Statistics
Measures of Dispersion: Range, Mean Deviation about Mean & Median (ungrouped & grouped), Variance, Standard Deviation, shortcut method & scale-shift laws.
Topic 13.1 – 13.2: Why Central Tendency Alone Isn't Enough (Measures of Dispersion)
Two school cricketers in Kochi — Arjun and Vishnu — check their last 10 match scores. Arjun scored
30, 91, 0, 64, 42, 80, 30, 5, 117, 71, while Vishnu scored 53, 46, 48, 50, 53, 53, 58, 60, 57, 52. Surprise! BOTH players have the EXACT same Mean (53) and Median (53)! Yet Arjun swings wildly from a duck (0) to a century (117), while Vishnu stays steady every match. To measure that spread, we need Measures of Dispersion!CENTRAL TENDENCYMean (
x̄), Median (M), Mode tell the middle — blind to spread!DISPERSIONMeasures how scattered or bunched data values are around the centre
NCERT TOOLKITThree rationalized tools: Range, Mean Deviation, Variance & S.D.
🧠 Memory Trick: A river with an average depth of 1 metre can still drown you if one end is 10 cm and the middle is a 3 m pit! Always check spread!
💡 Simple Kid-Check: Two classrooms can both score a class average of
50/100. In Class A, every student scored between 47 and 53. In Class B, half got 0 and half got 100! Averages match, yet the spread is night and day!- 1Same Mean, Different Consistency: Smaller scatter around the mean means higher consistency (Vishnu beats Arjun in reliability!).
- 2Zero Dispersion Case: If every single value equals the mean (say
50, 50, 50, 50), all measures of dispersion equal0. - 3Why Not Simple Deviations? Adding signed gaps
(xᵢ − x̄)ALWAYS cancels to0! Positive gaps abovex̄balance negative gaps belowx̄.
| Player | Mean (x̄) | Median (M) | Min – Max | Consistency |
|---|---|---|---|---|
| Batsman A (Arjun) | 53 | 53 | 0 – 117 | Wildly scattered! |
| Batsman B (Vishnu) | 53 | 53 | 46 – 60 | Tightly bunched & steady |
Watch out for 1-mark conceptual traps: (1) Saying "equal Mean and Median implies equally consistent data" — FALSE! You MUST inspect dispersion. (2) Trying to average raw deviations
Σ(xᵢ − x̄)n — that sum is ALWAYS identically 0!Topic 13.3 – 13.4.1: Range & Mean Deviation about Mean (Ungrouped Data)
Five friends in Kozhikode compare daily pocket money:
₹20, ₹24, ₹25, ₹26, ₹30 (Mean = ₹25). Simply taking Max − Min gives Range = ₹10 — fast, yet blind to the middle kids! Worse, if a cousin joins with ₹500, Range rockets to ₹480! To include EVERY value fairly, we average all unsigned distances |xᵢ − x̄| from the mean — called Mean Deviation M.D.(x̄)!RANGE
Range = L − S (Largest minus Smallest). Uses ONLY 2 extremes!MEAN DEVIATION M.D.(a)
M.D.(a) = 1n Σ |xᵢ − a| (average distance from a)ABOUT MEAN
M.D.(x̄) = 1n Σ |xᵢ − x̄| where x̄ = Σxᵢn🧠 Memory Trick: Absolute bars
| | act like a distance meter — walking 4 m left (−4) or 4 m right (+4) from home both count as +4 m!💡 Simple Kid-Check: How do you prevent positive and negative differences from cancelling out? Put every deviation inside modulus bars
|xᵢ − x̄| so every gap turns non-negative, then divide the total gap by n!- 1Find the Mean: Compute
x̄ = (x₁ + x₂ + ... + xₙ) / n. - 2Find Absolute Deviations: Compute
|xᵢ − x̄|for every observation (drop minus signs!). - 3Average Them: Divide the sum by
n:M.D.(x̄) = Σ|xᵢ − x̄|n.
Core Formulae:
Range = Max − Min | M.D.(x̄) = 1n Σ |xᵢ − x̄|
🧒 NCERT Worked Example: Find
1) Mean:
2) Raw deviations
3) Absolute deviations
4)
M.D.(x̄) for 6, 7, 10, 12, 13, 4, 8, 12 (n = 8):1) Mean:
x̄ = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9.2) Raw deviations
(xᵢ − 9): −3, −2, 1, 3, 4, −5, −1, 3 (check: sums to 0!).3) Absolute deviations
|xᵢ − 9|: 3, 2, 1, 3, 4, 5, 1, 3 → Sum = 22.4)
M.D.(x̄) = 228 = 2.75 (while Range = 13 − 4 = 9).
Two classic exam errors: (1) Forgetting the modulus bars
| | — if you sum (xᵢ − x̄) with signs, you get 0/n = 0! (2) Thinking Range measures scatter around the mean — Range completely ignores all middle points and reacts violently to a single outlier!Topic 13.4.1 – 13.4.2: Mean Deviation about Median & Grouped Frequency Tables
Imagine placing ONE school bus stop along a straight Alappuzha road where 5 students live at
km 1, 2, 3, 9, 25. Where should the stop sit so total walking distance Σ|xᵢ − a| is smallest? NOT at the Mean (km 8), which gets dragged right by the kid at km 25 — placing it at the Median (km 3) ALWAYS minimizes total distance!UNGROUPED M.D.(M)Sort data ascending first! Find Median
M, then 1n Σ|xᵢ − M|DISCRETE / GROUPEDMultiply by frequency
fᵢ: M.D.(M) = 1N Σ fᵢ|xᵢ − M| (N = Σfᵢ)CONTINUOUS MEDIAN
M = l + (N/2 − C)f × h (then use class midpoints xᵢ!)🧠 Memory Trick: Median Minimizes Mean Deviation! For ANY dataset and ANY number
a, M.D.(M) ≤ M.D.(a) — Median always beats Mean on absolute distance!💡 Simple Kid-Check: For frequency tables, a mark with frequency
fᵢ = 10 appears 10 times in class — so its gap |xᵢ − M| counts 10 times (fᵢ|xᵢ − M|), and total count is ALWAYS N = Σfᵢ!- 1Ungrouped Median
M: Sort ascending! Oddn→((n+1)/2)ᵗʰitem; Evenn→ mean of(n/2)ᵗʰand(n/2 + 1)ᵗʰitems. - 2Discrete Frequency Median: Build cumulative frequency (
c.f.) column; locateN/2(evenN: mean ofN/2andN/2 + 1; oddN:(N+1)/2). - 3Continuous Class Intervals: Find median class where
c.f. ≥ N/2, computeM = l + N/2 − Cf × h, then use class midpointsxᵢforΣ fᵢ|xᵢ − M| / N.
📊 Grouped M.D.(M) Quick Walkthrough: Classes
•
•
• Midpoints
0-10(6), 10-20(7), 20-30(15), 30-40(16), 40-50(4), 50-60(2):•
N = 50 ⇒ N/2 = 25. Cumulative frequencies: 6, 13, 28, 44, 48, 50 → Median class is 20-30.•
l = 20, C = 13, f = 15, h = 10 ⇒ M = 20 + 25 − 1315 × 10 = 28.• Midpoints
xᵢ = 5, 15, 25, 35, 45, 55 ⇒ |xᵢ − 28| = 23, 13, 3, 7, 17, 27 ⇒ Σfᵢ|xᵢ − 28| = 508 ⇒ M.D.(M) = 508/50 = 10.16!
Watch THREE classic DHSE traps: (1) Finding Median of ungrouped numbers WITHOUT sorting them in ascending order first! (2) Dividing by the number of table rows instead of total frequency
N = Σfᵢ! (3) Using discontinuous classes like 16-20, 21-25 directly — make them continuous (15.5-20.5, 20.5-25.5) so l and h = 5 are right!Topic 13.5: Variance & Standard Deviation Formulae
Absolute value bars
|x − x̄| have sharp corners that make higher algebra clumsy. How else can you wipe out minus signs? SQUARE every deviation (xᵢ − x̄)²! Averaging those squared gaps gives Variance (σ²) — though its units become squared (like cm² or runs²). Taking the positive square root brings the original unit back as Standard Deviation (σ)!VARIANCE (σ²)
σ² = 1n Σ(xᵢ − x̄)² | Grouped: 1N Σfᵢ(xᵢ − x̄)²STANDARD DEVIATION (σ)
σ = √(σ²) ≥ 0 (same units as original data!)DIRECT SHORTCUT
σ² = Σxᵢ²n − (x̄)² ("Mean of squares − Square of mean"!)🧠 Memory Trick: Remember
σ² = Mean(x²) − (Mean(x))²! For a frequency table: σ = 1N √[ N Σfᵢxᵢ² − (Σfᵢxᵢ)² ] — zero decimals during intermediate steps!💡 Simple Kid-Check: Why do we need BOTH Variance (
σ²) and Standard Deviation (σ)? Think of finding the side of a square tile: you compute the average area first (σ² = 9.25), then take √9.25 ≈ 3.04 to get back to ordinary length units!
Key Formulae (Ungrouped & Frequency):
• Ungrouped:
• Discrete / Continuous:
• Bonus NCERT Result (First
• Ungrouped:
σ² = Σ(xᵢ − x̄)²n = Σxᵢ²n − x̄²• Discrete / Continuous:
σ² = 1N² [ N Σfᵢxᵢ² − (Σfᵢxᵢ)² ] | σ = √(σ²)• Bonus NCERT Result (First
n natural numbers 1, 2, ..., n): x̄ = n + 12, σ² = n² − 112
🧮 Two Ways on
• Definition way:
• Shortcut way:
• Standard Deviation:
6, 7, 10, 12, 13, 4, 8, 12 (n = 8, x̄ = 9):• Definition way:
(xᵢ − 9)² = 9, 4, 1, 9, 16, 25, 1, 9 → Sum = 74 ⇒ σ² = 74/8 = 9.25.• Shortcut way:
Σxᵢ² = 36+49+100+144+169+16+64+144 = 722 ⇒ σ² = 722/8 − 9² = 90.25 − 81 = 9.25!• Standard Deviation:
σ = √9.25 ≈ 3.04!
Watch TWO classic blunders: (1) Stopping after computing
σ² when the question asks for Standard Deviation σ (always take √ at the end!), OR forgetting to square σ when given S.D. and asked for Variance! (2) Confusing (Σxᵢ)²/n with Σxᵢ² — squaring BEFORE summing is ALWAYS bigger than or equal to squaring AFTER averaging!Topic 13.5.4: Shortcut Step-Deviation Method for Grouped Distributions
Imagine squaring class midpoints like
95² × 14 or 107.5² × 22 in an exam hall without a calculator — one tiny multiplication slip ruins 5 marks! By shifting zero to a convenient Assumed Mean A in the middle of the table and dividing by class width h via yᵢ = (xᵢ − A)/h, numbers like 35, 45, 55, 65, 75, 85, 95 shrink into tiny integers −3, −2, −1, 0, 1, 2, 3!STEP-DEVIATION yᵢ
yᵢ = xᵢ − Ah (A = assumed mean, h = class width)SHORTCUT MEAN
x̄ = A + ΣfᵢyᵢN × hSHORTCUT σ² & σ
σ² = h²N² [N Σfᵢyᵢ² − (Σfᵢyᵢ)²] | σ = hN √[...]🧠 Memory Trick: Notice
A vanishes completely from σ² and σ! Shifting all values by A never changes spread — only scaling by h scales Variance by h² and S.D. by h!💡 Simple Kid-Check: Why does
A disappear from Variance? Because shifting every score on a number line left or right by A moves the whole crowd together without stretching or squishing how spread out they are!- 1Find Midpoints
xᵢ: Pick the central midpoint as Assumed MeanAand note common class widthh. - 2Shrink to
yᵢ: Writeyᵢ = (xᵢ − A)/h(clean integers:..., −2, −1, 0, 1, 2, ...). - 3Build Two Columns: Compute
fᵢyᵢANDfᵢyᵢ² = yᵢ × (fᵢyᵢ)cleanly! - 4Plug into Formula:
σ = hN √[ N Σfᵢyᵢ² − (Σfᵢyᵢ)² ](check inside the root is always≥ 0!).
⚡ NCERT Example 12 (7 Classes,
Suppose table totals give
• Mean:
• Variance:
• Standard Deviation:
N = 50): Classes 30-40..90-100, take A = 65, h = 10.Suppose table totals give
Σfᵢyᵢ = −15 and Σfᵢyᵢ² = 105:• Mean:
x̄ = 65 + (−1550)×10 = 65 − 3 = 62.• Variance:
σ² = 10²50² [50(105) − (−15)²] = 1002500 [5250 − 225] = 502525 = 201.• Standard Deviation:
σ = √201 = 14.18!
Top three 4–5 mark calculation traps in the shortcut table: (1) Writing
(fᵢyᵢ)² instead of fᵢyᵢ² = yᵢ · (fᵢyᵢ) — remember, only yᵢ gets squared, NOT fᵢ! (2) Multiplying S.D. σ by h² instead of h (or forgetting h outside the root!). (3) For inclusive classes like 33–36, 37–40, convert to continuous bounds 32.5–36.5 (h = 4, NOT 3!) first — though midpoints stay the same (34.5, 38.5), class width is 4!Topic 13.5 & Misc: Team Consistency, Scale-Shift Laws & Corrected Observations
Suppose every student in a Palakkad class gets
+5 grace marks. The class average climbs by +5, yet the gap between topper and last rank stays identical — Variance & S.D. do NOT change! Now suppose marks are doubled from /50 to /100 (×2): every gap doubles, so σ doubles (×2) and σ² quadruples (×4)!SHIFT (yᵢ = xᵢ + b)
ȳ = x̄ + b | σ_y = σ_x & σ_y² = σ_x² (unchanged!)SCALE (yᵢ = k xᵢ)
ȳ = k x̄ | σ_y = |k| σ_x & σ_y² = k² σ_x²!CONSISTENCY (SAME x̄)For two sets with equal Mean, smaller
σ = more consistent!🧠 Memory Trick: Spread ignores Shifts (
+b), scales by |k|! Multiplying by −3 gives σ_new = |−3|σ = 3σ (NEVER negative!), and Var_new = (−3)²σ² = 9σ²!💡 Simple Kid-Check: Picture 5 friends standing on an escalator: moving everyone up by
+5 m doesn't pull them apart or push them closer (σ unchanged!). Zooming a photograph to 3× width stretches every distance by 3 (σ becomes 3σ) and every square area by 9 (σ² becomes 9σ²)!
Complete Linear Transform Rules (
• Mean:
• S.D.:
yᵢ = k xᵢ + b):• Mean:
ȳ = k x̄ + b | Range: Range(y) = |k| Range(x) | M.D.: M.D.(y) = |k| M.D.(x)• S.D.:
σ_y = |k| σ_x | Variance: σ_y² = k² σ_x²
- 1Correcting Wrongly Read Data (NCERT Misc Ex): Recover wrong totals first:
Σxᵢ = n · x̄_wrongandΣxᵢ² = n · (σ_wrong² + x̄_wrong²). - 2Patch Totals: Subtract the wrong entry (
−w,−w²) and add the true entry (+c,+c²). - 3Recompute:
x̄_new = Σx_newnandσ_new² = Σx_new²n − x̄_new², then take√!
🔎 Quick Exam Check: For 6 numbers,
• New Mean:
• New S.D.:
x̄ = 8 and σ = 4. If each is multiplied by −3 and 5 is added (y = −3x + 5):• New Mean:
ȳ = −3(8) + 5 = −19.• New S.D.:
σ_y = |−3| × 4 = 12 (always positive!). New Variance: σ_y² = 12² = 144!
Watch TWO sneaky DHSE traps: (1) Multiplying observations by a negative factor
k = −2 and writing σ_new = −2σ — IMPOSSIBLE! Standard Deviation is ALWAYS non-negative, so always use |k|σ! (2) Multiplying Variance σ² by k instead of k²! Remember: σ scales by |k|, while Variance σ² scales by k²!Solved Previous Year Questions (Kerala DHSE & NCERT)
Q1 (Kerala HSE March • 1 Marks): If the variance of 20 observations is 5 and each observation is multiplied by 2, find the new variance.
Key (1 Mark): Scaling each observation by
New Variance =
k = 2 multiplies Variance by k² = 2² = 4.New Variance =
4 × 5 = 20.Q2 (Kerala HSE Say • 1 Marks): The standard deviation of 10 observations is 4.5. If 6 is subtracted from each observation, what is the new standard deviation?
Key (1 Mark): Standard Deviation is independent of change of origin (subtracting 6 leaves gaps unchanged).
New S.D. =
New S.D. =
4.5.Q3 (Kerala HSE March • 2 Marks): Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
n = 8, Mean x̄ = 4+7+8+9+10+12+13+178 = 808 = 10.|xᵢ − 10| = 6, 3, 2, 1, 0, 2, 3, 7 ⇒ Sum = 24.M.D.(x̄) = 248 = 3.Q4 (Kerala HSE Model • 2 Marks): Find the mean deviation about the median for: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Step 1 (Sort!):
Step 2: Median
Step 3:
3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 (n = 11, odd).Step 2: Median
M = 6ᵗʰ term = 9.Step 3:
|xᵢ − 9| = 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12 ⇒ Sum = 58.M.D.(M) = 5811 ≈ 5.27.Q5 (Kerala HSE March • 3 Marks): Find the mean and variance of the first 10 multiples of 3.
Observations:
• Mean:
• Variance of
• Scaling by
3, 6, 9, ..., 30 (i.e. 3k for k = 1..10).• Mean:
x̄ = 3 × 10 + 12 = 16.5.• Variance of
1..10 is 10² − 112 = 9912 = 8.25.• Scaling by
3: σ² = 3² × 8.25 = 9 × 8.25 = 74.25.Q6 (NCERT / DHSE March • 3 Marks): The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Let remaining two be
1)
2)
Sum of 6 squares =
3)
Solving gives remaining two observations:
x, y.1)
Σxᵢ = 8 × 9 = 72 ⇒ 60 + x + y = 72 ⇒ x + y = 12.2)
σ² = Σxᵢ²8 − 9² = 9.25 ⇒ Σxᵢ² = 8 × 90.25 = 722.Sum of 6 squares =
36+49+100+144+144+169 = 642 ⇒ x² + y² = 80.3)
(x + y)² = 144 ⇒ 2xy = 64 ⇒ (x − y)² = 80 − 64 = 16 ⇒ x − y = ±4.Solving gives remaining two observations:
4 and 8.Q7 (NCERT / Kerala HSE • 3 Marks): Mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. Find the correct mean and standard deviation.
1) Wrong totals:
2) Correct totals (replace 50 with 40):
•
•
3)
Hence
Σx = 100 × 40 = 4000.Σx² = 100(5.1² + 40²) = 100(26.01 + 1600) = 162601.2) Correct totals (replace 50 with 40):
•
Correct Σx = 4000 − 50 + 40 = 3990 ⇒ Correct Mean = 39.9.•
Correct Σx² = 162601 − 50² + 40² = 161701.3)
Correct σ² = 161701100 − (39.9)² = 1617.01 − 1592.01 = 25.Hence
Correct S.D. = √25 = 5.Q8 (Kerala HSE March • 4 Marks): Find the mean, variance and standard deviation using the shortcut method: Classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.
Midpoints
•
•
•
• Mean:
• Variance:
• S.D.:
xᵢ = 5, 15, 25, 35, 45. Choose A = 25, h = 10, so yᵢ = −2, −1, 0, 1, 2.•
N = 5 + 8 + 15 + 16 + 6 = 50.•
fᵢyᵢ = −10, −8, 0, 16, 12 ⇒ Σfᵢyᵢ = 10.•
fᵢyᵢ² = 20, 8, 0, 16, 24 ⇒ Σfᵢyᵢ² = 68.• Mean:
x̄ = 25 + (1050)×10 = 27.• Variance:
σ² = 10²50² [50(68) − 10²] = 125 [3300] = 132.• S.D.:
σ = √132 ≈ 11.49.Q9 (NCERT / DHSE Model • 4 Marks): Calculate the mean deviation about the median for: Class 0–10 (6), 10–20 (7), 20–30 (15), 30–40 (16), 40–50 (4), 50–60 (2).
N = 50 ⇒ N/2 = 25. Cumulative frequencies: 6, 13, 28, 44, 48, 50.Median class is
20–30 (l = 20, C = 13, f = 15, h = 10).Median M = 20 + 25 − 1315 × 10 = 28.Midpoints
xᵢ = 5, 15, 25, 35, 45, 55 ⇒ |xᵢ − 28| = 23, 13, 3, 7, 17, 27.Σfᵢ|xᵢ − 28| = 6(23) + 7(13) + 15(3) + 16(7) + 4(17) + 2(27) = 508.M.D.(M) = 50850 = 10.16.