Three averages, three different answers — and one of them breaks easily.
Twenty quiz scores
Score 6 × 2 students · 7 × 5 · 8 × 8 · 9 × 3 · 10 × 2
Find the mean.
Multiply each score by its frequency and total: 12 + 35 + 64 + 27 + 20 = 158. Divide by the number of students — 20, not the 5 different scores: 158 ÷ 20 = 7.9. C divides by 5, the classic error here. B is the median and the mode.
Frequency means “how many students got that score”.
| Score | Frequency | Score × freq | Running total |
|---|---|---|---|
| 6 | 2 | 12 | 2 |
| 7 | 5 | 35 | 7 |
| 8 | 8 | 64 | 15 |
| 9 | 3 | 27 | 18 |
| 10 | 2 | 20 | 20 |
| total | 20 | 158 | — |
There are five different scores but twenty students. Dividing by 5 instead of 20 is the most common error in this topic.
6(2) + 7(5) + 8(8) + 9(3) + 10(2)
score × frequency
12 + 35 + 64 + 27 + 20 = 158
the sum of all scores
158 ÷ 20 = 7.9
divide by 20 students
The mean need not be a score anybody actually got — no one scored 7.9.
With 20 values, the middle falls between the 10th and 11th.
positions 10 and 11
20 ÷ 2 = 10, so 10th and 11th
running total reaches 7 at score 7
not far enough
reaches 15 at score 8
so positions 8–15 are all 8s
median = 8
both middle values are 8
The running-total column does this work for you — find the first row where it passes the middle position.
A set can have two modes, or none at all if every value appears equally often. Mean and median always exist.
Suppose one more student sits the quiz and scores 0.
| Average | Before | After adding a 0 |
|---|---|---|
| mean | 7.9 | 7.52 |
| median | 8 | 8 |
| mode | 8 | 8 |
Only the mean moved. It uses every value, so one extreme score drags it.
Median, when the data has outliers — house prices, salaries, one disastrous test score. Mean, when the values are evenly spread. Mode, when you need the most common category rather than a number.
before: 6 × 15 = 90
total = count × mean
after: 5 × 14 = 70
removed: 90 − 70 = 20
A missing value works the same way: if 3, 7, 9 and x have mean 8, the total is 4 × 8 = 32, so x = 32 − 19 = 13.
15 − 14 = 1 is not the removed number. Means cannot be subtracted; totals can.
| Change | Mean | Median |
|---|---|---|
| add c to every value | + c | + c |
| multiply every value by k | × k | × k |
| add d to one value | + d ÷ n | usually no change |
+3 on all ten: mean +3
every value moves 3
+20 on one value: mean + 2010 = +2
shared over all ten values
mean rises by 5
3 + 2
median rises by 3 only
the largest value is not the middle one
The other two measures: the range (largest − smallest) changes whenever an end value changes, and the mode changes only if the changed value was, or becomes, the most common one. In 2, 4, 4, 6, 9, raising the 9 to 13 moves the mean (by 4 ÷ 5) and the range (7 to 11), but the median stays 4 and the mode stays 4.
Unsorted, with negatives. Find the mean, median and mode of −3, 5, 0, −1, 8, 2.
sorted: −3, −1, 0, 2, 5, 8
sort before the median
median = (0 + 2) ÷ 2 = 1
six values: average the 3rd and 4th
mean = 11 ÷ 6 ≈ 1.83
−3 − 1 + 0 + 2 + 5 + 8 = 11
mode: none
every value appears once
What is different: the median (1) is not in the list, the zero still counts as a value, and the negatives subtract from the total.
The 3rd and 4th values as written are 0 and −1, which gives −0.5. Sort first, every time.
The middle falls between two values. Scores 1, 2, 3, 4 have frequencies 3, 2, 4, 1. Find the median.
n = 3 + 2 + 4 + 1 = 10
positions 5 and 6
running totals 3, 5, 9, 10
the 5th is a 2, the 6th is a 3
median = (2 + 3) ÷ 2 = 2.5
average the two middle values
What is different: the running total lands exactly on 5, so the two middle positions sit in different rows.
Two modes. Find the mode of 3, 5, 5, 7, 8, 8, 10.
5 appears twice, 8 appears twice
a tie for the most
modes: 5 and 8
bimodal
What is different: a tie gives two modes; do not average them into 6.5. The mode is always a value from the data.
total = 5 × 8 = 40
smallest possible others: 1, 2, 7, 8
below the median as small as possible; above it, the next integer
largest = 40 − (1 + 2 + 7 + 8) = 22
Check: 1, 2, 7, 8, 22 — all different, median 7, total 40 ✓.
“Different” rules out 1, 1 or 7, 7. Check whether the values must be distinct before shrinking them.
Expect totals worked backwards (a missing or removed value), a value changed and the effect on each average, a constant added to every value, or the largest possible value under a given mean and median. Turn every mean into a total first. Ten-second check: the mean must sit between the smallest and largest values.
Fifteen students, scores recorded as
5 × three students, 6 × four, 7 × six, 9 × two
Find the mean.
5(3) + 6(4) + 7(6) + 9(2) = 15 + 24 + 42 + 18 = 99, then divide by the 15 students: 6.6. C averages the four distinct scores, (5 + 6 + 7 + 9) ÷ 4 = 6.75, ignoring how many students got each. D is the median and mode — three different averages, three different answers.
