Central tendency tells you where a data set is centred. Dispersion tells you how spread out the data is around that centre. You can have two completely different data sets with the same mean — dispersion is what separates them.
Take these two sets:
Same mean, wildly different behaviour. Set A has zero spread. Set B has high spread. Any exam question that says "compare the two distributions" is almost certainly asking you to compute a measure of dispersion.
The analogy that works: think of the mean as the anchor of a boat, and dispersion as the length of the chain. A short chain means the boat barely moves. A long chain means the boat swings all over the place. The anchor position is the same — what changes is how far the boat can drift.
SSC CGL tests five measures of dispersion in practice:
Each measure has a specific use case and a specific vulnerability to outliers. The exam tests both the computation and the conceptual property — so you need both.
One more thing to know upfront: absolute measures of dispersion (Range, MD, QD, SD, Variance) are in the same units as the data. Relative measures (CV, coefficient of quartile deviation) are percentages — unit-free — and are used when comparing two different distributions.
Formula: Range = Maximum value − Minimum value
That is it. Fast, but fragile — one outlier wrecks it.
Coefficient of Range (relative version):
Coefficient of Range = (Max − Min) / (Max + Min)
About the mean:
MD (about mean) = (1/n) × Σ|xᵢ − x̄|
About the median:
MD (about median) = (1/n) × Σ|xᵢ − Median|
Key property: MD is minimum when calculated about the median. If an exam question asks "about which value is MD least?", the answer is always the median.
Coefficient of MD (relative version):
Coefficient of MD = MD / Mean (when computed about mean)
QD = (Q3 − Q1) / 2
Also called the semi-interquartile range. It ignores the top 25% and bottom 25% of data, making it the least affected measure by outliers — the direct opposite of range.
Coefficient of QD:
Coefficient of QD = (Q3 − Q1) / (Q3 + Q1)
Population Variance:
σ² = (1/n) × Σ(xᵢ − x̄)²
Standard Deviation:
σ = √[Σ(xᵢ − x̄)² / n]
An equivalent and often faster formula:
σ² = (Σxᵢ²/n) − (x̄)²
In words: variance = mean of squares minus square of mean. This avoids computing deviations one by one when the mean is a decimal.
Effect of linear transformation on SD and Variance:
This is where SSC CGL loves to set traps.
k: SD gets multiplied by |k|, and Variance gets multiplied by k².So if each observation is multiplied by 3, new Variance = 9 × original Variance. This is tested directly.
CV = (σ / x̄) × 100
CV is the go-to tool when you need to compare variability of two distributions that have different means or different units of measurement. The distribution with higher CV is more variable / less consistent.
Typical exam phrasing:
This appears in the SSC CGL Statistics paper and connects dispersion to skewness.
Using mode:
Skₚ = (Mean − Mode) / SD
Using median (more stable):
Skₚ = 3(Mean − Median) / SD
| Measure | Uses All Values | Affected by Outliers | Unit-Free | |---|---|---|---| | Range | No (only 2) | Most affected | No | | QD | No (middle 50%) | Least affected | No | | MD | Yes | Moderately affected | No | | SD / Variance | Yes | Highly affected | No | | CV | Yes | Highly affected | Yes |
When you see "CV = X%, mean = Y, find SD" — don't re-derive. Rearrange the formula once and burn it in: SD = (CV × Mean) / 100. That's it. No division step-by-step, just two multiplications and a division by 100 (which is just a decimal shift). Example: CV = 40%, Mean = 25 → SD = (40 × 25)/100 = 1000/100 = 10. Standard method (re-deriving the formula): 30 seconds. This direct substitution: 6 seconds.
The exam will always give you a "multiply by k" transformation and ask what happens to variance. The pattern: addition/subtraction → zero effect on variance; multiplication/division → variance scales by k². Visualise it as: SD is a length, variance is an area. Doubling every length quadruples the area. If k=3, variance × 9; if k=5, variance × 25. Standard method (computing new variance from scratch): 6-8 steps. Pattern recognition: 1 step.
Remember the order of outlier sensitivity from most to least: R-S-MD-QD (Range, SD, Mean Deviation, Quartile Deviation). Any conceptual question asking "which is most/least affected by outliers" is answered directly from this chain. Most affected = Range (only uses the two extreme values). Least affected = QD (completely ignores the outer 25% on both sides). Standard method (reasoning from definitions): 20-25 seconds. Memorised chain: 3 seconds.
If asked "about which average is mean deviation least?", the answer is always median. This is a proven mathematical result — not a coincidence. Contrast: variance is minimum about the mean (not the median). So: MD → minimised at median; Variance → minimised at mean. One line to remember, zero computation needed. Exam questions on this type: answered in under 5 seconds vs. 40 seconds of second-guessing.
Karl Pearson's formula Skₚ = 3(Mean − Median)/SD means the sign of skewness is just the sign of (Mean − Median). Mean > Median → positive skew (right tail). Mean < Median → negative skew (left tail). Mean = Median → zero skew (symmetric). You often don't need to compute the full formula — just check the numerator's sign first, then compute the magnitude only if asked. Saves 15-20 seconds on sign-check questions.
When you see a dispersion question in the exam hall, run this decision tree:
Step 1 — What is being asked?
3(Mean − Median)/SD if median is given; (Mean − Mode)/SD if mode is given.Step 2 — Verify your mean/median calculation first. A wrong central value cascades into every subsequent step.
Step 3 — For CV comparison questions, never just compare raw SDs. Always divide by respective means. The series with the larger SD is not automatically more variable if its mean is also much larger.
Step 4 — Check units. If the question asks for a coefficient (coefficient of variation, coefficient of MD), the answer should be a ratio or percentage, not in the original units.
Why this question: The most basic dispersion formula — tests if you know exactly what range means.
Solving path: Scan the data set {4, 7, 13, 2, 19, 11}. Maximum = 19, Minimum = 2. Range = 19 − 2 = 17. Option D.
Why this question: Tests the full MD computation pipeline — mean first, then absolute deviations, then average.
Solving path: Data = {2, 4, 6, 8, 10}. Mean = 30/5 = 6. Absolute deviations: 4, 2, 0, 2, 4. Sum = 12. MD = 12/5 = 2.4. Option B.
Why this question: Direct application of CV formula in reverse — given CV and mean, find SD.
Solving path: CV = (SD/Mean) × 100 → SD = (40 × 25)/100 = 10. Option C. No re-derivation needed if you've burned the rearranged form into memory.
Why this question: Tests whether you can identify the most outlier-sensitive measure — a pure conceptual question.
Solving path: Range depends only on the maximum and minimum values. Change one extreme value by any amount and range changes completely. QD ignores the outer 25% on each side — it is the least sensitive. Answer: Range. Option C.
Why this question: Classic transformation property — directly tests the k² rule for variance.
Solving path: Multiplication by k → SD scales by k, variance scales by k². Here k = 3, so new variance = 3² × original = 9 times original. Option C.
Why this question: Tests Karl Pearson's skewness formula using median — a formula you must have ready in the exam hall.
Solving path: Formula = 3(Mean − Median)/SD = 3(50 − 48)/15 = 6/15 = 0.4. Positive value → positive skew. Option D.
Why this question: CV comparison across two series — tests whether you resist the trap of comparing raw SDs.
Solving path: CV(A) = (4/30) × 100 = 13.33%. CV(B) = (6/20) × 100 = 30%. Series A has larger mean, which absorbs the larger SD. Series B is more variable. Option B. Note: if you had just compared SDs (4 vs 6), you'd still get B correct here — but the mean difference is why CV is the right tool, and this setup is designed to show that.
Comparing SDs directly without checking means. Raw SD tells you nothing about relative variability. Always use CV when the means differ. Two students scoring 4/30 and 6/20 on different tests — the second is more variable even though their absolute SD is only slightly higher.
Forgetting to take absolute values in MD. MD is always Σ|xᵢ − x̄| / n, not Σ(xᵢ − x̄) / n. The latter always equals zero (deviations around the mean cancel). Every time.
Applying the k² rule to SD instead of Variance. If every value is multiplied by 3: new SD = 3 × old SD (not 9× old SD). New Variance = 9 × old Variance. Many students flip this and select the wrong option.
Using mode instead of median in the Pearson formula when median is given. The formula has two versions. The median version uses the factor of 3: 3(Mean − Median)/SD. The mode version has no factor of 3: (Mean − Mode)/SD. Mixing them up gives a wrong answer even with correct arithmetic.
Confusing "least affected by outliers" with "least accurate". QD is least affected by outliers because it ignores the tails — but that is by design, not a weakness. Exam conceptual questions sometimes frame this ambiguously; QD deliberately discards extreme values.
Ignoring the sign of the skewness coefficient. After computing 3(Mean − Median)/SD, many students report the magnitude but not the sign. SSC CGL questions will give answer choices with both positive and negative versions of the same magnitude. Check: Mean > Median → positive; Mean < Median → negative.