IB Maths AI SL Topic 4 — Statistics Toolkit Paper 1 & 2 Mean, median, mode ~7 min read

Measures of Central Tendency

Three “averages” describe where the centre of a data set sits: the mode (most common), the median (middle), and the mean (sum ÷ count). Each behaves differently — especially when there is an outlier — so always think about which one best represents the data.

📘 What you need to know

The three averages

Mode — just look for the most common value. Useful for qualitative data (“most popular flavour”) where mean and median don’t apply.

Median — sort the data, then pick the middle.
• Odd count n: median is the value in position n + 12.
• Even count: median is the mean of the two middle values, in positions n2 and n2 + 1.

Mean — sum every value, divide by how many. Uses every piece of data, which is why a single huge outlier can pull it sideways.

Same data, three different averages data: 2, 3, 4, 4, 5, 6, 7, 9, 14 0246810121416 mean = 6 median = 5 mode = 4 outlier the 14 pulls the mean to the right
For the data 2, 3, 4, 4, 5, 6, 7, 9, 14 the three averages all land on different values. Mode (red) sits at the tallest stack, median (orange) at the middle value, mean (teal) is the balance point — pulled right by the outlier 14.
The mean formula = 1n Σ xi  =  sum of all valuesn

🧭 Recipe — find mode, median, mean by hand

  1. Mode: scan the list; pick whichever value appears most. If two appear the same maximum number of times, list both (bimodal).
  2. Order the data from smallest to largest. Essential before finding the median.
  3. Median: if n is odd, the middle entry; if even, the average of the two middle entries.
  4. Mean: add every value, then divide by n.
  5. Sanity check: all three averages should sit between the smallest and largest values.
Spotting outliers: a single value far from the rest can drag the mean noticeably while leaving the median untouched. When a question features an obvious extreme value, expect to discuss this.

Worked examples

WE 1

Mode, median, mean from a small set

Eight students took a test (max 20). Their marks were:

12,  18,  14,  20,  18,  16,  18,  14

Find the mode, median, and mean.

Step 1 — order 12, 14, 14, 16, 18, 18, 18, 20 Mode = most common value 18 appears 3 times → mode = 18 Median: n = 8 (even) → mean of 4th & 5th (16 + 18) / 2 = 17 Mean: sum / n sum = 130, n = 8 x̄ = 130 / 8 = 16.25 mode = 18 · median = 17 · mean = 16.25 all three sit between 12 (min) and 20 (max), so the sanity check passes.
WE 2

A data set with two modes

The number of customers visiting a small café each day for a week was:

9,  12,  9,  15,  12,  18,  20

Find the mode, median, and mean.

Order 9, 9, 12, 12, 15, 18, 20 Mode — check most frequent 9 appears 2 times 12 appears 2 times → two modes (bimodal): 9 and 12 Median: n = 7 (odd) → 4th value median = 12 Mean xÌ„ = 95 / 7 ≈ 13.6 modes 9 & 12 · median 12 · mean ≈ 13.6 if two (or more) values tie for “most common”, quote them all. Saying “no mode” or picking one only is wrong.
WE 3

Median of an even-count data set

The monthly rainfall (mm) of a town for 8 months was:

12,  15,  11,  14,  18,  13,  16,  11

Find the median and mean.

Order 11, 11, 12, 13, 14, 15, 16, 18 Median: n = 8 (even) → mean of 4th & 5th (13 + 14) / 2 = 13.5 Mean sum = 110, n = 8 xÌ„ = 110 / 8 = 13.75 median = 13.5 mm · mean = 13.75 mm for even n, the median lies between two data values — it doesn’t have to be a value that actually appears in the data. Keep the same units.
WE 4

Finding a missing value given the mean

The mean of six numbers is 14. Five of the numbers are:

10,  18,  12,  9,  20

Find the sixth number.

Use: mean × n = total sum total = 14 × 6 = 84 Sum of the five known values 10 + 18 + 12 + 9 + 20 = 69 Sixth value = total − known sum 84 − 69 = 15 sixth number = 15 “reverse mean” problems use mean × n = total. This trick comes up a lot — memorise it.
WE 5

Outliers — choose the right average

A small bakery has seven employees. Their weekly hours worked are:

20,  22,  24,  21,  23,  22,  80

The 80 is the owner. (a) Find the mean and median. (b) Which better represents a “typical employee”?

(a) Order 20, 21, 22, 22, 23, 24, 80 Median: n = 7 → 4th value median = 22 Mean sum = 212, n = 7 xÌ„ = 212 / 7 ≈ 30.3 mean ≈ 30.3 h · median = 22 h (b) The 80 is an outlier it pulls the mean up to 30.3 (above every staff value!) the median = 22 sits with the rest median better represents a typical employee whenever ONE value is far from the rest, the median is the safer “centre”. The mean tells you about the total workload, not the typical person.
WE 6

Combined mean of two groups

Class A has 20 students with a mean test mark of 14. Class B has 30 students with a mean test mark of 18. Find the mean mark of all 50 students combined.

Use: total = mean × n for each class Class A total: 20 × 14 = 280 Class B total: 30 × 18 = 540 Combined sum & count total sum = 280 + 540 = 820 total count = 20 + 30 = 50 Combined mean x̄ = 820 / 50 = 16.4 combined mean = 16.4 the answer is NOT (14 + 18) / 2 = 16 — that ignores the different group sizes. Always convert each mean back to a total first, then add and divide.

💡 Top tips

âš  Common mistakes

Next up: Measures of Dispersion — range, interquartile range, standard deviation, and variance. Knowing the centre is only half the picture; you also need to describe how spread out the data is. Same data, different spreads, very different stories.

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