Data, Averages, Charts and Diagrams at Extended Level
Frequency tables, grouped means, pie charts and back-to-back stems
Quick recap — types of data. Categorical data are words or labels (colour of a combi, favourite subject). Numerical data are numbers, and they come in two kinds: discrete data are counted and take separate values (number of goats, number of people in a car); continuous data are measured and can take any value in a range (mass, time, height).
The three averages and the spread. Mean = total ÷ number of values. Median = middle value when the data are in order. Mode = most common value. Range = largest − smallest. For a list of \(n\) ordered values, the lower quartile (LQ) is the \(\dfrac{n+1}{4}\)th value, the upper quartile (UQ) is the \(\dfrac{3(n+1)}{4}\)th value, and the interquartile range IQR = UQ − LQ. The IQR measures the spread of the middle half, so one extreme value does not spoil it.
From a frequency table:
\(\text{mean} = \dfrac{\sum fx}{\sum f}\)
The median is the \(\dfrac{n+1}{2}\)th value — find it with a running total of the frequencies. The mode is the value with the highest frequency (not the frequency itself).
Grouped data: you do not know the exact values, so use the midpoint of each class as \(x\). The answer is only an estimate of the mean. The class with the highest frequency is the modal class.
Charts. In a pie chart the whole circle (360°) stands for the total frequency, so angle = \(\dfrac{\text{frequency}}{\text{total}} \times 360°\). In a bar chart the bar heights are frequencies (bars for categories have gaps). A pictogram needs a key (for example, one symbol = 4 cattle). A stem-and-leaf diagram keeps every value and needs a key; a back-to-back one compares two sets, with the left-hand leaves read from right to left.
Comparing distributions: always make TWO comparisons — one about the average (median or mean) and one about the spread (IQR or range) — and say what they mean in the context. Be careful: a small sample, or one from only one place, may not support a general conclusion.
A survey in Maun asked 35 households how many children aged under 5 live there.
| Number of children \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency \(f\) | 4 | 9 | 12 | 7 | 3 |
Find the mean, median and mode.
\(\sum fx = 0 + 9 + 24 + 21 + 12 = 66\), so mean \(= \dfrac{66}{35} = 1.885\ldots = 1.89\) (3 s.f.).
Median: the \(\dfrac{35+1}{2} = 18\)th value. Running totals: 4, 13, 25 … so the 14th to 25th values are all 2. Median = 2.
Mode = 2 (frequency 12 is the highest — the mode is 2, not 12).
The times, \(t\) minutes, of 50 combi journeys in Gaborone are recorded.
| Time \(t\) (min) | \(0 < t \le 10\) | \(10 < t \le 20\) | \(20 < t \le 30\) | \(30 < t \le 40\) | \(40 < t \le 50\) |
|---|---|---|---|---|---|
| Frequency | 5 | 12 | 18 | 10 | 5 |
(a) Calculate an estimate of the mean. (b) Write down the modal class and the class containing the median.
(a) Midpoints 5, 15, 25, 35, 45. \(\sum fx = 25 + 180 + 450 + 350 + 225 = 1230\). Estimate of mean \(= \dfrac{1230}{50} = 24.6\) minutes.
(b) Modal class \(20 < t \le 30\). The median is between the 25th and 26th values; running totals 5, 17, 35, so it lies in \(20 < t \le 30\) too.
It is an estimate because the midpoint stands in for every value in its class.
(a) The mean of 8 numbers is 12. A ninth number is added and the mean becomes 13. Find the ninth number.
(b) Class A has 20 students with mean mark 64. Class B has 30 students with mean mark 54. Find the mean mark of all 50 students.
(a) Old total \(= 8 \times 12 = 96\). New total \(= 9 \times 13 = 117\). Ninth number \(= 117 – 96 = 21\).
(b) Totals: \(20 \times 64 = 1280\) and \(30 \times 54 = 1620\). Combined mean \(= \dfrac{1280 + 1620}{50} = \dfrac{2900}{50} = 58\).
Notice that 58 is NOT the halfway value 59 — the bigger class pulls the mean towards its own mean.
72 students said how they travel to school: walk 25, combi 30, car 12, taxi 5. Work out the pie chart angles. Then: in a pictogram where one symbol = 4 students, how many symbols show the car group?
Each student gets \(\dfrac{360°}{72} = 5°\).
walk \(25 \times 5 = 125°\), combi \(30 \times 5 = 150°\), car \(12 \times 5 = 60°\), taxi \(5 \times 5 = 25°\)
Check: \(125 + 150 + 60 + 25 = 360°\) ✔. Pictogram: \(12 \div 4 = 3\) symbols for the car group (taxi would be \(1\dfrac{1}{4}\) symbols — a quarter of a symbol).
The back-to-back stem-and-leaf diagram shows the masses (kg) of 7 goats from Farm P and 7 goats from Farm Q. Compare the two distributions.
| Farm P | Farm Q | |
|---|---|---|
| 8 4 1 | 2 | 6 9 |
| 7 3 0 | 3 | 1 1 4 5 8 |
| 1 | 4 |
Key: 4 | 3 | 1 means 34 kg for Farm P and 31 kg for Farm Q
Farm P in order: 21, 24, 28, 30, 33, 37, 41. With \(n = 7\): LQ = 2nd value, median = 4th, UQ = 6th. So LQ = 24, median = 30, UQ = 37, IQR = 13.
Farm Q in order: 26, 29, 31, 31, 34, 35, 38. LQ = 29, median = 31, UQ = 35, IQR = 6.
Comparison: Farm Q’s goats are slightly heavier on average (median 31 kg > 30 kg), and their masses are much more consistent (IQR 6 kg < 13 kg). With only 7 goats from each farm, these are observations about the samples — not proof about every goat on each farm.
Q1. Classify each type of data as categorical, discrete or continuous.
(a) The number of cattle at a cattle post. (b) The mass of a newborn calf. (c) The colour of a combi.
👀 Show answer
(b) continuous ✓ B1
(c) categorical ✓ B1
Q2. A farmer records the number of eggs laid by each of 30 hens in one week.
| Number of eggs | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|
| Frequency | 2 | 5 | 9 | 8 | 6 |
(a) Calculate the mean. (b) Find the median.
👀 Show answer
161 ÷ 30 ✓ M1 = 5.37 (5.366…) ✓ A1
(b) 15th and 16th values: running totals 2, 7, 16 → both are 5 → median = 5 ✓ B1
Q3. The masses, \(m\) kg, of 40 calves at a cattle post near Serowe are recorded.
| Mass \(m\) (kg) | \(20 < m \le 30\) | \(30 < m \le 40\) | \(40 < m \le 50\) | \(50 < m \le 60\) | \(60 < m \le 70\) |
|---|---|---|---|---|---|
| Frequency | 6 | 11 | 14 | 7 | 2 |
(a) Calculate an estimate of the mean mass. (b) Explain why your answer is only an estimate. (c) Write down the modal class.
👀 Show answer
6×25 + 11×35 + 14×45 + 7×55 + 2×65 = 150 + 385 + 630 + 385 + 130 = 1680 ✓ M1
1680 ÷ 40 = 42 kg ✓ A1
(b) the exact masses are not known — midpoints are used ✓ B1
(c) 40 < m ≤ 50 ✓ B1
Q4. The mean mass of 6 parcels is 4.5 kg. A seventh parcel is added and the mean mass becomes 5 kg. Find the mass of the seventh parcel.
👀 Show answer
7 × 5 = 35 ✓ M1
35 − 27 = 8 kg ✓ A1
Q5. In a test, the mean mark of 15 boys was 62 and the mean mark of 25 girls was 70. Calculate the mean mark of all 40 students.
👀 Show answer
930 + 175040 = 268040 ✓ M1 = 67 ✓ A1
Q6. The pie chart shows the main animal kept by each of 90 farmers in a village near Kanye.
(a) How many farmers keep goats? (b) What fraction of the farmers keep chickens? Give your answer in its simplest form.
(c) In a second village, the goats sector has angle 75° and stands for 15 farmers. How many farmers are there in the second village?
👀 Show answer
(b) 80360 = 29 ✓ B1
(c) 75 ÷ 15 = 5° per farmer ✓ M1 360 ÷ 5 = 72 farmers ✓ A1
Q7. The bar chart shows the number of books read last month by each student in a class of 25.
(a) Calculate the mean number of books read. (b) Find the median.
👀 Show answer
47 ÷ 25 ✓ M1 = 1.88 ✓ A1
(b) 13th value: running totals 3, 10, 18 → median = 2 ✓ B1
Q8. The back-to-back stem-and-leaf diagram shows the test marks of 11 students in Class A and 11 students in Class B.
| Class A | Class B | |
|---|---|---|
| 8 2 | 1 | |
| 7 5 3 | 2 | 1 6 8 |
| 8 6 4 1 | 3 | 0 3 3 5 7 9 |
| 5 2 | 4 | 0 1 |
Key: 1 | 3 | 0 means 31 marks for Class A and 30 marks for Class B
Find the median and interquartile range for each class, and make two comparisons between the classes.
👀 Show answer
B: 21 26 28 30 33 33 35 37 39 40 41 → median 33, IQR 39 − 28 = 11 ✓ B1 ✓ B1
Class B did better on average (higher median) ✓ B1
Class B’s marks were more consistent (smaller IQR) ✓ B1
Q9. The monthly wages of six workers at a small business are P3200, P3400, P3500, P3600, P3800 and P25 000.
(a) Calculate the mean and the median. (b) Which average better represents a typical wage? Give a reason.
👀 Show answer
median = 3500 + 36002 = P3550 ✓ B1
(b) median — the mean is pulled up by the extreme value P25 000 ✓ B1
Q10. The table shows the number of people in some cars passing through Lobatse. The mean number of people per car is 2.2. Find \(x\).
| Number of people | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Frequency | 5 | \(x\) | 8 | 2 |
👀 Show answer
37 + 2x = 33 + 2.2x ✓ M1
4 = 0.2x → x = 20 ✓ A1
- Dividing by the number of columns: in a frequency table, divide \(\sum fx\) by \(\sum f\) (the number of people or items), never by how many columns the table has.
- Giving the frequency as the mode: the mode is the VALUE with the highest frequency (2 children), not the frequency (12).
- Using class boundaries instead of midpoints: for \(20 < m \le 30\) use 25, not 20 or 30, and say the answer is an estimate.
- Averaging two means: combined means need totals — mean × number in each group — before dividing.
- Reading the left side of a back-to-back diagram the wrong way: the left leaves read outwards from the stem, so “8 2 | 1” means 12 and 18.
- Comparing with numbers only: “median 31, median 33” earns nothing. Say which group is higher or more consistent, in context.
- Paper 4 (calculator) usually has a full statistics question: frequency table or grouped table → mean, median, mode, modal class, often worth 6–10 marks in total.
- Paper 2 (no calculator) favours short reverse-mean, combined-mean and pie-chart angle questions worth 2–3 marks each.
- Command words: calculate, calculate an estimate of, write down, find, compare, explain, complete the pie chart / bar chart.
- “Compare the distributions” questions expect one comparison of average and one of spread, written in context — 1 mark each.
(then move on to S-E0b Scatter Diagrams at Extended Level)