Histograms (Frequency Density) – Grade 10 IGCSE Maths

Welcome to Histograms!

A histogram looks like a bar chart — but it is fundamentally different. In a histogram, area represents frequency, not height. This matters especially when class widths are unequal. Frequency density on the y-axis corrects for different widths so comparisons remain fair.

Frequency Density = Frequency ÷ Class Width | Frequency = FD × Class Width

Key fact: The y-axis of a histogram MUST be labelled "Frequency Density" — not "Frequency". Labelling it "Frequency" is an automatic error on every exam mark scheme.

FD Concept

Why area = frequency

Drawing & Reading

From table to histogram and back

Estimating Statistics

Mean, median, quartiles

Worked Examples

6 fully solved problems

Histogram Builder

Enter data, see histogram

Formula Sheet

All key formulas at a glance

Learn 1 — The Frequency Density Concept

Why Area = Frequency?

Consider two classes in a survey of journey times: "0–10 minutes" (frequency 20) and "10–30 minutes" (frequency 20). If we draw bars of equal height, it looks like they contain the same "concentration" of data — but the second class covers twice the range. The data is actually half as dense in the 10–30 group.

Frequency Density (FD) = Frequency ÷ Class Width

For "0–10 min", f=20, cw=10: FD = 20/10 = 2.0
For "10–30 min", f=20, cw=20: FD = 20/20 = 1.0

Now the bar for "10–30" is half the height, correctly showing the data is less concentrated there.

Area = Frequency — The Key Relationship

Area of bar = FD × Class Width = Frequency

This means: to find the frequency from a histogram, you multiply the height (FD) by the class width. You do NOT just read the height.

Memory trick: FD is the "rate" of data per unit. Just like speed = distance ÷ time, FD = frequency ÷ width. To reverse, multiply: frequency = FD × width.

Y-Axis Label

The y-axis MUST say "Frequency Density". Writing "Frequency" or "Number of people" is incorrect and will lose a mark. Bars must also be adjacent (no gaps) — gaps between bars means it is a bar chart, not a histogram.

Five-Class Worked Table

Class	Class Width	Frequency (f)	FD = f ÷ cw
0–5	5	10	2.0
5–10	5	25	5.0
10–20	10	40	4.0
20–35	15	30	2.0
35–50	15	15	1.0

Total frequency = 10+25+40+30+15 = 120. The modal class is 5–10 (highest FD = 5.0).

Learn 2 — Drawing and Reading Histograms

From Table to Histogram

Steps:
1. For each class, calculate FD = f ÷ class width
2. Draw the x-axis with continuous class boundaries (no gaps)
3. Draw each bar with height = FD (not frequency)
4. Label y-axis "Frequency Density" and x-axis with variable name
5. Each bar width proportional to class width

From Histogram to Table

Steps:
1. Read the height (FD) from the y-axis for each bar
2. Identify the class width from the x-axis boundaries
3. Calculate frequency: f = FD × class width
4. Check: do all frequencies sum to total n?

Finding a Missing Bar

If total frequency is given and one bar's FD is missing:
1. Calculate all known frequencies (FD × width for each known bar)
2. Missing f = total − Σ(known frequencies)
3. Missing FD = missing f ÷ class width of that bar

Example: Full Worked Table → Histogram Description

Age (years)	Frequency	Class Width	FD
10–20	18	10	1.8
20–25	30	5	6.0
25–30	40	5	8.0
30–40	24	10	2.4
40–60	16	20	0.8

Total = 128. Modal class = 25–30 (FD = 8.0). Note: 20–25 has higher FD than 10–20 even though both look "similar" on a raw count chart.

Important: Modal Class

The modal class is the class with the HIGHEST FREQUENCY DENSITY, not the highest raw frequency or the greatest bar area. Check the FD column, not the f column.

Learn 3 — Estimating Statistics from Histograms

Total Frequency

Total n = Σ(FD × class width) = Σ(area of each bar)

Estimated Mean

Mean ≈ Σ(midpoint × frequency) ÷ Σfrequency

1. Find midpoint of each class: (lower + upper) ÷ 2
2. Calculate f = FD × class width for each class
3. Multiply: midpoint × f for each class
4. Sum all (midpoint × f) values
5. Divide by total frequency

Estimated Median

Median ≈ L + ((n/2 − F) ÷ f) × cw

Where: L = lower class boundary of median class, n = total frequency, F = cumulative frequency BEFORE the median class, f = frequency OF the median class, cw = class width of median class.

Steps:
1. Build cumulative frequency table
2. Median is at the (n/2)th value
3. Find which class contains this value (the "median class")
4. Apply the interpolation formula above

Quartiles — Same Method

Lower Quartile (Q1) at the (n/4)th value. Upper Quartile (Q3) at the (3n/4)th value.
Apply same interpolation formula with the appropriate class.
IQR = Q3 − Q1

Worked Median Example

Data: n=80. Classes: 0–10 (f=8), 10–20 (f=20), 20–30 (f=28), 30–40 (f=16), 40–50 (f=8).
Cumulative: 8, 28, 56, 72, 80. Median at 40th value → in class 20–30 (cumulative goes 28 to 56).
L=20, n/2=40, F=28 (before class), f=28, cw=10.
Median ≈ 20 + ((40−28)/28) × 10 = 20 + (12/28) × 10 = 20 + 4.29 ≈ 24.3

Example 1 — Compute FD Table

Time (minutes): 0–4 (f=12), 4–6 (f=18), 6–10 (f=24), 10–20 (f=30), 20–30 (f=10). Find the FD for each class.

Class widths: 4, 2, 4, 10, 10

FD: 12/4=3.0, 18/2=9.0, 24/4=6.0, 30/10=3.0, 10/10=1.0

Modal class = 4–6 (highest FD = 9.0). Total = 94.

Example 2 — Read Frequency from Histogram

From histogram: class 15–20 has FD=4.0. Class 20–30 has FD=2.5. Class 30–50 has FD=1.2. Find the frequencies.

f = FD × cw: 4.0×5=20, 2.5×10=25, 1.2×20=24

Frequencies: 20, 25, 24. Total = 69.

Example 3 — Find Total from Histogram (Missing Bar)

Total frequency = 120. Known bars: 10–20 FD=3 (f=30), 20–30 FD=4 (f=40), 30–50 FD=? (cw=20). Find the missing FD.

Known total = 30 + 40 = 70. Missing f = 120 − 70 = 50.

Missing FD = 50 ÷ 20 = 2.5

Example 4 — Estimated Mean

Heights (cm): 150–160 (f=10), 160–170 (f=25), 170–180 (f=35), 180–190 (f=20), 190–200 (f=10). Estimate the mean.

Midpoints: 155, 165, 175, 185, 195

f×mid: 155×10=1550, 165×25=4125, 175×35=6125, 185×20=3700, 195×10=1950

Σ(f×mid) = 17450. Σf = 100. Mean = 17450/100 = 174.5 cm

Example 5 — Estimated Median

n=100. Classes: 0–10(8), 10–20(22), 20–40(35), 40–60(25), 60–80(10). Estimate the median.

Cumulative: 8, 30, 65, 90, 100. Median at 50th value → in 20–40 class (30→65).

L=20, F=30, f=35, cw=20: Median ≈ 20 + ((50−30)/35)×20 = 20 + 11.43 = 31.4

Example 6 — Full Analysis from Histogram

From histogram: 0–5 FD=2, 5–10 FD=6, 10–20 FD=4, 20–30 FD=3, 30–50 FD=1. Find total, mean, and median.

Frequencies: 2×5=10, 6×5=30, 4×10=40, 3×10=30, 1×20=20. Total n=130.

Midpoints: 2.5, 7.5, 15, 25, 40. Σ(f×mid)=25+225+600+750+800=2400. Mean=2400/130≈18.5

Median at 65th value. Cumul: 10,40,80. Class 10–20 spans 40→80. L=10, F=40, f=40, cw=10. Median≈10+(25/40)×10=16.25

Common Mistakes to Avoid

Mistake 1 — "Frequency" on Y-Axis

Writing "Frequency" or "Number of students" on the y-axis. It MUST say "Frequency Density". This is worth an automatic mark on IGCSE: "label y-axis appropriately" is always a mark point. Even if your bars are correct, the missing label costs a mark.

Mistake 2 — Dividing the Wrong Way

Writing FD = f × cw instead of FD = f ÷ cw. Frequency density is a rate: data per unit of class width. Higher class width means data is more spread out → lower density. If class width increases, FD should decrease for the same frequency.

Mistake 3 — Modal Class by Area (Not FD)

Choosing the modal class as the one with the most area (largest frequency) rather than the tallest bar (highest FD). The modal class is strictly defined as the class with the highest frequency DENSITY — the tallest bar, regardless of its width.

Mistake 4 — Median Without Cumulative Frequency

Trying to estimate the median by inspection or from the middle class without building a cumulative frequency table. Always: build the table, find n/2, identify the class, then interpolate. Skipping the cumulative step almost always gives the wrong class.

Mistake 5 — Wrong Interpolation (Using f+F Instead of F)

In the formula Median = L + ((n/2 − F)/f) × cw, F is the cumulative frequency BEFORE the median class, not including it. Students sometimes use the cumulative frequency AT the end of the median class. Double-check: F is the running total up to (but not including) the median class.

Key Formulas — Histograms

Frequency Density = Frequency ÷ Class Width

Frequency = FD × Class Width (= Area of bar)

Total n = Σ(FD × class width)

Estimated Mean = Σ(midpoint × f) ÷ Σf

Median ≈ L + ((n/2 − F) ÷ f) × cw

Q1 ≈ L + ((n/4 − F) ÷ f) × cw | Q3 ≈ L + ((3n/4 − F) ÷ f) × cw

Key rules:
• Modal class = class with highest FD (tallest bar)
• Bars are adjacent (no gaps), width proportional to class width
• Y-axis: "Frequency Density" (never just "Frequency")
• Midpoint = (lower boundary + upper boundary) ÷ 2
• For interpolation: L = lower boundary, F = cumulative f BEFORE the class

Histogram Builder

Enter class boundaries and frequencies. The tool computes FD and draws the histogram.

Class 1

Class 2

Class 3

Class 4

Class 5

Enter class data above and click Draw Histogram.

Exercise 1 — Compute Frequency Density

Calculate FD = frequency ÷ class width. Give your answer to 1 decimal place where needed.

1. Class 0–5, frequency = 15. FD = ?

2. Class 5–10, frequency = 20. FD = ?

3. Class 10–20, frequency = 40. FD = ?

4. Class 20–30, frequency = 30. FD = ?

5. Class 30–50, frequency = 24. FD = ?

6. Class 0–10, frequency = 35. FD = ?

7. Class 10–15, frequency = 25. FD = ?

8. Class 15–25, frequency = 30. FD = ?

Exercise 2 — Read Frequency from Histogram Bars

Given the FD and class width, calculate the frequency (f = FD × class width).

1. FD = 3.0, class width = 5. Find f.

2. FD = 4.0, class width = 10. Find f.

3. FD = 2.5, class width = 8. Find f.

4. FD = 6.0, class width = 5. Find f.

5. FD = 1.5, class width = 20. Find f.

6. FD = 0.8, class width = 25. Find f.

7. FD = 3.2, class width = 5. Find f.

8. FD = 1.6, class width = 15. Find f.

Exercise 3 — Find Total or Missing Frequency

Use totals and FD relationships to find missing values.

1. Classes: 0–10 FD=2, 10–20 FD=3, 20–40 FD=1.5. Total frequency = ?

2. Total = 80. Known: 0–5 FD=4 (f=20), 5–15 FD=3 (f=30). Missing class 15–25: f = ?

3. Total = 100. Class A f=25, class B f=35, class C f=?

4. Classes: 0–5(f=10), 5–10(f=15), 10–20(f=?), 20–30(f=10), 30–50(f=8). Total = 60. Find missing f.

5. Total = 120. Classes sum to 120 except class 20–30 (cw=10) with FD=2.4. Its frequency = ?

6. FD = 3.6, class width = 5. Frequency = ?

7. Total n = 150. Five classes with f: 20, 35, 50, 30, ?. Find missing f.

8. Classes: 0–10 FD=2.5, 10–30 FD=3.0, 30–50 FD=1.5. Total = ?

Exercise 4 — Estimate the Mean

Estimate the mean using Σ(midpoint × f) ÷ Σf. Round to 1 dp where needed.

1. Classes: 0–10 (f=10), 10–20 (f=20), 20–30 (f=10). Midpoints: 5, 15, 25. Mean = Σ(mid×f)/40 = ?

2. Classes: 0–4 (f=8), 4–8 (f=12), 8–12 (f=20). Midpoints: 2, 6, 10. Mean = ?

3. Classes: 10–20 (f=5), 20–30 (f=15), 30–40 (f=20), 40–50 (f=10). Estimate mean.

4. Classes: 0–5 (f=10), 5–10 (f=30), 10–20 (f=40), 20–30 (f=20). Estimate mean [Σ(mid×f)=10(2.5)+30(7.5)+40(15)+20(25)].

5. From histogram: 0–10 FD=2, 10–20 FD=4, 20–30 FD=2. Total=80. Estimate mean.

6. Classes: 0–20 (f=10), 20–40 (f=30), 40–60 (f=20), 60–80 (f=10). Midpoints 10,30,50,70. Estimate mean.

7. Classes: 5–10 (f=20), 10–15 (f=40), 15–20 (f=30), 20–25 (f=10). Estimate mean.

8. Histogram: 0–5 FD=4, 5–15 FD=3, 15–25 FD=2. Compute frequencies then mean.

Exercise 5 — Estimate Median and Quartiles

Use interpolation: Median = L + ((n/2 − F)/f) × cw. Round to 1 dp.

1. n=40. Classes: 0–10(f=8), 10–20(f=16), 20–30(f=12), 30–40(f=4). Cumul:8,24,36,40. Median at 20th value in class 10–20. L=10,F=8,f=16,cw=10. Median=?

2. n=60. Cumul reach 30th value in class 20–30. L=20,F=22,f=20,cw=10. Median=?

3. n=80. 40th value in class 30–40. L=30,F=30,f=25,cw=10. Median≈?

4. Same data as Q3 (n=80). Q1 at 20th value. Cumul: 10,30 — 20th value in class 20–30. L=20,F=10,f=20,cw=10. Q1=?

5. Same data. Q3 at 60th value. Cumul reaches 60th in class 40–50. L=40,F=55,f=15,cw=10. Q3=?

6. IQR = Q3 − Q1 using Q1=25, Q3=43.3. IQR=?

7. n=100. Median at 50th value in class 15–25. L=15,F=35,f=30,cw=10. Median=?

8. n=120. 60th value in class 20–40. L=20,F=48,f=36,cw=20. Median=?

Practice — 25 Questions

Mixed practice on all histogram skills.

1. FD = f ÷ cw. f=24, cw=6. FD=?

2. f = FD × cw. FD=3.5, cw=4. f=?

3. FD=2.4, cw=5. f=?

4. f=36, cw=9. FD=?

5. f=50, cw=20. FD=?

6. FD=4.5, cw=10. f=?

7. Total frequency from classes: 0–10(FD=3), 10–20(FD=5), 20–30(FD=2). Total=?

8. Modal class has highest FD. FDs: 2.0, 5.0, 3.0, 1.0. Which number is the modal FD?

9. Classes: 0–5(f=10), 5–10(f=25), 10–20(f=40). Total=?

10. From histogram: bar heights 4.0, 6.0, 2.5, 1.0 with cw 5,5,10,20. Total frequency=?

11. Midpoint of class 20–30 = ?

12. Midpoint of class 15–25 = ?

13. Σ(mid×f): classes 0–10(mid=5,f=10), 10–20(mid=15,f=20). Σ=?

14. Mean = Σ(mid×f)/n. Σ(mid×f)=450, n=30. Mean=?

15. Median interpolation: L=10,F=12,f=20,cw=10,n=40. Median=?

16. Median interpolation: L=20,F=15,f=25,cw=5,n=50. Median=?

17. n=60. Q1 at 15th value. L=5,F=10,f=20,cw=5. Q1=?

18. n=80. Q3 at 60th value. L=30,F=55,f=15,cw=10. Q3=?

19. IQR = Q3−Q1 = 42−18 = ?

20. FD=0.6, cw=30. f=?

21. f=18, cw=3. FD=?

22. Total=90. Known f: 20+30+25=75. Missing f=?

23. Missing FD: missing f=12, cw=4. FD=?

24. Three classes: 0–10(FD=3), 10–20(FD=5), 20–40(FD=1). Which is modal class? Enter the FD value of the modal class.

25. Estimated mean: classes 0–10(f=20,mid=5), 10–30(f=40,mid=20), 30–50(f=20,mid=40). n=80. Mean=?

Challenge — 12 Questions

Harder histogram problems. Round to 1 dp unless exact.

1. From histogram: bars 0–4(FD=5), 4–8(FD=8), 8–16(FD=4), 16–24(FD=2), 24–40(FD=0.5). Find total frequency.

2. Same data. Find the modal class FD value.

3. Same data. Estimate the mean (use midpoints 2,6,12,20,32).

4. Same data. Total=148. Median at 74th value. Cumul: 20,52,84. 74th in class 8–16. L=8,F=52,f=32,cw=8. Median=?

5. Total=200. Q1 at 50th value. Cumul:30,80. 50th in class 10–20. L=10,F=30,f=50,cw=10. Q1=?

6. Same. Q3 at 150th value. Cumul:30,80,130,180. 150th in class 30–40. L=30,F=130,f=50,cw=10. Q3=?

7. IQR from ch5 and ch6 (Q1=14, Q3=34). IQR=?

8. FD of missing bar: total=160, known classes have f: 24+40+36+32=132. Missing class width=7. Missing FD=?

9. Classes: 0–5(FD=6), 5–10(FD=10), 10–20(FD=4), 20–30(FD=3), 30–50(FD=1). Estimate mean using midpoints 2.5,7.5,15,25,40.

10. n=120. Cumul: 15,45,90,110,120. Median at 60th value in class 20–40 (L=20,F=45,f=45,cw=20). Median=?

11. Mean from histogram: 0–10(FD=3), 10–20(FD=5), 20–30(FD=3), 30–50(FD=1). Total=?

12. From ch11: estimate mean using mid=5,15,25,40. Σ(mid×f) then divide by total from ch11.

Exam Style — 5 Questions

IGCSE-style multi-part questions. Enter the final numerical answer.

[3 marks] Class 10–30 has FD = 2.5. Class width = 20. Find the frequency.

[4 marks] Histogram: 0–5(FD=4), 5–10(FD=8), 10–20(FD=3), 20–40(FD=1.5). Find total frequency.

[5 marks] n=100. Cumul: 0–20(f=15), 20–30(f=30), 30–40(f=35), 40–60(f=20). Estimate median. (n/2=50, 50th in 30–40, L=30,F=45,f=35,cw=10)

[5 marks] Classes: 0–10(f=20,mid=5), 10–20(f=35,mid=15), 20–40(f=30,mid=30), 40–60(f=15,mid=50). n=100. Estimate mean.

[4 marks] Total = 80. Classes: 0–5(f=16), 5–15(f=32), 15–25(FD=?), 25–35(f=8). Find the missing FD for class 15–25.