9 is in position 3 of the list as written — without sorting first.
You found a missing value by averaging the numbers you could see, instead of computing total = mean × n and then subtracting.
You divided $\Sigma(f \times x)$ by the number of rows in a frequency table instead of by $\Sigma f$ .
You picked one of the two middle values for an even-sized data set instead of averaging the two middle values.

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Nov24 P3 Q12 shape):

The values 7, 2, 9, 4, 5 are given in this order. What is the median?

The misconception answer is $9$ — the value in position 3 of the list as written. But the median is the middle of the sorted list.

Sort: $2, 4, 5, 7, 9$ . Five values — the middle is position 3. Median = $5$ .

Why students fall for this

The word “median” activates the idea of “middle,” and students read “middle” as “the one in the middle of the page as written.” The sorting step is invisible — it is a structural requirement the recipe never announces.

The reverse-mean trap has the same root: students know the mean recipe runs $\text{total} \div n = \text{mean}$ , but when a value is missing they reach for the visible inputs and run the recipe on what they can see — averaging the three or four known values. The step “multiply mean × n to get the total” is not in their toolkit, because they have only ever run the recipe forwards.

In a frequency table the denominator confusion is structural: the table has rows, and dividing by the number of rows feels natural. But n in the mean formula refers to the total number of data points — which is $\Sigma f$ , not the number of categories.

The fix: Sort first; total = mean × n; divide by Σf

Median: sort the data first, then pick the middle. For $7, 2, 9, 4, 5$ : sorted list is $2, 4, 5, 7, 9$ . Five values — middle is position 3. Median = $5$ . For an even count, average the two middle values.

Reverse mean: total = mean × n, then subtract the known sum. Four numbers, mean 10, three known values 5, 8, 9: total = 10 × 4 = 40; known sum = 5 + 8 + 9 = 22; fourth value = 40 − 22 = 18. Never average the values you can see.

Frequency table mean: divide $\Sigma(f \times x)$ by $\Sigma f$ . $\Sigma f$ is the total number of data points, not the number of rows. For x = 0, 1, 2, 3 with frequencies 2, 5, 8, 5: $\Sigma f = 20$ , $\Sigma(f \times x) = 36$ , mean = 36 ÷ 20 = 1.8.

Worked example

Find the median of $7, 2, 9, 4, 5$ .

Sort smallest to largest.
$2, \; 4, \; 5, \; 7, \; 9$
Count the values. There are 5 values, so the middle is position 3.
Pick the middle value.
$\text{Median} = 5$

The trap answer $9$ comes from reading position 3 of the original list. The original order is irrelevant — only the sorted order defines the median.

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Common questions

How do I find the median of an even number of values?: Sort the values, then average the two middle ones. For six values 3, 5, 7, 9, 12, 15: the two middle values are position 3 and position 4, which are $7$ and $9$ . Median = $(7 + 9) \div 2 = 8$ .
Why is it wrong to average the values I can see when finding the missing value in a reverse-mean question?: Averaging the known values gives the mean of those values — not the value that is missing. You need the grand total first: total = mean × n. Then subtract the known sum. For four numbers with mean 10 and known values 5, 8, 9: total = 40; known sum = 22; missing value = 18. Averaging 5, 8, 9 gives 7.33 — a completely different number.
What is Σf and why do I divide by it?: $\Sigma f$ is the sum of all the frequencies — the total number of data points in the study. When you calculate the mean you always divide by n, the total count. In a frequency table, n = Σf. The number of rows (categories) is irrelevant. For x = 0, 1, 2, 3 with frequencies 2, 5, 8, 5: $\Sigma f = 20$ , not 4.
Does it matter what order I write the values in for the median?: The order you write them in has no effect on the median — only the sorted order matters. Whether the data is given as 7, 2, 9, 4, 5 or as 9, 7, 5, 4, 2, the sorted list is the same (2, 4, 5, 7, 9) and the median is the same (5).

Related misconceptions

Inverse operations: why solving an equation means undoing each stepThe same forwards-only thinking that traps students on reverse-mean questions also traps them when solving equations — both need you to run an operation in reverse.
Fraction of an amount: why 2/5 of 1020 is 408, not 40A related recipe-application error: copying a label instead of performing the operation the recipe demands.
Combining unlike terms: why 9x + y − 6x + y is 3x + 2y, not 5xyAnother case where the structure of the operation (sort by type, collect within each type) must be applied before the arithmetic — the numbers alone are not enough.

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