Theory & Formulas
1. Mean (Arithmetic Average)
The mean is the sum of all values divided by the total number of values. Sensitive to outliers.
Direct: x̄ = Σx / n
Shortcut: x̄ = A + (Σfd / Σf)
Step Deviation: x̄ = A + h(Σfu / Σf)
Grouped: x̄ = Σ(f×mid) / Σf
2. Median (Middle Value)
Middle value in sorted data. Not affected by outliers.
Odd n: ((n+1)/2)th term
Even n: avg of (n/2)th & (n/2+1)th
Grouped: L + [(n/2−cf)/f]×h
3. Mode (Most Frequent Value)
Raw: Value with highest frequency
Grouped: L + [(f₁−f₀)/(2f₁−f₀−f₂)]×h
4. Empirical Relationship
Mode = 3×Median − 2×Mean
Beginner Problems (1–5)
Problem 1 — Simple Mean
Data: 4, 7, 13, 2, 1, 8, 5, 10
Solution
n=8, Σx=50, Mean=50/8=6.25
Answer: Mean = 6.25
Problem 2 — Median (Odd n)
Data: 3, 9, 5, 1, 7, 11, 6
Solution
Sorted: 1,3,5,6,7,9,11. 4th term=6
Answer: Median = 6
Problem 3 — Median (Even n)
Data: 12, 4, 8, 20, 16, 6
Solution
Sorted: 4,6,8,12,16,20. (8+12)/2=10
Answer: Median = 10
Problem 4 — Mode
Data: 5,3,7,3,9,5,3,8,7,3
Solution
3 appears 4 times
Answer: Mode = 3
Problem 5 — Missing Value
Mean of 5 numbers = 18. Four are: 14,20,25,16.
Solution
Total=90, Known=75, Missing=15
Answer: 5th Number = 15
Intermediate Problems (6–10)
Problem 6 — Mean from Frequency Table
| x | f | fx |
| 2 | 3 | 6 |
| 4 | 5 | 20 |
| 6 | 8 | 48 |
| 8 | 3 | 24 |
| 10 | 1 | 10 |
| Total | 20 | 108 |
Mean = 108/20 = 5.4
Answer: Mean = 5.4
Problem 7 — Median from Frequency Table
| x | f | cf |
| 1 | 4 | 4 |
| 2 | 6 | 10 |
| 3 | 10 | 20 |
| 4 | 8 | 28 |
| 5 | 2 | 30 |
N=30, N/2=15. First cf≥15 at x=3
Answer: Median = 3
Problem 8 — Shortcut Method
x: 10,20,30,40,50. f: 2,3,5,4,1. A=30.
| x | f | d=x−30 | fd |
| 10 | 2 | −20 | −40 |
| 20 | 3 | −10 | −30 |
| 30 | 5 | 0 | 0 |
| 40 | 4 | 10 | 40 |
| 50 | 1 | 20 | 20 |
| Total | 15 | — | −10 |
Mean = 30+(−10/15) = 29.33
Answer: Mean = 29.33
Problem 9 — Mode (Grouped)
Classes 0–10 to 40–50, f: 3,7,15,8,2.
Modal class=20–30. Mode=20+[8/15]×10=25.33
Answer: Mode = 25.33
Problem 10 — Median (Grouped)
| Class | f | cf |
| 0–10 | 5 | 5 |
| 10–20 | 10 | 15 |
| 20–30 | 20 | 35 |
| 30–40 | 10 | 45 |
| 40–50 | 5 | 50 |
Median=20+[(25−15)/20]×10=25
Answer: Median = 25
Advanced Problems (11–16)
Problem 11 — Step Deviation
A=35, h=10. f: 5,12,20,8,5.
| Class | Mid | f | u | fu |
| 10–20 | 15 | 5 | −2 | −10 |
| 20–30 | 25 | 12 | −1 | −12 |
| 30–40 | 35 | 20 | 0 | 0 |
| 40–50 | 45 | 8 | 1 | 8 |
| 50–60 | 55 | 5 | 2 | 10 |
| Total | — | 50 | — | −4 |
Mean=35+10(−4/50)=34.2
Answer: Mean = 34.2
Problem 12 — Combined Mean
A: 40 students, mean 65. B: 60 students, mean 70.
=(2600+4200)/100=68
Answer: Combined Mean = 68
Problem 13 — Empirical Formula
Mean=42.5, Median=41.2.
Mode=3(41.2)−2(42.5)=38.6
Answer: Mode = 38.6
Problem 14 — Missing Frequency
Median=28. f: 5,8,f,10,7.
Solving: f=7
Answer: f = 7
Problem 15 — Bimodal
Data: 2,4,6,4,8,6,2,4,6,10. Both 4 & 6 appear 3 times.
Answer: Mode = 4 and 6 (Bimodal)
Problem 16 — Weighted Mean
Maths 72(4), Physics 65(3), Chemistry 80(3).
=(288+195+240)/10=72.3
Answer: Weighted Mean = 72.3
Extra Practice Problems (17–26)
Additional problems for deeper understanding.
Problem 17 — Mean of Large Dataset
Find the mean of: 15, 22, 30, 18, 25, 28, 35, 40, 12, 20
Solution
n=10, Σx=245, Mean=245/10=24.5
Answer: Mean = 24.5
Problem 18 — Median with Repeated Values
Data: 5, 8, 8, 10, 12, 12, 12, 15, 18
Solution
n=9, 5th term=12
Answer: Median = 12
Problem 19 — No Mode (Uniform)
Data: 3, 5, 7, 9, 11. Each appears once.
Answer: No Mode exists
Problem 20 — All Three Measures
Data: 2, 3, 4, 4, 5, 5, 5, 6, 7
Solution
Mean=41/9=4.56, Median=5th term=5, Mode=5 (3 times)
Answer: Mean=4.56, Median=5, Mode=5
Problem 21 — Combined Mean (Three Groups)
A: 20 students, mean 55. B: 30, mean 60. C: 50, mean 70.
Solution
=(1100+1800+3500)/100=6400/100=64
Answer: Combined Mean = 64
Problem 22 — Correcting a Wrong Value
Mean of 20 values is 45. One value 55 is replaced by 35.
Solution
New Σx=900−55+35=880, New Mean=880/20=44
Answer: New Mean = 44
Problem 23 — Median (Unequal Class Widths)
| Class | f | cf |
| 0–10 | 4 | 4 |
| 10–20 | 6 | 10 |
| 20–40 | 12 | 22 |
| 40–70 | 8 | 30 |
| 70–100 | 5 | 35 |
Solution
N=35, N/2=17.5. Median class=20–40. L=20,cf=10,f=12,h=20
Median=20+[(7.5)/12]×20=20+12.5=32.5
Answer: Median = 32.5
Problem 24 — Correcting Wrong Observation
Mean of 30 students is 20. One mark was wrongly recorded as 35 instead of 25. Find correct mean.
Solution
Wrong Σx=30×20=600. Correct Σx=600−35+25=590
Correct Mean=590/30=19.67
Answer: Correct Mean = 19.67
Problem 25 — Find Median via Empirical Formula
Mean=36, Mode=30.
Solution
30=3×Median−72 → Median=102/3=34
Answer: Median = 34
Problem 26 — Effect of Adding a Constant
Data: 10,15,20,25,30. Add 5 to each.
Solution
Original: Mean=20, Median=20. New: Mean=25, Median=25.
Answer: New Mean=25, New Median=25
Key: Adding constant k shifts Mean, Median, Mode each by k.
Bonus — Verify Empirical Relationship
Data: 10,20,20,30,30,30,40,50
Solution
Mean=28.75, Median=30, Mode=30
Empirical: 3(30)−2(28.75)=32.5≠30. Formula is approximate.
Answer: Mean=28.75, Median=30, Mode=30
Quick Formula Summary
| Measure | Type | Formula |
| Mean | Raw | x̄=Σx/n |
| Mean | Shortcut | x̄=A+(Σfd/Σf) |
| Mean | Step Dev. | x̄=A+h(Σfu/Σf) |
| Mean | Combined | (n₁x̄₁+n₂x̄₂)/(n₁+n₂) |
| Median | Odd n | ((n+1)/2)th term |
| Median | Even n | Avg (n/2)th & (n/2+1)th |
| Median | Grouped | L+[(n/2−cf)/f]×h |
| Mode | Raw | Most frequent value |
| Mode | Grouped | L+[(f₁−f₀)/(2f₁−f₀−f₂)]×h |
| Empirical | Any | Mode=3×Median−2×Mean |
Exam Tips
- Always check if n is odd or even before finding median.
- For grouped data, clearly identify L, cf, f, h first.
- Use empirical formula when only two of three measures are given.
- Step deviation saves time — choose A as middle class mid-value.
- Combined mean requires total sum, not average of means.
- Adding constant k shifts Mean, Median, Mode each by k.
- Multiplying by k scales Mean, Median, Mode each by k.
Swathika B is an MBA graduate in Finance & Business Analytics , the founder of The Commerce Lab. With a strong academic foundation in B.Com BFSI and hands-on experience in financial analysis, data analytics, and business studies, she created this platform to make Commerce and Accountancy simple, practical, and exam-ready for students across India.