📌 Where it comes in CGL: Statistics made up about 9.6% of the questions in the 8 SSC CGL Tier 2 Paper I (Mathematical Abilities) PYQ papers (2022–2025). All questions here are practice questions prepared by SF Prep (not PYQs). JSO (Paper II) level: Chapter 25 — JSO PYQs are not yet available on SF Prep.
📑 Topics on this page (Jump to Topic):
📌 1. Arithmetic Mean
Mean = sum of all values ÷ number of values. It depends on every value, so one very large or very small value (outlier) pulls it.
📘 Formulas:
- Individual: x̄ = Σx/n; frequency: x̄ = Σfx/Σf; weighted: x̄ = Σwx/Σw
- Combined mean = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)
- Multiply every value by a and add b → new mean = a·x̄ + b
- Mean of the first n natural numbers = (n + 1)/2; of the first n odd numbers = n; of the first n even numbers = n + 1
- A value misread: correct mean = old mean + (correct − wrong)/n
✏️ Worked Example: The mean of 10 numbers is 25. Later it is found that one number, 40, was read as 20. The correct mean?
→ Wrong total = 250; correct total = 250 − 20 + 40 = 270
→ Correct mean = 270/10 = 27 (shortcut: 25 + 20/10)
→ Wrong total = 250; correct total = 250 − 20 + 40 = 270
→ Correct mean = 270/10 = 27 (shortcut: 25 + 20/10)
✏️ Worked Example: 30 students of class A average 60 and 20 students of class B average 70. Combined average?
→ (30 × 60 + 20 × 70)/50 = (1800 + 1400)/50
→ = 3200/50 = 64 (closer to the larger group, not 65)
→ (30 × 60 + 20 × 70)/50 = (1800 + 1400)/50
→ = 3200/50 = 64 (closer to the larger group, not 65)
✏️ Worked Example: x: 2, 4, 6, 8 and f: 3, 5, 8, 4. Mean?
→ Σf = 20; Σfx = 6 + 20 + 48 + 32 = 106
→ x̄ = 106/20 = 5.3
→ Σf = 20; Σfx = 6 + 20 + 48 + 32 = 106
→ x̄ = 106/20 = 5.3
⚠️ Exam Traps:
- For a combined mean do not simply average the two means — weight them by n.
- The sum of deviations from the mean, Σ(x − x̄), is always 0.
📌 2. Median and Mode
📘 Formulas:
- Median: arrange in ascending order; n odd → the (n + 1)/2-th term; n even → average of the n/2-th and (n/2 + 1)-th terms
- Discrete frequency: build the cumulative frequency (cf) and find the (N + 1)/2-th item
- Mode = the value that occurs most often; two such values → bimodal; all equally often → no mode
- Empirical relation: Mode = 3 Median − 2 Mean
- The median is not affected by very large/small values — suitable for skewed data such as incomes.
- The mode also works for qualitative data (e.g. the most popular colour, shoe size).
✏️ Worked Example: Median of 12, 7, 15, 9, 20, 11?
→ Order: 7, 9, 11, 12, 15, 20 (n = 6, even)
→ Median = (11 + 12)/2 = 11.5
→ Order: 7, 9, 11, 12, 15, 20 (n = 6, even)
→ Median = (11 + 12)/2 = 11.5
✏️ Worked Example: x: 10, 20, 30, 40 and f: 4, 6, 7, 3. Median?
→ N = 20; cf = 4, 10, 17, 20; the (N + 1)/2 = 10.5-th item
→ 10th item = 20, 11th item = 30 → median = 25
→ N = 20; cf = 4, 10, 17, 20; the (N + 1)/2 = 10.5-th item
→ 10th item = 20, 11th item = 30 → median = 25
✏️ Worked Example: A distribution has mean 40 and median 38. Mode?
→ Mode = 3 × 38 − 2 × 40 = 114 − 80 = 34
→ Mode = 3 × 38 − 2 × 40 = 114 − 80 = 34
⚠️ Exam Traps:
- Do not forget to arrange the data in order before finding the median.
📌 3. Grouped Data
📘 Formulas:
- Class mark x = (lower limit + upper limit)/2; mean = Σfx/Σf
- Median = L + [(N/2 − cf)/f] × h (cf = cumulative frequency before the median class)
- Mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h (f₁ = the highest frequency)
✏️ Worked Example: Classes 0–20, 20–40, 40–60, 60–80; frequencies 6, 10, 14, 10. Mean, median and mode?
→ x = 10, 30, 50, 70; Σfx = 60 + 300 + 700 + 700 = 1760; N = 40 → mean = 44
→ cf = 6, 16, 30, 40; N/2 = 20 → class 40–60: median = 40 + (20 − 16)/14 × 20 ≈ 45.71
→ Modal class 40–60 (f₁ = 14, f₀ = 10, f₂ = 10): 40 + (4/8) × 20 = 50
→ x = 10, 30, 50, 70; Σfx = 60 + 300 + 700 + 700 = 1760; N = 40 → mean = 44
→ cf = 6, 16, 30, 40; N/2 = 20 → class 40–60: median = 40 + (20 − 16)/14 × 20 ≈ 45.71
→ Modal class 40–60 (f₁ = 14, f₀ = 10, f₂ = 10): 40 + (4/8) × 20 = 50
Next: the step-deviation method, quartiles, deciles, percentiles and JSO-level questions — Statistics (Advanced) — Unit 2.
📌 4. Dispersion: Range, Mean Deviation, Variance, SD
📘 Formulas:
- Range = maximum − minimum; coefficient of range = (L − S)/(L + S)
- Mean deviation MD = Σ|x − A|/n (A = mean or median)
- Variance σ² = Σ(x − x̄)²/n = Σx²/n − x̄²; standard deviation σ = √variance
- Coefficient of variation CV = σ/x̄ × 100% — lower CV = more consistent
- Variance of the first n natural numbers = (n² − 1)/12; SD of an AP with common difference d = d√[(n² − 1)/12]
✏️ Worked Example: SD of 6, 8, 10, 12, 14?
→ Mean = 10; (x − 10)² = 16, 4, 0, 4, 16 → sum 40
→ σ² = 40/5 = 8 → σ = √8 = 2√2 ≈ 2.83
→ Mean = 10; (x − 10)² = 16, 4, 0, 4, 16 → sum 40
→ σ² = 40/5 = 8 → σ = √8 = 2√2 ≈ 2.83
✏️ Worked Example: n = 5, Σx = 30, Σx² = 220. Variance?
→ x̄ = 30/5 = 6
→ σ² = 220/5 − 6² = 44 − 36 = 8
→ x̄ = 30/5 = 6
→ σ² = 220/5 − 6² = 44 − 36 = 8
✏️ Worked Example: The SD of some data is 3. Each value is multiplied by 2 and then 5 is subtracted. New SD and variance?
→ Subtracting/adding does not change the SD; multiplying doubles it → SD = 6
→ Variance = 6² = 36 (i.e. 2² × 9)
→ Subtracting/adding does not change the SD; multiplying doubles it → SD = 6
→ Variance = 6² = 36 (i.e. 2² × 9)
⚠️ Exam Traps:
- The unit of variance is the square of the data unit (cm²); the SD has the data unit (cm).
- If all values are equal, SD = 0; the SD is never negative.
📌 5. Tables and Graphs
| Graph | When | How to recognise |
|---|---|---|
| Bar diagram | Separate categories (brands, years) | Equal widths, gaps between bars |
| Histogram | Continuous class intervals | Bars touch; area ∝ frequency |
| Frequency polygon | Comparing two distributions | Joins the mid-points of bar tops; closed at the axis at both ends |
| Ogive | Finding the median/quartiles graphically | The "less than" and "more than" curves meet at the median |
| Pie chart | Share of a whole (budget) | Angle = value/total × 360° |
✏️ Worked Example: In a pie chart an item has an angle of 72° and the total spending is ₹50,000. The item’s percentage and amount?
→ Percentage = 72/360 × 100 = 20%
→ Amount = 20% × 50,000 = ₹10,000
→ Percentage = 72/360 × 100 = 20%
→ Amount = 20% × 50,000 = ₹10,000
✏️ Worked Example: In a histogram the other classes have width 10, but the class 10–30 (width 20) has frequency 16. Height of its rectangle?
→ Unequal width → height = frequency × (standard width/class width)
→ = 16 × 10/20 = 8
→ Unequal width → height = frequency × (standard width/class width)
→ = 16 × 10/20 = 8
⚠️ Exam Traps:
- Taking the frequency directly as the height when class widths are unequal is the most common mistake.
📌 6. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The mean of the first 15 odd natural numbers is: Practice
Correct answer: (C)
Solution: The sum of the first n odd numbers is n² = 225; mean = 225/15 = 15.
Solution: The sum of the first n odd numbers is n² = 225; mean = 225/15 = 15.
Q2. The mean of 20 numbers is 30. One number, 45, was wrongly read as 25. The correct mean is: Practice
Correct answer: (A)
Solution: Correct total = 600 − 25 + 45 = 620; correct mean = 620/20 = 31.
Solution: Correct total = 600 − 25 + 45 = 620; correct mean = 620/20 = 31.
Q3. 40 boys average 50 marks and 60 girls average 55. The average of the whole class is: Practice
Correct answer: (D)
Solution: (40 × 50 + 60 × 55)/100 = (2000 + 3300)/100 = 53.
Solution: (40 × 50 + 60 × 55)/100 = (2000 + 3300)/100 = 53.
Q4. The mean of some data is 18. If each value is multiplied by 2 and then 6 is subtracted, the new mean is: Practice
Correct answer: (B)
Solution: New mean = 2 × 18 − 6 = 30.
Solution: New mean = 2 × 18 − 6 = 30.
Q5. The mean of x: 1, 2, 3, 4 with f: 4, 6, 7, 3 is: Practice
Correct answer: (B)
Solution: Σf = 20; Σfx = 4 + 12 + 21 + 12 = 49; mean = 49/20 = 2.45.
Solution: Σf = 20; Σfx = 4 + 12 + 21 + 12 = 49; mean = 49/20 = 2.45.
Q6. The values x: 10, 20, 30 have frequencies 3, k, 2 and the mean is 19. The value of k is: Practice
Correct answer: (D)
Solution: Σfx = 30 + 20k + 60 = 90 + 20k; Σf = 5 + k → 90 + 20k = 19(5 + k) = 95 + 19k → k = 5.
Solution: Σfx = 30 + 20k + 60 = 90 + 20k; Σf = 5 + k → 90 + 20k = 19(5 + k) = 95 + 19k → k = 5.
Q7. The median of 15, 6, 22, 9, 18, 11, 30, 14 is: Practice
Correct answer: (A)
Solution: Order: 6, 9, 11, 14, 15, 18, 22, 30; n = 8 → average of the 4th and 5th terms = (14 + 15)/2 = 14.5.
Solution: Order: 6, 9, 11, 14, 15, 18, 22, 30; n = 8 → average of the 4th and 5th terms = (14 + 15)/2 = 14.5.
Q8. The values x: 5, 10, 15, 20 have frequencies 3, 8, 6, 4. The median is: Practice
Correct answer: (C)
Solution: N = 21 → the (N + 1)/2 = 11th item; cf = 3, 11, 17, 21 → the 11th item lies in the group x = 10.
Solution: N = 21 → the (N + 1)/2 = 11th item; cf = 3, 11, 17, 21 → the 11th item lies in the group x = 10.
Q9. The mode of 3, 5, 5, 7, 8, 8, 9 is: Practice
Correct answer: (A)
Solution: 5 and 8 each occur twice and the rest once — so both are modes.
Solution: 5 and 8 each occur twice and the rest once — so both are modes.
Q10. A distribution has mode 24 and mean 30. By the empirical relation, the median is: Practice
Correct answer: (D)
Solution: Mode = 3 Median − 2 Mean → 24 = 3M − 60 → M = 84/3 = 28.
Solution: Mode = 3 Median − 2 Mean → 24 = 3M − 60 → M = 84/3 = 28.
Q11. The classes 0–10, 10–20, 20–30 have frequencies 2, 5, 3. The mean is: Practice
Correct answer: (C)
Solution: Class marks 5, 15, 25; Σfx = 10 + 75 + 75 = 160; N = 10 → mean = 16.
Solution: Class marks 5, 15, 25; Σfx = 10 + 75 + 75 = 160; N = 10 → mean = 16.
Q12. The classes 0–10, 10–20, 20–30, 30–40 have frequencies 5, 10, 15, 10. The median is: Practice
Correct answer: (B)
Solution: N/2 = 20; cf = 5, 15, 30 → median class 20–30: 20 + (20 − 15)/15 × 10 = 20 + 3.33 = 23.33.
Solution: N/2 = 20; cf = 5, 15, 30 → median class 20–30: 20 + (20 − 15)/15 × 10 = 20 + 3.33 = 23.33.
Q13. The classes 10–20, 20–30, 30–40, 40–50 have frequencies 4, 12, 8, 6. The mode is: Practice
Correct answer: (D)
Solution: Modal class 20–30: f₁ = 12, f₀ = 4, f₂ = 8 → 20 + (12 − 4)/(24 − 4 − 8) × 10 = 20 + 8/12 × 10 = 26.67.
Solution: Modal class 20–30: f₁ = 12, f₀ = 4, f₂ = 8 → 20 + (12 − 4)/(24 − 4 − 8) × 10 = 20 + 8/12 × 10 = 26.67.
Q14. The coefficient of range of 12, 30, 18, 45, 25 is: Practice
Correct answer: (B)
Solution: L = 45, S = 12 → (45 − 12)/(45 + 12) = 33/57 = 11/19.
Solution: L = 45, S = 12 → (45 − 12)/(45 + 12) = 33/57 = 11/19.
Q15. The mean deviation about the mean of 2, 4, 6, 8, 10 is: Practice
Correct answer: (A)
Solution: Mean = 6; |x − 6| = 4, 2, 0, 2, 4 → sum 12; MD = 12/5 = 2.4.
Solution: Mean = 6; |x − 6| = 4, 2, 0, 2, 4 → sum 12; MD = 12/5 = 2.4.
Q16. For 10 observations Σx = 50 and Σx² = 290. The standard deviation is: Practice
Correct answer: (C)
Solution: x̄ = 5; σ² = 290/10 − 5² = 29 − 25 = 4 → σ = 2.
Solution: x̄ = 5; σ² = 290/10 − 5² = 29 − 25 = 4 → σ = 2.
Q17. The standard deviation of the AP 3, 6, 9, 12, 15 is: Practice
Correct answer: (C)
Solution: SD of 1, 2, 3, 4, 5 = √[(25 − 1)/12] = √2; every term is 3 times → SD = 3√2.
Solution: SD of 1, 2, 3, 4, 5 = √[(25 − 1)/12] = √2; every term is 3 times → SD = 3√2.
Q18. A series has mean 25 and standard deviation 5. Its coefficient of variation (CV) is: Practice
Correct answer: (B)
Solution: CV = σ/x̄ × 100 = 5/25 × 100 = 20%.
Solution: CV = σ/x̄ × 100 = 5/25 × 100 = 20%.
Q19. Player A: average 60, SD 12; player B: average 45, SD 6. Who is more consistent? Practice
Correct answer: (A)
Solution: CV(A) = 12/60 × 100 = 20%; CV(B) = 6/45 × 100 ≈ 13.3% → B, with the lower CV, is more consistent.
Solution: CV(A) = 12/60 × 100 = 20%; CV(B) = 6/45 × 100 ≈ 13.3% → B, with the lower CV, is more consistent.
Q20. The variance of some data is 9. If each value is multiplied by 3 and 4 is added, the new variance is: Practice
Correct answer: (D)
Solution: Var(3x + 4) = 3² × 9 = 81; the added 4 does not change the variance.
Solution: Var(3x + 4) = 3² × 9 = 81; the added 4 does not change the variance.
Q21. The mean of 9 observations is 20. A new observation, 30, is added. The new mean is: Practice
Correct answer: (B)
Solution: New total = 180 + 30 = 210; 10 observations → mean = 21.
Solution: New total = 180 + 30 = 210; 10 observations → mean = 21.
Q22. In a pie chart an item has a central angle of 54°. Its percentage share of the total is: Practice
Correct answer: (C)
Solution: Percentage = 54/360 × 100 = 15%.
Solution: Percentage = 54/360 × 100 = 15%.
Q23. In a histogram the frequency of a class is proportional to the rectangle’s: Practice
Correct answer: (D)
Solution: In a histogram, area ∝ frequency; only when class widths are equal is the height also proportional to the frequency.
Solution: In a histogram, area ∝ frequency; only when class widths are equal is the height also proportional to the frequency.
Q24. The x-coordinate of the point where the "less than" and "more than" ogives intersect is the: Practice
Correct answer: (A)
Solution: Both ogives meet at N/2; the x-value of that point is the median.
Solution: Both ogives meet at N/2; the x-value of that point is the median.
Q25. The most suitable graph for comparing the sales of four brands is a: Practice
Correct answer: (B)
Solution: Brands are separate categories (not continuous classes), so a bar diagram; histograms and ogives are for continuous data.
Solution: Brands are separate categories (not continuous classes), so a bar diagram; histograms and ogives are for continuous data.