📈 Cumulative Frequency & Quartiles
Running totals reveal the shape of a distribution — where the middle 50% of readings actually lie, and how to spot outliers using a box plot, without a single squared deviation in sight.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
📈Cumulative FrequencyRunning totals & ogives
📦Quartiles & Box PlotsIQR, outliers
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Build cumulative frequency tables and plot the ogive; read off quartiles and the median graphically or by interpolation; calculate the interquartile range; and construct box plots to compare distributions and flag outliers.
💡 The interquartile range (Q3−Q1) is the "range of the middle half" — it's a spread measure that, like the median, ignores extreme outliers entirely.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
📈
Cumulative Frequency & Ogives
A running total of frequency, plotted as a smooth S-shaped curve.
🔧Cumulative Frequency BuilderEnter frequencies (comma-separated) to build the running total
Enter the frequency of each class in order, then press Build to see the cumulative frequency running total…
📖 Key Vocabulary
Cumulative frequency — the running total of frequencies up to and including a given class.
Ogive — the S-shaped curve obtained by plotting cumulative frequency against upper class boundary.
Upper class boundary — the value plotted on the x-axis for each cumulative frequency point.
n — the total frequency (final cumulative frequency value); used to locate quartiles.
👁️ Visual aid: the ogive can never go down — each point only adds to the running total, giving the characteristic S-shape.
✏️Worked Example 1 — Building a cumulative frequency tableType: Building CF
Question Classes 0–10,10–20,20–30,30–40 have frequencies 5,12,18,5. Build the cumulative frequency table.
Running total
5, 5+12=17, 17+18=35, 35+5=40Check Final CF (40) equals n (total data items) ✓
Answer CF: 5, 17, 35, 40 at boundaries 10, 20, 30, 40
📊Worked Example 2 — Reading the ogive for a value below a boundaryType: Reading the ogive
Question Using the table from Example 1, estimate how many items are below 25 (interpolating within the 20–30 class).
Position within class 25 is halfway through 20–30 (which spans CF 17 to 35)
Interpolate
17 + 0.5×(35−17) = 17+9Answer Approximately 26 items lie below 25.
🔍Worked Example 3 — Estimating the median from the ogiveType: Median from CF
Question Using the same table (n=40), estimate the median using the ogive.
Find n/2
40/2 = 20Locate on ogive CF=20 falls in the 20–30 class (CF was 17 at boundary 20, reaches 35 at boundary 30)
Answer Median ≈ 20 + [(20−17)/18]×10 = 20+1.67 ≈ 21.7 — matching the interpolation method from the Central Tendency file.
⚙️Worked Example 4 — Engineering context (percentage passing spec)Type: Engineering
Question 80 cable samples are tested for breaking load; the ogive shows CF=68 at a load of 500 N. What percentage of cables fail below 500 N?
Interpret 68 out of 80 cables broke at or below 500 N
Convert to percentage
(68/80)×100Answer 85% of cables fail below 500 N — directly useful for setting a minimum-load specification with confidence.
🧠 Common slip: plotting cumulative frequency against the class midpoint instead of the upper class boundary — cumulative frequency only makes sense as "up to and including this exact boundary".
📦
Quartiles, IQR & Box Plots
Dividing the data into quarters reveals the spread of the middle 50% and flags potential outliers.
🔧Quartile & Outlier FinderEnter sorted or unsorted data
Enter a list of numbers, then press Calculate to find Q1, median, Q3, IQR and any outliers…
📖 Key Vocabulary
Quartile — one of three values (Q1, Q2, Q3) dividing sorted data into four equal parts.
Interquartile range (IQR) — Q3 − Q1; the range of the central 50% of the data.
Box plot — a diagram showing minimum, Q1, median, Q3, maximum, with outliers marked separately.
Outlier — a value unusually far from the rest, formally flagged using the 1.5×IQR rule.
👁️ Visual aid: the box spans the middle 50% (Q1 to Q3), the whiskers reach to the min/max within 1.5×IQR, and points beyond that are plotted separately as outliers.
✏️Worked Example 1 — Finding quartiles from raw dataType: Quartiles
Question Find Q1, median, Q3 for 3,7,8,12,13,14,18,21 (n=8, already sorted).
Median average of 4th & 5th:
(12+13)/2 = 12.5Q1 median of lower half (3,7,8,12):
(7+8)/2 = 7.5Q3 median of upper half (13,14,18,21):
(14+18)/2 = 16Answer Q1=7.5, Median=12.5, Q3=16
📏Worked Example 2 — Calculating the IQRType: IQR
Question Using Q1=7.5 and Q3=16 from Example 1, find the IQR.
Apply formula
IQR = Q3 − Q1 = 16 − 7.5Answer IQR = 8.5 — the middle 50% of the data spans a range of 8.5.
🚩Worked Example 3 — Identifying an outlierType: Outlier test
Question Using Q1=7.5, Q3=16, IQR=8.5, test whether the value 35 is an outlier.
Upper fence
Q3 + 1.5×IQR = 16 + 12.75 = 28.75Compare
35 > 28.75Answer Yes — 35 is an outlier, and would be plotted as a separate point beyond the box plot's whisker.
⚙️Worked Example 4 — Engineering context (comparing two suppliers)Type: Engineering
Question Supplier A's delivery times have IQR=2 days; Supplier B's have IQR=6 days, similar medians. What does this tell an engineer choosing a supplier?
Interpret A smaller IQR means the middle 50% of deliveries are far more tightly bunched around the typical time.
Answer Supplier A (IQR=2) is far more predictable/reliable for planning purposes, even if both suppliers have the same "typical" delivery time — consistency matters as much as the average.
🧠 Common slip: confusing "outlier" with "just an unusually large value" — the 1.5×IQR rule gives an objective, repeatable test rather than a subjective judgement call.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔗 Cable Testing — Percentage Passing a Load Spec
Using cumulative frequency to read off what fraction of a batch meets or fails a required breaking load.
% passing = (n − CF at spec) ÷ n × 100
📟 Try it — Pass Rate Calculator
Enter total tested and cumulative frequency at the spec, then press Calculate…
🚚 Delivery Times — Quartiles & IQR
Comparing IQR between suppliers or shifts reveals which is more consistent, independent of any outliers.
IQR = Q3 − Q1
📟 Try it — Quartiles from Data
Enter delivery times, then press Calculate…
🚨 Outlier Detection — Sensor Anomaly Flag
Automated monitoring systems flag readings beyond the 1.5×IQR fences as potential sensor faults or genuine anomalies.
Upper fence = Q3 + 1.5×IQR; Lower fence = Q1 − 1.5×IQR
📟 Try it — Fence Calculator
Enter Q1, Q3 and a value to test, then press Calculate…
📐 Median from Grouped Data — Interpolation
The same interpolation formula used for the ogive-based median, applied directly when only a grouped table is available.
Median = L + [(n/2 − F)/f_m] × w
📟 Try it — Median Interpolator
Enter L, n, F, f_m and w, then press Calculate…
🎮
Quick-Fire Cumulative Frequency Game
12 questions on ogives, quartiles and box plots. How many can you get?
⚡ Cumulative Frequency Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Classes with frequencies 6,10,9,5 — find the cumulative frequencies.
6, 16, 25, 30
Question 2 🥉2 marks
For sorted data 4,8,9,12,15,20 (n=6), find the median.
Middle two: 9,12 → (9+12)/2 = 10.5
Question 3 🥉2 marks
Q1=10 and Q3=22. Find the IQR.
22 − 10 = 12
Question 4 🥉2 marks
State what a box plot's box represents.
The middle 50% of the data (Q1 to Q3)
Question 5 🥈3 marks
For sorted data 2,5,7,9,11,14,16,19 (n=8), find Q1 and Q3.
Lower half: 2,5,7,9 → Q1=(5+7)/2=6
Upper half: 11,14,16,19 → Q3=(14+16)/2=15
Q1=6, Q3=15
Upper half: 11,14,16,19 → Q3=(14+16)/2=15
Q1=6, Q3=15
Question 6 🥈4 marks
Using Q1=6, Q3=15 (IQR=9) from the previous question, test if 32 is an outlier.
Upper fence = 15+1.5(9) = 15+13.5 = 28.5
32 > 28.5
Yes, 32 is an outlier
32 > 28.5
Yes, 32 is an outlier
Question 7 🥈3 marks
n=60 items; cumulative frequency reaches 45 at a value of 30. What percentage of items are ≤ 30?
(45/60)×100 = 75%
Question 8 🥈4 marks
Grouped data has L=30, n=50, F=18, f_m=20, w=10. Estimate the median.
Median = 30+[(25−18)/20]×10
= 30+3.5
= 33.5
= 30+3.5
= 33.5
Question 9 🥇4 marks
80 castings tested; CF=62 at the minimum required strength. Find the percentage failing (below spec).
(62/80)×100 = 77.5% pass or fail below?
CF represents "at or below" — if spec is a minimum, these 62 FAIL (77.5% failure rate) — a serious quality issue.
CF represents "at or below" — if spec is a minimum, these 62 FAIL (77.5% failure rate) — a serious quality issue.
Question 10 🥇5 marks
Delivery times for two suppliers both have median 5 days. Supplier X: IQR=1; Supplier Y: IQR=4. Explain which is preferable for tight scheduling and why.
Same median, but X's IQR is much smaller
Supplier X is preferable — the middle 50% of deliveries cluster much more tightly, making arrival dates far more predictable.
Supplier X is preferable — the middle 50% of deliveries cluster much more tightly, making arrival dates far more predictable.
Question 11 🥇5 marks
Sorted data: 8,10,12,14,15,17,19,60. Find Q1, median, Q3, IQR, and test whether 60 is an outlier.
Median=(14+15)/2=14.5
Q1=(10+12)/2=11, Q3=(17+19)/2=18
IQR=18−11=7
Upper fence=18+1.5(7)=28.5
60>28.5 → 60 IS an outlier
Q1=(10+12)/2=11, Q3=(17+19)/2=18
IQR=18−11=7
Upper fence=18+1.5(7)=28.5
60>28.5 → 60 IS an outlier
Question 12 🥇5 marks
Grouped data: classes 0–20(f=8),20–40(f=22),40–60(f=15),60–80(f=5), n=50. Find the cumulative frequencies and estimate the median (interpolate).
CF: 8, 30, 45, 50
n/2=25, falls in class 20–40 (CF was 8, reaches 30)
Median = 20+[(25−8)/22]×20
= 20+15.45
≈ 35.5
n/2=25, falls in class 20–40 (CF was 8, reaches 30)
Median = 20+[(25−8)/22]×20
= 20+15.45
≈ 35.5