1. Chapter at a Glance
Statistics summarises data with a few numbers: measures of central tendency (mean, median, mode) tell us the typical value, and measures of dispersion (range, mean deviation, variance, standard deviation) tell us how spread out the data are. JEE Main usually has one short, calculation-based question — often on variance, the effect of changing data, or finding missing observations.
- Mean and variance of a set, including finding missing values from given mean and variance.
- Effect of adding a constant or multiplying by a constant on mean and variance.
- Combined mean and variance of two groups.
- Mean deviation about mean or median.
2. Measures of Central Tendency
- Mean: x̄ = Σxi/n (ungrouped) or Σfixi/Σfi (frequency data).
- Median: the middle value after arranging in order — the ((n + 1)/2)-th value if n is odd, the average of the (n/2)-th and (n/2 + 1)-th values if n is even. For grouped data: Median = l + ((N/2 − C)/f) × h.
- Mode: the most frequent value. For moderately skewed data: Mode ≈ 3 Median − 2 Mean.
Combined mean
3. Measures of Dispersion
- Range = largest − smallest value.
- Mean deviation about a value A: Σ|xi − A|/n. It is least when A is the median.
- Variance σ² = Σ(xi − x̄)²/n = (Σxi²/n) − x̄². Standard deviation σ = √variance.
- For frequency data: σ² = Σfixi²/N − x̄².
- Coefficient of variation CV = (σ/x̄) × 100. The series with the smaller CV is more consistent.
| Change applied to every observation | Mean | Variance | SD |
|---|---|---|---|
| Add k | increases by k | no change | no change |
| Multiply by k | multiplied by k | multiplied by k² | multiplied by |k| |
| y = ax + b | a x̄ + b | a²σ² | |a|σ |
Standard results
- Variance of the first n natural numbers: (n² − 1)/12.
- Variance of an AP with common difference d and n terms: d²(n² − 1)/12.
- Combined variance of two groups: σ² = [n1(σ1² + d1²) + n2(σ2² + d2²)]/(n1 + n2), where di = x̄i − x̄.
4. Formula Master Table
| Quantity | Formula |
|---|---|
| Mean | Σxi/n or Σfixi/N |
| Median (grouped) | l + ((N/2 − C)/f) h |
| Empirical relation | Mode = 3 Median − 2 Mean |
| Mean deviation | Σ|xi − A|/n |
| Variance | Σxi²/n − x̄² |
| Variance of 1, 2, …, n | (n² − 1)/12 |
| Coefficient of variation | (σ/x̄) × 100 |
| Combined mean | (n1x̄1 + n2x̄2)/(n1 + n2) |
| Linear transformation | Var(ax + b) = a² Var(x) |
5. Mind Map
- Mean, median, mode
- Grouped formulas
- Combined mean
- Range
- Mean deviation
- Variance, SD
- +k: mean shifts only
- ×k: variance × k²
- Coefficient of variation
- Consistency
6. Tips and Tricks
- To find two missing observations from the mean and variance: first get their sum (from the mean), then their sum of squares (from Σx² = n(σ² + x̄²)), then solve.
- Shift the data (subtract a convenient number) before computing variance — variance does not change.
- If one wrong observation is corrected, recompute Σx and Σx² by subtracting the wrong value and adding the correct one.
- Thinking that adding a constant changes the variance.
- Multiplying the variance by k instead of k².
- Forgetting to arrange data in order before finding the median.
7. Important One-liners
- The sum of deviations of all observations from their mean is zero.
- The sum of squared deviations is least when taken about the mean.
- Mean deviation is least when taken about the median.
- Variance is never negative; it is zero only when all observations are equal.
- Standard deviation has the same unit as the data; variance has the square of that unit.
- For a symmetric distribution, mean = median = mode.
8. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
Find the variance of the first 10 natural numbers.
Show solution
Variance = (n² − 1)/12 = (100 − 1)/12 = 99/12 = 8.25.
Answer: 8.25
If each observation of a data set is multiplied by 3 and then 2 is added, the standard deviation of the new data is:
- The same as before
- 3 times the old SD
- 3 times the old SD plus 2
- 9 times the old SD
Show solution
SD(ax + b) = |a| SD(x). Adding 2 has no effect; multiplying by 3 multiplies the SD by 3.
Answer: (B)
Find the mean deviation about the mean of 3, 5, 7, 9, 11.
Show solution
Mean = 35/5 = 7. Absolute deviations: 4, 2, 0, 2, 4, with sum 12.
Mean deviation = 12/5 = 2.4.
Answer: 2.4
The variance of 2, 4, 6, 8, 10 is:
- 4
- 8
- 6
- 10
Show solution
Mean = 6. Squared deviations: 16, 4, 0, 4, 16, total 40. Variance = 40/5 = 8.
(Shortcut: AP with d = 2, n = 5: d²(n² − 1)/12 = 4 × 24/12 = 8.)
Answer: (B)
The mean of a group of 10 students is 20 and that of another group of 15 students is 30. The combined mean is:
- 25
- 26
- 24
- 27
Show solution
Combined mean = (10 × 20 + 15 × 30)/25 = (200 + 450)/25 = 26.
Answer: (B)
The mean and variance of five observations are 5 and 4. Three of the observations are 2, 5 and 8. The absolute difference between the other two observations is:
- 1
- 2
- 3
- 4
Show solution
Sum of all five = 25, so the other two add to 25 − 15 = 10.
Σx² = n(σ² + x̄²) = 5(4 + 25) = 145. Known squares: 4 + 25 + 64 = 93, so x² + y² = 52.
2xy = (x + y)² − (x² + y²) = 100 − 52 = 48 ⇒ (x − y)² = 52 − 48 = 4 ⇒ |x − y| = 2. (The observations are 4 and 6.)
Answer: (B)
Find the median of 3, 7, 1, 9, 5, 11.
Show solution
In order: 1, 3, 5, 7, 9, 11. With 6 values, the median is the average of the 3rd and 4th: (5 + 7)/2 = 6.
Answer: 6
Two batsmen have mean scores 50 and 40 with standard deviations 10 and 6 respectively. Which statement is correct?
- The first batsman is more consistent
- The second batsman is more consistent
- Both are equally consistent
- Consistency cannot be compared
Show solution
CV1 = 10/50 × 100 = 20% and CV2 = 6/40 × 100 = 15%. The smaller CV means greater consistency.
Answer: (B)
9. Quick Revision Summary
- σ² = Σx²/n − x̄²; Σx² = n(σ² + x̄²) for missing-value problems.
- +k shifts the mean only; ×k scales the mean by k and the variance by k².
- Variance of 1 to n = (n² − 1)/12; AP: d²(n² − 1)/12.
- Combined mean = weighted average; smaller CV = more consistent.
- Mean deviation is least about the median.