SAT Math Review · Problem-Solving and Data Analysis
Mean, median and mode from data displays
Three averages, three different answers. The SAT shows you a table or a plot and then names the one it wants.
What is the median?
A teacher recorded the number of days absent for each of the 25 students in a class. Nine students were absent 0 days, seven were absent 1 day, five were absent 2 days, three were absent 3 days, and one student was absent 12 days. What is the median number of days absent?
There are 25 students, so the median is the 13th value once the 25 numbers are in order. Nine 0s fill positions 1–9 and seven 1s fill positions 10–16, so the 13th value is 1 day. A is the mode — 0 is the most common value, but it is not the middle one. C is the mean, 38 ÷ 25 = 1.52, which the single 12-day absence drags upward; the question asked for the median. D reads the middle of the list of different values — 0, 1, 2, 3, 12 has 2 in the middle — instead of the middle of the 25 students.
1
A frequency column means the value repeats that many times.
The left column is the value. The right column is how many students had it. The data set is 25 numbers long, not 5.
Days absent
Frequency
Days × frequency
Running total
0
9
0
9
1
7
7
16
2
5
10
21
3
3
9
24
12
1
12
25
total
25
38
—
Written out, the data set is 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 12. Every question in this lesson is a question about that list; the table is only a shorter way to write it.
There are 25 students, not 5 rows
Dividing by the number of rows is the single most common error in this topic. The size of the data set, n, is always the sum of the frequency column. Add that column first and write the total down before you do anything else.
2
Mean: total everything, then divide by how many.
With a frequency table you never have to write the list out. Multiply each value by its frequency, add, divide by n.
0(9) + 1(7) + 2(5) + 3(3) + 12(1)
value × frequency, row by row
0 + 7 + 10 + 9 + 12 = 38
the total number of days missed
n = 9 + 7 + 5 + 3 + 1 = 25
the frequency column, not the row count
mean = 38 ÷ 25 = 1.52
days per student
The mean need not be a value anybody actually had — no student was absent 1.52 days. It also need not be a whole number, and on the Digital SAT a decimal answer such as 1.52 is perfectly normal: type it straight into Desmos as 3825 if you would rather not divide by hand.
sum = mean × nmean = sum ÷ nn = sum of frequencies
3
Median: the middle position, found with the running total.
Odd n: the median is the value in position n + 12. Even n: average the values in positions n2 and n2 + 1.
n = 25, so the middle is position 25 + 12 = 13
12 values below it, 12 above
running total at “0 days” is 9
positions 1–9 — not far enough
running total at “1 day” is 16
so positions 10–16 are all 1s
median = 1 day
position 13 lands inside that block
An even-sized set has no single middle value, so you average the two in the middle. Six repair times, in minutes, in order:
479121516
n = 6, so positions 3 and 4
6 ÷ 2 = 3, then the next one
median = (9 + 12) ÷ 2 = 10.5
a value that is not in the set at all
Order the list first, every time
The median is defined on the sorted list. A stem that hands you 15, 9, 4, 16, 7, 12 in that order is hoping you will average the middle two as written and report 4 + 162 = 10. A frequency table and a dot plot are already sorted for you — a bare list is not.
4
A dot plot is a frequency table drawn sideways.
One dot is one member of the data set. The height of a stack is that value's frequency, so everything from the last three steps works unchanged.
running totals 1, 6, 8 — the 6th is a 1, the 7th is a 2
median = (1 + 2) ÷ 2 = 1.5
the two middle dots disagree
mode = 1
the tallest stack, 5 dots high
mean 2 · median 1.5 · mode 1
The mode is the value the tallest stack sits above — 1 club — not the height of that stack, which is 5. A set can have two modes if two stacks tie, and no mode at all if every stack is the same height. The mean and the median always exist.
5
One extreme value moves the mean and leaves the median alone.
Five friends watched 3, 5, 6, 6 and 10 movies last month. A sixth friend, who watched 24, joins the group.
Measure
3, 5, 6, 6, 10
3, 5, 6, 6, 10, 24
n
5
6
sum
30
54
mean
6
9
median
6 (3rd value)
6 (3rd and 4th: 6 + 62)
mode
6
6
Only the mean moved, and it moved a long way: 6 to 9, half again as large, on the strength of one friend. The mean uses every value, so one extreme value drags it. The median only cares about position, and the mode only cares about which value repeats.
“The median never changes” is not quite the rule
Adding a value changes n, and the median position moves with it. Here the list went from 5 values to 6, so the median switched from the 3rd value to the average of the 3rd and 4th — and only stayed at 6 because both of those are 6. Add an outlier to the list 2, 4, 7, 9, 11 and the median slides from 7 to 7 + 92 = 8. The honest rule is that the median shifts by at most one position, while the mean can move without limit.
6
Work backwards from a mean with sum = mean × n.
Every “what must the last value be” item is this one identity read from right to left. Find the total you need, find the total you have, subtract.
4 tests, mean 82 → sum = 82 × 4 = 328
the points already earned
5 tests, mean 84 → sum = 84 × 5 = 420
the points needed in total
fifth score = 420 − 328 = 92
the gap between the two totals
The same move works on a frequency table. Take the 25 absences from step 1, where the total was 38, and suppose a 26th student joins and the class mean becomes exactly 1.5 days.
new sum = 1.5 × 26 = 39
mean × n, with the new n
26th value = 39 − 38 = 1 day
and the mean does fall, 1.52 → 1.5
Where the calculator earns its keep
Desmos will do the whole thing in one line. Type mean(0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,2,2,2,2,2,3,3,3,12) and it returns 1.52; swap in median(…) or total(…) on the same list. For a short data set that is faster and safer than adding by hand, and you can edit one entry to see instantly which measures move.
✓
Check: is each answer the right kind of number?
the count9 + 7 + 5 + 3 + 1 = 25the frequency column is n ✓
the mean0 ≤ 1.52 ≤ 12inside the range of the data ✓
the medianposition 13 of 25a value from the list, or the average of two ✓
the shapemedian 1 < mean 1.52the one long absence pulls the mean up ✓
Remember the flow
n = sum of frequencies · mean = sum ÷ n · median = middle position · mode = tallest stack
An outlier moves the mean and barely touches the other two.
Now you try
Practice problem
Which statement is true?
The annual salaries, in thousands of dollars, of the 7 employees at a firm are 38, 41, 44, 44, 47, 50 and 53. The owner, whose annual salary is 250 thousand dollars, is then included in the data set. Which of the following describes how the mean, the median and the mode change?
The 7 salaries total 317, so the mean is 3177 ≈ 45.3; with the owner the total is 567 and the mean is 5678 = 70.875, a large jump. The count also changes from 7 to 8, so the median moves from the 4th value, 44, to the average of the 4th and 5th, 44 + 472 = 45.5 — an increase, but a small one. The mode is 44 either way, since 44 is still the only repeated value. So the mean and the median increase and the mode is unchanged. A applies “only the mean moves” without checking n: that rule holds when a value is replaced, not when one is added to an odd-sized set. C forgets that adding 250 once cannot make it repeat, so the mode stays at 44. D has the median going the wrong way — adding a value larger than every other one can never pull the middle down.
Hint: write the new sorted list out and count it before you decide anything. Two things changed at once — the values in the set and how many of them there are — and the median depends on both.
