Dispersion and range
Two batters can have the same average score; yet one always scores around 40 while the other scores 0 one day and 100 another. The number that tells how widely the data is spread around the average is dispersion. Building on classes 9–10 : In classes 9–10 we learnt the mean, median and mode; they tell the "centre" of the data. Now we learn to measure its "spread".
The difference between the largest and the smallest value is the range. Range ÷ (largest + smallest) is the coefficient of range. The more spread out the data, the greater the dispersion.
Rule 1: Range and its coefficient — Range = L − S; coefficient of range = (L − S) ÷ (L + S). L is the largest value and S the smallest. For grouped data L is the upper limit of the last class and S the lower limit of the first class.
In the figure : Both sets have the same mean, but B has the larger range.
Rule 2: Limitation of the range — The range looks only at the two extreme values; the middle values do not change it at all, but one extreme value changes it completely.
How is the formula obtained? : When the same number is added to every value, all values shift by the same amount; the largest and the smallest both grow by k. So (L + k) − (S + k) = L − S and the range stays the same.
- Easy : Data 12, 7, 19, 4, 15: L = 19, S = 4. Range = 19 − 4 = 15. Coefficient = 15/23.
- Medium : Classes 0–10, 10–20, 20–30, 30–40, 40–50: L = 50, S = 0. Range = 50 − 0 = 50.
- Exam-type : Smallest value 6 and range 14: L = 6 + 14 = 20. Coefficient = 14/26 = 7/13.
The range has a unit (the same as the data); the coefficient of range has no unit. So data in different units is compared through the coefficient.
Writing (L − S) ÷ (L − S) or L ÷ S for the coefficient. The denominator is always the sum (L + S).
Adding or subtracting the same number to all values leaves the range unchanged; multiplying all by k makes the range |k| times. Memory clue : "add or subtract, spread stays; multiply, spread grows".
When CET says "5 is added to every value", the answer is the same range without any calculation.
CET-style example : The data is 3, 8, 12, 20. What is the range after 5 is added to every value? (A) 17 (B) 22 (C) 12 (D) 5
Answer : (A). The original range is 20 − 3 = 17 and it does not change. Time-saving method : Adding does not change the range; just 20 − 3.
Spot the difference
| Point | Range | Coefficient of range |
|---|---|---|
| Formula | L − S | (L − S) ÷ (L + S) |
| Unit | unit of the data | no unit |
| Use | spread of one set | comparing two sets |
Practice questions
- 1Find the range of the data 9, 3, 14, 6.
- 2Find the range of the data 25, 40, 31, 18, 36.
- 3The largest value is 48 and the smallest is 30. Find the coefficient of range.
- 4Find the range of the classes 10–20, 20–30, …, 60–70.
- 5Find the coefficient of range of the data 4, 10, 16.
- 6The range is 25 and the smallest value is 15. Find the largest value.
- 7Every value of the data 4, 9, 13 is multiplied by 3. Find the new range.
- 8Every value of the data 5, 8, 12, 20 is multiplied by 2 and then 2 is subtracted. Find the new range.
Show answers (try them yourself first)
- 114 − 3 = 11.
- 240 − 18 = 22.
- 318/78 = 3/13.
- 470 − 10 = 60.
- 5(16 − 4)/(16 + 4) = 12/20 = 3/5.
- 6L = 15 + 25 = 40.
- 7Original range 13 − 4 = 9; new range 3 × 9 = 27.
- 8Original range 20 − 5 = 15; new range 2 × 15 = 30 (subtraction does not change it).
Variance and standard deviation: ungrouped data
Runs in the overs of two bowlers: one gave 6, 6, 6 and the other 1, 6, 11. The average is the same, but the second bowler’s numbers lie far from the average. The average measure of how far each number is from the mean is the standard deviation. Building on classes 9–10 : In classes 9–10 we learnt to find the mean and to square and take square roots; these are the two supports needed here: first squares of the differences, then the square root.
Find the difference of every value from the mean and square it; the mean of those squares is the variance (σ²). The square root of the variance is the standard deviation (σ), which has the same unit as the data.
Rule 1: Two formulas — Variance σ² = Σ(x − x̄)² ÷ n. Quick form: σ² = Σx² ÷ n − (x̄)². Standard deviation σ = √σ².
In the figure : The difference between each value and the mean is shown by segments; the mean of the squares of these differences is the variance.
Rule 2: Properties of variance — Variance is never negative. If all values are equal, σ = 0. The larger the σ, the more spread out the data.
How is the formula obtained? : Σ(x − x̄)² = Σ(x² − 2x̄x + x̄²) = Σx² − 2x̄Σx + n x̄². Since Σx = n x̄, the middle term is −2n x̄². What remains is Σx² − n x̄². Divide by n: σ² = Σx²/n − x̄².
- Easy : Values 4, 8: mean 6; squared differences 4, 4; variance 8/2 = 4; σ = 2.
- Medium : 1 to 7: mean 4; squared differences 9, 4, 1, 0, 1, 4, 9 = 28; variance 28/7 = 4; σ = 2.
- Exam-type : n = 10, Σx = 50, Σx² = 290: x̄ = 5; σ² = 290/10 − 25 = 4; σ = 2.
The unit of variance is the square of the unit of the data (cm²); the unit of the standard deviation is the same as the data (cm). So σ is used to state the spread.
Using (Σx)² instead of Σx². Σx² means square each value first, then add; (Σx)² means add first, then square.
If the mean is a whole number, find the differences and square them; if the mean is a fraction, use Σx²/n − x̄². Memory clue : "mean of squares minus square of the mean".
CET gives Σx and Σx²; find the mean and Σx²/n and subtract at once. Do not forget the square root (read whether the question asks σ or σ²).
CET-style example : For 20 observations the mean is 8 and Σx² = 1300. The standard deviation is? (A) 1 (B) 2 (C) 65 (D) 4
Answer : (A). σ² = 1300/20 − 8² = 65 − 64 = 1; σ = 1. Time-saving method : Directly 65 − 64; no need to square the differences.
Spot the difference
| Point | Variance σ² | Standard deviation σ |
|---|---|---|
| Formula | Σ(x − x̄)² ÷ n | √variance |
| Unit | square of the unit | same as the data |
| Value | never negative | never negative |
Practice questions
- 1Find the variance of 2 and 6.
- 2What is the variance of 7, 7, 7, 7?
- 3Find the variance of 1, 2, 3, 4, 5.
- 4Find the variance of 4, 6, 6, 8.
- 5If n = 8, Σx = 40 and Σx² = 240, find the variance.
- 6The mean of 5 observations is 6 and Σx² = 190. Find the variance.
- 7Find the variance of the first 11 natural numbers.
- 8The mean of 9 observations is 7 and the variance is 4. One more value, equal to the mean 7, is added. Find the new variance.
Show answers (try them yourself first)
- 1Mean 4; squares 4 + 4 = 8; 8/2 = 4.
- 2All values equal the mean; variance 0.
- 3Mean 3; squares 4 + 1 + 0 + 1 + 4 = 10; 10/5 = 2.
- 4Mean 6; squares 4 + 0 + 0 + 4 = 8; 8/4 = 2.
- 5x̄ = 5; σ² = 240/8 − 25 = 5.
- 6σ² = 190/5 − 36 = 38 − 36 = 2.
- 7Mean 6; σ² = (121 − 1)/12 = 10.
- 8Original Σ(x − 7)² = 9 × 4 = 36. The new value 7 has difference 0; new n = 10 and the mean is the same. Variance 36/10 = 3.6.
Variance and standard deviation: frequency distributions
Marks of 40 students: some got 6, some 7, some 8. Instead of writing every mark separately we write "6 marks — 12 students" in a table. The spread can be measured from the table too. Building on classes 9–10 : In classes 9–10 we learnt the frequency table and the class mark (middle value of a class); we use the same here.
When a value x is given with its frequency f, each value is counted f times. N = Σf is the total number of observations. In grouped data x is the class mark.
Rule 1: Formulas for a frequency distribution — x̄ = Σfx ÷ N; σ² = Σfx² ÷ N − (x̄)²; σ = √σ².
In the figure : The bars show frequencies; the mean is the balance point of the set and σ is one step of the spread.
Rule 2: Step-deviation method — If the class marks are large, take u = (x − A) ÷ h (A is an assumed mean, h the class width). Then x̄ = A + h × Σfu ÷ N and σ = h × √(Σfu² ÷ N − (Σfu ÷ N)²).
How is the formula obtained? : We have x = A + hu. The values are multiplied by h and shifted by A, so the mean is x̄ = A + h ū and the standard deviation is σ_x = h σ_u (shifting does not change the spread, multiplying makes it h times). σ_u² = Σfu²/N − (Σfu/N)².
- Easy : x: 3, 5, 7; f: 1, 2, 1. N = 4, Σfx = 20, x̄ = 5. Σfx² = 9 + 50 + 49 = 108; σ² = 27 − 25 = 2.
- Medium : x: 2, 4, 6, 8; f: 1, 3, 4, 2. N = 10, Σfx = 54, x̄ = 5.4. Σfx² = 4 + 48 + 144 + 128 = 324; σ² = 32.4 − 29.16 = 3.24; σ = 1.8.
- Exam-type : Classes 0–10, 10–20, 20–30, 30–40; f: 1, 3, 5, 1. Class marks 5, 15, 25, 35; A = 25, h = 10: u = −2, −1, 0, 1. Σfu = −4, Σfu² = 8, N = 10. x̄ = 25 + 10 × (−0.4) = 21. σ = 10 × √(0.8 − 0.16) = 10 × 0.8 = 8.
In the variance N = Σf. For grouped data always take the class mark as x, not the class limit.
Taking (fx)² for Σfx². Correct: f × x². For example x = 6, f = 4: 4 × 36 = 144, not (24)² = 576.
If the values are large take u = (x − A)/h; at the end multiply σ by h. If A is chosen near the mean, Σfu is small. Memory clue : "shrink, find, expand again (multiply by h)".
When CET gives σ_u the answer is h × σ_u. No need to work with A at all.
CET-style example : Taking u = (x − 30)/5 we got Σfu = 0, Σfu² = 80, N = 20. σ = ? (A) 10 (B) 2 (C) 20 (D) 50
Answer : (A). σ_u² = 80/20 − 0 = 4, σ_u = 2; σ = 5 × 2 = 10. Time-saving method : Since Σfu = 0, just the square root of Σfu²/N, then multiply by h.
Spot the difference
| Point | Direct method | Step-deviation |
|---|---|---|
| Use | small x | large x or class marks |
| Mean | Σfx ÷ N | A + h × Σfu ÷ N |
| σ | √(Σfx²/N − x̄²) | h × √(Σfu²/N − (Σfu/N)²) |
Practice questions
- 1x: 1, 3, 5; f: 1, 2, 1. Find the variance.
- 2x: 0, 2; f: 1, 1. Find the variance.
- 3x: 10, 20, 30; f: 1, 2, 1. Find the variance.
- 4x: 2, 4, 6, 8; f: 2, 3, 3, 2. Find the variance.
- 5x: 10, 20, 30, 40; f: 2, 4, 3, 1. Taking A = 20, h = 10, find σ.
- 6If N = 20, Σfx = 200 and Σfx² = 2100, find the variance.
- 7Classes 0–10, 10–20, 20–30, 30–40; f: 2, 4, 3, 1. Find the mean and σ.
- 8x: 1, 2, 3, 4, 5; f: 1, 2, 4, 2, 1. Find the variance.
Show answers (try them yourself first)
- 1x̄ = 3; Σfx² = 1 + 18 + 25 = 44; σ² = 11 − 9 = 2.
- 2Mean 1; (1 + 1)/2 = 1.
- 3Mean 20; (100 + 0 + 0 + 100)/4 = 50.
- 4N = 10, Σfx = 50, x̄ = 5; Σfx² = 8 + 48 + 108 + 128 = 292; σ² = 29.2 − 25 = 4.2.
- 5Σfu = 3, Σfu² = 9; σ = 10 × √(0.9 − 0.09) = 10 × 0.9 = 9.
- 6x̄ = 10; σ² = 2100/20 − 100 = 5.
- 7Class marks 5, 15, 25, 35; x̄ = 18, σ² = 405 − 324 = 81, σ = 9.
- 8N = 10, x̄ = 3; Σf(x − 3)² = 4 + 2 + 0 + 2 + 4 = 12; σ² = 12/10 = 1.2.
Change of origin and scale
If a teacher adds 5 marks to every student, everyone’s marks move 5 ahead, but who is ahead of whom, and by how much, stays the same. If the marks are doubled, the gaps double. Building on classes 9–10 : In classes 9–10 we learnt the line y = ax + b; here we see how the mean and σ change when every x is turned into y = ax + b.
Adding or subtracting the same number to every value is a change of origin. Multiplying or dividing every value by the same number is a change of scale.
Rule 1: For y = ax + b — ȳ = a x̄ + b; σ_y = |a| σ_x; σ_y² = a² σ_x². b has no effect on σ.
In the figure : All points moved by 4: the mean moved but the gaps are the same; all gaps doubled: σ doubled.
How is the formula obtained? : If y = ax + b then ȳ = a x̄ + b. The difference y − ȳ = a(x − x̄) (b cancels). Squared: a²(x − x̄)². Taking the mean gives σ_y² = a² σ_x². In the square root a negative a becomes |a| because σ is never negative.
- Easy : Mean 8, σ = 3. 10 is added to every value: the new mean is 18; σ = 3 stays.
- Medium : Mean 8, σ = 3 and y = 2x − 5: ȳ = 16 − 5 = 11; σ_y = 2 × 3 = 6; σ_y² = 36.
- Exam-type : For u = (x − A)/10, σ_u = 1.2: σ_x = 10 × 1.2 = 12.
Adding/subtracting: the mean changes, σ stays. Multiplying/dividing: both change (σ is multiplied by |a| and the variance by a²).
Writing a negative σ (−12) for a negative a. σ is never negative; use |a|.
Square a for the variance, take the absolute value of a for σ. Ignore b completely. Memory clue : "b disappears, square the a".
When CET gives y = 1 − 3x, use b for the mean but leave b out for σ and take |−3| = 3.
CET-style example : The mean of x is 20 and σ = 4. If y = 1 − 3x, what is the σ of y? (A) 12 (B) −12 (C) 4 (D) 13
Answer : (A). σ_y = |−3| × 4 = 12 (the mean comes to −59, but it is not asked). Time-saving method : Just |a| × σ; no need to read b or the mean.
Spot the difference
| Change | Mean | σ |
|---|---|---|
| x + b | x̄ + b | same |
| a x | a x̄ | σ × abs(a) |
| a x + b | a x̄ + b | σ × abs(a) |
Practice questions
- 1Mean 12, σ = 2. 5 is added to every value. State the new mean and σ.
- 2σ = 3 and every value is multiplied by 4. Find the new σ.
- 3The variance is 9 and 7 is subtracted from every value. Find the new variance.
- 4σ_x = 5 and y = 3x + 2; find σ_y.
- 5y = (x − 50)/10 and σ_y = 2.4; find σ_x.
- 6Mean 30, σ = 6. Find the mean and σ of y = (x − 30)/6.
- 7The variance of x is 16. Find the variance of y = 3 − 2x.
- 8Mean 10 and σ = 5. All values are multiplied by 3. What is the coefficient of variation (in %)?
Show answers (try them yourself first)
- 1The mean is 17; σ = 2 stays.
- 2σ = 4 × 3 = 12.
- 3The variance stays 9.
- 4σ_y = 3 × 5 = 15.
- 5σ_x = 10 × 2.4 = 24.
- 6Mean (30 − 30)/6 = 0; σ = 6/6 = 1.
- 7(−2)² × 16 = 64.
- 8Original 5/10 × 100 = 50%; new mean 30, σ = 15; 15/30 × 100 = 50%.
Combined standard deviation
The marks of two divisions are known separately, but we need the overall average and spread of the school. We do not re-check every answer paper; the number, mean and σ of each division are enough. Building on classes 9–10 : In classes 9–10 we learnt the weighted average; the combined mean is the same idea. The σ formula is built on it.
If two sets have sizes n₁, n₂, means x̄₁, x̄₂ and standard deviations σ₁, σ₂, we can find the combined mean and the combined standard deviation of all the observations.
Rule 1: Combined formulas — x̄ = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂). d₁ = x̄₁ − x̄, d₂ = x̄₂ − x̄. σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)] ÷ (n₁ + n₂).
In the figure : The two set means lie at distances d₁ and d₂ from the combined mean; these distances add to σ.
How is the formula obtained? : For each set Σ(x − x̄)² = n σ². The difference from the combined mean: x − x̄ = (x − x̄₁) + d₁. When squared and added, the middle term is zero (the differences from a set’s own mean add to 0). What remains is n₁σ₁² + n₁d₁². The same for the second set; add both and divide by N.
- Easy : Two divisions: 30 students with mean 40 and 20 students with mean 50. Combined mean = (1200 + 1000)/50 = 44.
- Medium : 30 students (mean 40, σ 3) and 20 students (mean 50, σ 7): x̄ = 44; d₁ = −4, d₂ = 6. σ² = [30(9 + 16) + 20(49 + 36)]/50 = (750 + 1700)/50 = 49; σ = 7.
- Exam-type : 20 students (mean 50, σ 2) and 30 students (mean 70, σ 2): x̄ = 62; d₁ = −12, d₂ = 8. σ² = [20(4 + 144) + 30(4 + 64)]/50 = 5000/50 = 100; σ = 10.
The combined σ can be larger than the average of the two σ, because if the set means are far apart the spread increases (the d term).
Taking σ = (σ₁ + σ₂)/2. The correct way: add n(σ² + d²) for each set and divide by N.
First the combined mean, then d₁ and d₂. If d is zero, σ² is just the weighted mean of σ₁² and σ₂². Memory clue : "tie σ² and d² together, multiply by n".
When CET gives very different n₁ and n₂ the combined mean lies close to the larger set’s mean; use this to check your answer.
CET-style example : 10 observations (mean 40, σ 3) and 40 observations (mean 35, σ 2). The combined σ is? (A) 3 (B) 2.5 (C) 9 (D) 5
Answer : (A). x̄ = 36; d₁ = 4, d₂ = −1; σ² = (10 × 25 + 40 × 5)/50 = 9; σ = 3. Time-saving method : σ₁² + d₁² = 25 and σ₂² + d₂² = 5; (250 + 200)/50.
Spot the difference
| Point | Combined mean | Combined σ² |
|---|---|---|
| Built from | n × x̄ | n × (σ² + d²) |
| Divide by | N | N |
| Uses d | no | yes |
Practice questions
- 1The mean of 40 observations is 10 and of 60 observations is 15. Find the combined mean.
- 2The mean of 20 students is 50 and of 30 students is 60. Find the combined mean.
- 3Two sets of 25 and 25 observations have means 8 and 12. Find the combined mean.
- 420 observations (mean 40, σ 2) and 30 observations (mean 60, σ 2). Find the combined mean and σ.
- 520 observations (mean 50, σ 2) and 30 observations (mean 40, σ 8). Find the combined mean and σ.
- 630 observations (mean 40, σ 3) and 20 observations (mean 30, σ 7). Find the combined mean and σ.
- 7One set has 10 values all equal to 10 and another has 10 values all equal to 30. Find the combined σ.
- 840 observations (mean 60, σ 2) and 60 observations (mean 50, σ 8). Find the combined mean and σ.
Show answers (try them yourself first)
- 1(400 + 900)/100 = 13.
- 2(1000 + 1800)/50 = 56.
- 3(200 + 300)/50 = 10.
- 4x̄ = 52; d₁ = −12, d₂ = 8; σ² = [20 × 148 + 30 × 68]/50 = 100; σ = 10.
- 5x̄ = 44; d₁ = 6, d₂ = −4; σ² = [20 × 40 + 30 × 80]/50 = 64; σ = 8.
- 6x̄ = 36; d₁ = 4, d₂ = −6; σ² = [30 × 25 + 20 × 85]/50 = 49; σ = 7.
- 7x̄ = 20; d₁ = −10, d₂ = 10; σ² = (10 × 100 + 10 × 100)/20 = 100; σ = 10.
- 8x̄ = 54; d₁ = 6, d₂ = −4; σ² = [40 × 40 + 60 × 80]/100 = 64; σ = 8.
Coefficient of variation
One town has a daily temperature around 30 degrees varying by 3 degrees; another has around 6 degrees varying by 3 degrees. The same change is felt much more in the second place. So the spread is measured relative to the mean. Building on classes 9–10 : In classes 9–10 we learnt percentages; the coefficient of variation is σ as a percentage of the mean.
Dividing σ by the mean and multiplying by 100 gives the coefficient of variation (CV). It is in percent and has no unit. The set with the smaller CV is more consistent.
Rule 1: Formula of CV — CV = (σ ÷ x̄) × 100 %.
In the figure : A has the larger σ, but relative to its mean B is more spread out. So the comparison is done through CV.
Rule 2: When to use it — When two sets have different means or units, σ cannot be compared directly; use CV. A smaller CV means more consistent, a larger one means more fluctuation.
How is the formula obtained? : σ and the mean have the same unit, so σ ÷ x̄ has no unit. Multiplying by 100 turns it into a percentage. If both the mean and σ are doubled the CV is the same, because the 2 cancels in the division.
- Easy : x̄ = 40, σ = 8: CV = 8/40 × 100 = 20%.
- Medium : A: x̄ = 50, σ = 6, CV = 12%. B: x̄ = 80, σ = 8, CV = 10%. B has the smaller CV so B is more consistent.
- Exam-type : CV = 25% and x̄ = 16: σ = 25/100 × 16 = 4.
CV is in percent; σ is in the unit of the data. If the two sets have different units (kg and cm) only CV can be compared.
Assuming the set with the larger σ is the more spread out. If the means differ, look at the CV.
To compare two sets find σ/x̄; no need to multiply by 100 since both get the same 100. Memory clue : "small CV = steady".
When CET asks "which is more consistent?" choose the one with the smaller CV, not the smaller σ.
CET-style example : A: x̄ = 60, σ = 9. B: x̄ = 40, σ = 8. Which set is more consistent? (A) A (B) B (C) both the same (D) cannot be said
Answer : (A). CV_A = 15%, CV_B = 20%; A has the smaller CV. Time-saving method : 9/60 = 0.15 and 8/40 = 0.2; compare directly.
Spot the difference
| Point | σ | CV |
|---|---|---|
| Unit | of the data | percent |
| Compare | when means are equal | any sets |
| Small value | less spread | more consistent |
Practice questions
- 1x̄ = 50, σ = 5: find the CV.
- 2x̄ = 20, σ = 3: find the CV.
- 3If CV = 12% and x̄ = 50, find σ.
- 4If CV = 20% and σ = 6, find the mean.
- 5A: x̄ = 25, σ = 5. B: x̄ = 40, σ = 6. Which is more consistent?
- 6Find the CV of the data 2, 4, 4, 4, 5, 5, 7, 9 (σ = 2).
- 7The mean is 40 and CV is 15%. 10 is added to every value. Find the new CV.
- 8A has CV 30%, mean 20; B has CV 45%, mean 30. Find both σ. Which has the larger spread?
Show answers (try them yourself first)
- 15/50 × 100 = 10%.
- 23/20 × 100 = 15%.
- 3σ = 12/100 × 50 = 6.
- 4x̄ = σ × 100 ÷ CV = 600/20 = 30.
- 5CV_A = 20%, CV_B = 15%; B is more consistent.
- 6Mean 5; CV = 2/5 × 100 = 40%.
- 7σ = 15/100 × 40 = 6; new mean 50; CV = 6/50 × 100 = 12%.
- 8σ_A = 6; σ_B = 13.5. B has the larger spread.
Lesson 8: Measures of Dispersion — quick revision
| Measure | Formula / point |
|---|---|
| Range | L − S; coefficient (L − S)/(L + S); adding changes nothing |
| Variance | σ² = Σ(x − x̄)²/n = Σx²/n − x̄² |
| Standard deviation | σ = √σ²; unit same as the data |
| Frequency distribution | σ² = Σfx²/N − x̄²; step deviation σ = h × σ_u |
| y = ax + b | ȳ = a x̄ + b; σ_y = abs(a) × σ_x |
| Combined | x̄ = (n₁x̄₁ + n₂x̄₂)/N; σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/N |
| Coefficient of variation | CV = σ/x̄ × 100%; smaller CV = more consistent |
🎯 Most important points for the exam
- The range looks only at the two extremes; σ looks at all values.
- σ has the unit of the data, σ² the squared unit.
- Adding/subtracting: σ stays; multiplying: multiply σ by |a|.
- Do not forget d² in the combined σ.
- Use CV for comparison; a smaller CV means more consistent.
"measure the spread: square, mean, root".