IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 ~6 min read

Measures of Central Tendency

The mean, median, and mode all give you a “centre” for a data set — but they capture different kinds of typical. The mean uses every value, the median resists outliers, and the mode finds the most common.

📘 What you need to know

Mean = sum of all values ÷ number of values: ⎯x = 1n Σ x_i.
Median = middle value when data is in order. For even n, take the average of the two middle values.
Mode = most common value. Can be unique, multiple, or none — but never “0” by default.
Same units as the original data (e.g., metres → metres).
Mean reacts to outliers; median doesn’t.
Symbol convention: ⎯x = sample mean, μ = population mean (Greek ‘mu’).
Use your GDC: 1-Variable Statistics gives all three at once.

Three averages, three uses

Mean (formula booklet) ⎯x = 1n Σ x_i

Measure	How to find it	Best when…
Mode	look for most common value	data is qualitative or you want most popular
Median	order data, take middle	data has outliers / is skewed
Mean	sum ÷ count	data is roughly symmetrical, no extreme values

Special cases for the mode

Unlike the mean and median, the mode can behave unusually. If every value appears the same number of times → no mode (don’t write “mode = 0”). If two or more values tie for most frequent → all of them are modes (the data set is bimodal, trimodal, etc.).

🧭 Recipe — find the three averages by hand

Sort the data in ascending order (key for both median and mode).
Mode: find the value (or values) appearing most often.
Median: if n is odd, take the middle one; if even, average the two middle ones.
Mean: sum every value, divide by n.
Sanity check: each measure should be within the range of your data.

Worked examples

WE 1

Find the mean, median, and mode

Find the mean, median, and mode of the following data set: 7, 12, 5, 9, 7, 14, 8, 11.

Step 1: Sort 5, 7, 7, 8, 9, 11, 12, 14 Step 2: Mode — 7 appears twice (most often) Mode = 7 Step 3: Median (n = 8, even → avg of 4th and 5th) Median = (8 + 9)/2 = 8.5 Step 4: Mean — Σx/n Sum = 73; mean = 73/8 = 9.125 Mean = 9.125; Median = 8.5; Mode = 7 all three values lie within [5, 14] — sanity check passes

WE 2

Bimodal data set

Find the mean, median, and mode of: 8, 10, 12, 14, 8, 16, 10, 18.

Step 1: Sort 8, 8, 10, 10, 12, 14, 16, 18 Step 2: Mode — both 8 and 10 appear twice Modes = 8 and 10 (bimodal) Step 3: Median (n = 8 → avg of 4th and 5th) Median = (10 + 12)/2 = 11 Step 4: Mean — sum = 96, n = 8 Mean = 96/8 = 12 Mean = 12; Median = 11; Modes = 8 and 10 when two values tie for most frequent, both are modes

WE 3

Find a missing value given the mean

The mean of five numbers is 14. Four of the numbers are 10, 12, 15, and 18. Find the fifth number.

Step 1: Total sum needed = n × mean Total = 5 × 14 = 70 Step 2: Sum of the four known values 10 + 12 + 15 + 18 = 55 Step 3: Fifth value = 70 − 55 Fifth number = 15 verify: (10 + 12 + 15 + 18 + 15)/5 = 70/5 = 14 ✓

WE 4

Effect of adding one more value on the mean

The mean of 8 numbers is 15. A 9th number is added and the new mean of all 9 numbers is 16. Find the 9th number.

Step 1: Original sum Old sum = 8 × 15 = 120 Step 2: New sum after adding the 9th New sum = 9 × 16 = 144 Step 3: 9th number = new sum − old sum 144 − 120 = 24 9th number = 24 since the mean went up from 15 to 16, the new value must be above 15 — and 24 fits

WE 5

Mean vs median when outliers are present

The annual salaries (in $1000s) of 7 employees at a small company are 35, 38, 40, 42, 45, 48, and 250. (a) Find the mean and the median. (b) Which is a more representative average? Justify.

(a) Mean Sum = 35 + 38 + 40 + 42 + 45 + 48 + 250 = 498 Mean = 498/7 ≈ 71.1 → about $71,100 Median (n = 7 odd → 4th value) Sorted: 35, 38, 40, 42, 45, 48, 250 → median = 42 → $42,000 (b) Compare $250k is an outlier (likely the boss); it pulls the mean up → the median ($42k) better reflects a typical employee’s salary Median is more representative when outliers are present classic salary scenario — always prefer median for income data

WE 6

Combining two data sets to find a new mean

A test out of 50 was taken by 12 students whose mean score was 36. Three additional students later took the test and scored 28, 42, and 35. Find the new mean for all 15 students.

Step 1: Original sum from 12 students Sum₁ = 12 × 36 = 432 Step 2: Sum of the 3 new scores 28 + 42 + 35 = 105 Step 3: Combined total Total sum = 432 + 105 = 537; total students = 12 + 3 = 15 Step 4: New mean Mean = 537/15 = 35.8 New mean = 35.8 when combining groups, work with sums (not means) — never average the means

💡 Top tips

Always sort first — saves time on both median and mode.
For combined data sets, work with sums (n × mean), not the means themselves.
Skewed or outlier data → use median; symmetrical data → use mean.
Keep units — averages have the same units as the original data.
Use your GDC’s 1-Var Stats — gives mean, median, and quartiles instantly.

⚠ Common mistakes

Forgetting to sort before finding the median or mode.
Saying “mode = 0” when no value repeats — the correct answer is “no mode”.
Averaging two means when combining groups — only works if groups are equal size; otherwise use weighted (total sum) approach.
Reporting the position of the median (e.g., “the median is the 4th”) — give the actual value.
Ignoring units in the answer.

Next: Measures of Dispersion. Knowing the centre is only half the story. Two data sets can share the same mean while one is tightly clustered and the other wildly spread out. The range, interquartile range, variance, and standard deviation all measure that spread — each in a slightly different way.

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