📚 Chapters

📊 Probability and Statistics

Unit 1 — Basic Statistics and Probability

📊 Chapter 1.1 — Central Tendency & Dispersion
📊 CHAPTER 1.1 — CENTRAL TENDENCY & DISPERSION MIND MAP
Mean → Direct(Σfx/Σf), Short-Cut(A+Σfd/Σf), Step-Deviation(A+(Σfd'/Σf)×h) — all give SAME answer
Median → L+[(N/2−cf)/f]×h; find Median Class where cumulative freq first reaches N/2
Mode → L+[(f1−f0)/(2f1−f0−f2)]×h; Empirical: Mode=3Median−2Mean
GM/HM → AM ≥ GM ≥ HM always (equal only if all values same)
Range/QD → QD=(Q3−Q1)/2; Q1,Q3 use same method as Median but N/4, 3N/4
Mean Deviation → Σf|x−x̄|/Σf — always use ABSOLUTE values
SD/CV → SD=√(Σ(x−x̄)²/n); CV=(SD/Mean)×100 — use CV to compare 2 datasets

1. Arithmetic Mean — Direct, Short-Cut & Step-Deviation Methods

📖 Arithmetic Mean (AM): The sum of all values divided by the number of values — the most common "average."
A. Direct Method — Individual (Ungrouped) Data:
Mean (x̄) = Σx / n
💡 Worked Example: Find the mean of 20, 22, 25, 28, 30.
Σx = 20+22+25+28+30 = 125, n = 5
Mean = 125/5 = 25
B. Direct Method — Discrete Frequency Data:
Mean (x̄) = Σfx / Σf
💡 Worked Example: Numbers: 8,10,15,20 with Frequency: 5,8,8,4

fx: 8×5=40,  10×8=80,  15×8=120,  20×4=80
Σfx = 40+80+120+80 = 320,  Σf = 5+8+8+4 = 25
Mean = 320/25 = 12.8
C. Continuous (Grouped) Data — All 3 Methods, Same Data:
Data: Class 0-10, 10-20, 20-30, 30-40, 40-50 with Frequency 7, 8, 20, 10, 5
Midpoints (x): 5, 15, 25, 35, 45
Method 1 — Direct: Mean = Σfx / Σf
Method 2 — Short-Cut: Mean = A + (Σfd / Σf),   where d = x − A
Method 3 — Step-Deviation: Mean = A + (Σfd′ / Σf) × h,   where d′ = (x−A)/h
💡 Method 1 (Direct):
fx: 7×5=35, 8×15=120, 20×25=500, 10×35=350, 5×45=225
Σfx = 35+120+500+350+225 = 1230, Σf = 50
Mean = 1230/50 = 24.6
💡 Method 2 (Short-Cut, taking A=25):
d = x−25: −20, −10, 0, 10, 20
fd: 7×(−20)=−140, 8×(−10)=−80, 20×0=0, 10×10=100, 5×20=100
Σfd = −140−80+0+100+100 = −20
Mean = 25 + (−20/50) = 25 − 0.4 = 24.6 ✓ (matches Method 1)
💡 Method 3 (Step-Deviation, A=25, h=10):
d′ = d/10: −2, −1, 0, 1, 2
fd′: 7×(−2)=−14, 8×(−1)=−8, 20×0=0, 10×1=10, 5×2=10
Σfd′ = −14−8+0+10+10 = −2
Mean = 25 + (−2/50)×10 = 25 − 0.4 = 24.6 ✓ (all 3 methods agree!)
💡 Exam Tip: Short-Cut and Step-Deviation ALWAYS give the SAME final answer as Direct method — they're just calculation shortcuts to avoid large numbers. Step-Deviation is fastest when class width (h) is constant. Choose ANY assumed mean (A) — usually the midpoint of the class with highest frequency, for smaller numbers.

2. Median

📖 Median: The MIDDLE value when data is arranged in ascending/descending order — divides the data into two equal halves.
A. Individual (Ungrouped) Data:
If n is ODD: Median = value at position (n+1)/2
If n is EVEN: Median = average of values at positions n/2 and (n/2)+1
💡 Worked Example: Find the median of 6, 8, 9, 10, 11, 12, 13.
Already sorted, n=7 (odd)
Position = (7+1)/2 = 4th value
Median = 10 (the 4th number in the sorted list)
B. Continuous (Grouped) Data:
Median = L + [(N/2 − cf) / f] × h
L = lower boundary of median class, N = Σf (total frequency), cf = cumulative frequency BEFORE median class, f = frequency of median class, h = class width
💡 Worked Example (same data as Mean section):
Class 0-10,10-20,20-30,30-40,40-50, Frequency 7,8,20,10,5

Step 1 — Cumulative Frequency: 7, 15, 35, 45, 50
Step 2 — N=50, N/2=25
Step 3 — Find Median Class: First cf ≥ 25 is 35, so Median Class = 20-30
Step 4 — L=20, cf(before)=15, f=20, h=10
Median = 20 + [(25−15)/20] × 10 = 20 + (10/20)×10 = 20 + 5 = 25
💡 Exam Tip: Always find the "Median Class" first by locating where the cumulative frequency FIRST reaches or exceeds N/2 — this class's boundaries, frequency, AND the cumulative frequency of the class BEFORE it (not including it) are all needed in the formula.

3. Mode & Empirical Relation Between Mean, Median, Mode

📖 Mode: The value that occurs MOST FREQUENTLY in a dataset.
A. Individual (Ungrouped) Data:
💡 Worked Example: Find the mode of 0,1,6,7,2,3,7,6,6,2,6,0,5,6,0

Count each value: 0→3 times, 1→1 time, 2→2 times, 3→1 time, 5→1 time, 6→5 times, 7→2 times
Mode = 6 (appears most often — 5 times)
B. Continuous (Grouped) Data:
Mode = L + [(f₁−f₀) / (2f₁−f₀−f₂)] × h
L = lower boundary of modal class, f₁ = frequency of modal class, f₀ = frequency of class BEFORE modal class, f₂ = frequency of class AFTER modal class, h = class width
💡 Worked Example (same data as Mean/Median sections):
Class 0-10,10-20,20-30,30-40,40-50, Frequency 7,8,20,10,5

Modal Class = class with HIGHEST frequency = 20-30 (frequency 20)
L=20, f₁=20, f₀=8, f₂=10, h=10
Mode = 20 + [(20−8)/(2×20−8−10)] × 10 = 20 + [12/22] × 10 = 20 + 5.45 = 25.45
C. Empirical Relationship (Quick Estimate):
Mode = 3 × Median − 2 × Mean
(used as an approximation when direct calculation isn't possible/needed — assumes a moderately skewed distribution)
💡 Verify with our data: Mean=24.6, Median=25 (from previous sections)
Mode ≈ 3×25 − 2×24.6 = 75 − 49.2 = 25.8
(Close to our directly-calculated 25.45 — small difference is normal since the empirical formula is an approximation)
MeasureAdvantageDisadvantage
MeanUses ALL data values, best for further mathHighly affected by extreme outliers
MedianNot affected by extreme outliersDoesn't use all values directly
ModeEasy to understand, works for categorical data tooMay not exist, or multiple modes possible
💡 Exam Tip: "Modal Class" for the formula = class with the SINGLE highest frequency (not cumulative!) — a common mistake is confusing this with the Median Class (which uses cumulative frequency).

4. Geometric Mean & Harmonic Mean

A. Geometric Mean (GM):
📖 Geometric Mean: The n-th root of the PRODUCT of n values — used for growth rates, ratios, and data that multiplies rather than adds.
GM = ⁿ√(x₁ × x₂ × ... × xₙ)
💡 Worked Example: Find the GM of 4, 8, 16.
GM = ³√(4 × 8 × 16) = ³√512
Since 8³ = 8×8×8 = 512:
GM = 8
B. Harmonic Mean (HM):
📖 Harmonic Mean: The reciprocal of the average of reciprocals — used for rates (speed, time-based averages).
HM = n / Σ(1/x)
💡 Worked Example: Calculate the HM of 4, 8, 16.
1/4 + 1/8 + 1/16 = 4/16 + 2/16 + 1/16 = 7/16
HM = 3 / (7/16) = 3 × 16/7 = 48/7
HM ≈ 6.857
RelationshipAlways True (for positive numbers)
AM ≥ GM ≥ HMEquality holds ONLY when all values are identical
💡 Exam Tip: Verify with our example: AM of (4,8,16) = 28/3 ≈ 9.33, GM = 8, HM ≈ 6.857 → 9.33 ≥ 8 ≥ 6.857 ✓ This ordering ALWAYS holds — a quick way to sanity-check your calculations.

5. Range & Quartile Deviation

A. Range:
📖 Range: The simplest measure of dispersion — the difference between the LARGEST and SMALLEST values.
Range = Maximum value − Minimum value
💡 Worked Example: Marks obtained by six students: 24, 12, 16, 11, 40, 42.
Maximum = 42, Minimum = 11
Range = 42 − 11 = 31
B. Quartile Deviation (QD):
📖 Quartiles: Q₁ (Lower Quartile) is the value below which 25% of data lies; Q₃ (Upper Quartile) is the value below which 75% of data lies.

📖 Quartile Deviation: Half the distance between Q₁ and Q₃ — a measure of dispersion unaffected by extreme values.
QD = (Q₃ − Q₁) / 2
For individual data: Q₁ position = (n+1)/4,   Q₃ position = 3(n+1)/4
💡 Worked Example: Calculate Range and QD of: 20, 25, 29, 30, 35, 39, 41, 48, 51, 60, 70 (n=11, already sorted)

Range = 70 − 20 = 50

Q₁ position = (11+1)/4 = 3rd value = 29
Q₃ position = 3×(11+1)/4 = 9th value = 51
QD = (51−29)/2 = 22/2 = 11
For Grouped Data:
Q₁ = L + [(N/4 − cf)/f] × h    (same logic as Median, but using N/4 instead of N/2)
Q₃ = L + [(3N/4 − cf)/f] × h
💡 Exam Tip: Finding Q₁ and Q₃ for grouped data uses the EXACT SAME method as finding the Median — just replace N/2 with N/4 (for Q₁) or 3N/4 (for Q₃) when locating the correct class and applying the formula.

6. Mean Deviation

📖 Mean Deviation (MD): The average of the ABSOLUTE differences between each value and the mean (or median) — measures how spread out the data is around the center.
MD (about Mean) = Σf|x − x̄| / Σf
💡 Worked Example: Find MD from mean for: x: 10,11,12,13,14   f: 3,12,18,12,3

Step 1 — Find Mean:
fx: 10×3=30, 11×12=132, 12×18=216, 13×12=156, 14×3=42
Σfx = 30+132+216+156+42 = 576,   N = 3+12+18+12+3 = 48
Mean = 576/48 = 12

Step 2 — Find |x − Mean|:
|10−12|=2,  |11−12|=1,  |12−12|=0,  |13−12|=1,  |14−12|=2

Step 3 — Multiply by frequency:
f|x−x̄|: 3×2=6, 12×1=12, 18×0=0, 12×1=12, 3×2=6
Σf|x−x̄| = 6+12+0+12+6 = 36

Step 4 — Divide by N:
MD = 36/48 = 0.75
💡 Exam Tip: Mean Deviation ALWAYS uses ABSOLUTE values |x−x̄| — never leave negative signs in, or your deviations will cancel out to zero (which is why we use absolute value instead of just Σ(x−x̄), which is always 0 by definition of the mean).

7. Variance, Standard Deviation & Coefficient of Variation

📖 Variance: The average of the SQUARED deviations from the mean.
📖 Standard Deviation (SD): The square root of variance — the MOST widely used measure of dispersion, in the SAME units as the original data.
Variance (σ²) = Σ(x−x̄)² / n
SD (σ) = √Variance
💡 Worked Example: Find SD of: 8, 10, 12, 14, 16, 18, 20, 22, 24, 26

Step 1 — Mean:
Σx = 8+10+12+14+16+18+20+22+24+26 = 170, n=10
Mean = 170/10 = 17

Step 2 — Deviations (x−x̄):
−9, −7, −5, −3, −1, 1, 3, 5, 7, 9

Step 3 — Square each deviation:
81, 49, 25, 9, 1, 1, 9, 25, 49, 81

Step 4 — Sum the squares:
Σ(x−x̄)² = 81+49+25+9+1+1+9+25+49+81 = 330

Step 5 — Variance:
Variance = 330/10 = 33

Step 6 — SD:
SD = √33 ≈ 5.745
Coefficient of Variation (CV):
📖 CV: Expresses SD as a PERCENTAGE of the mean — allows comparing the relative variability of TWO DIFFERENT datasets (even with different units/scales).
CV = (SD / Mean) × 100
💡 Using our example above:
CV = (5.745/17) × 100 = 33.79%
💡 Exam Tip: When comparing "which dataset is MORE consistent/less variable" between two different datasets, ALWAYS use CV (not raw SD) — the dataset with the SMALLER CV is more consistent. Raw SD alone can be misleading if the datasets have very different means or units.
📈 Chapter 1.2 — Moments, Skewness & Kurtosis
📈 CHAPTER 1.2 — MOMENTS, SKEWNESS & KURTOSIS MIND MAP
Moments → mr=Σ(x−A)^r/n (about A) vs μr=Σ(x−x̄)^r/n (about mean); μ1 ALWAYS = 0
Conversion → Mean=A+m1; μ2=m2−m1²; μ3=m3−3m1m2+2m1³
Skewness → Positive: Mean>Median>Mode | Negative: Mean<Median<Mode
Karl Pearson's → SKp=(Mean−Mode)/SD or 3(Mean−Median)/SD
Bowley's → SKb=(Q3+Q1−2Median)/(Q3−Q1); always between −1 and +1
Kurtosis → β2=μ4/μ2²; β2=3(Meso), β2>3(Lepto-peaked), β2<3(Platy-flat)

1. Moments & Central Moments

📖 Moments: Statistical measures that describe the SHAPE of a distribution — the r-th moment captures a specific characteristic (1st=location, 2nd=spread, 3rd=skewness, 4th=peakedness).

📖 Raw Moments (about a value A): mᵣ = Σ(x−A)ʳ / n
📖 Central Moments (about the MEAN): μᵣ = Σ(x−x̄)ʳ / n
💡 Worked Example — Find first 4 Central Moments for: 2, 4, 6, 8

Step 1 — Mean: (2+4+6+8)/4 = 20/4 = 5

Step 2 — Deviations (x−mean): −3, −1, 1, 3

μ₁ = Σ(x−x̄)/n = (−3−1+1+3)/4 = 0/4 = 0 (always zero!)
μ₂ = Σ(x−x̄)²/n = (9+1+1+9)/4 = 20/4 = 5 (= Variance)
μ₃ = Σ(x−x̄)³/n = (−27−1+1+27)/4 = 0/4 = 0
μ₄ = Σ(x−x̄)⁴/n = (81+1+1+81)/4 = 164/4 = 41
MomentMeaning
μ₁ (First central moment)ALWAYS equals 0 (by definition of the mean)
μ₂ (Second central moment)Equals the VARIANCE
μ₃ (Third central moment)Used to measure SKEWNESS (asymmetry)
μ₄ (Fourth central moment)Used to measure KURTOSIS (peakedness)
💡 Exam Tip: μ₁ = 0 ALWAYS — if you calculate it and get a non-zero answer, you've made an arithmetic error somewhere. This is a built-in self-check for every moments problem.

2. Relationship Between Moments and Central Moments

📖 Why We Need This: Sometimes moments are calculated about a CONVENIENT arbitrary point A (not the mean) to keep numbers small — these formulas convert those "raw moments about A" into the actual "central moments about the mean."
Conversion Formulas (m = raw moment about A, μ = central moment):
Mean = A + m₁
μ₂ = m₂ − m₁²
μ₃ = m₃ − 3m₁m₂ + 2m₁³
μ₄ = m₄ − 4m₁m₃ + 6m₁²m₂ − 3m₁⁴
💡 Worked Example: The first two moments of a distribution about the value 5 are 2 and 20. Find the Mean and Variance.

Given: A=5, m₁=2, m₂=20

Mean = A + m₁ = 5 + 2 = 7
Variance = μ₂ = m₂ − m₁² = 20 − (2)² = 20 − 4 = 16
💡 Exam Tip: This is one of the MOST common question types — "moments about value A are given, find mean/variance." Always remember: Mean = A + m₁ FIRST (this gets you the mean), THEN apply the μ₂ formula for variance. Don't skip finding the mean even if only variance is asked, since m₁ is needed in the μ₂ formula anyway.

3. Concept of Skewness & Karl Pearson's Coefficient

📖 Skewness: A measure of the ASYMMETRY of a distribution — how much it "leans" to one side compared to a perfectly symmetric (bell-shaped) distribution.
Negative Skew (left tail) Symmetric Positive Skew (right tail) ___ ___ ___ / \___ / \ ___/ \ / \ / \ / \ Mode Median Mean Mean=Median=Mode Mean Median Mode
Fig: Negative, Symmetric, and Positive Skewness
TypeTail DirectionRelationship
Positive SkewLonger tail on the RIGHTMean > Median > Mode
Symmetric (no skew)No tail biasMean = Median = Mode
Negative SkewLonger tail on the LEFTMean < Median < Mode
Karl Pearson's Coefficient of Skewness:
SKp = (Mean − Mode) / SD

Alternative form (when Mode is unstable/unclear):
SKp = 3(Mean − Median) / SD
💡 Worked Example (Reverse Calculation): Given: Pearson's measure of skewness = 0.4, Mean = 30, SD = 8. Find Median and Mode.

Finding Mode (using SKp = (Mean−Mode)/SD):
0.4 = (30 − Mode)/8
0.4 × 8 = 30 − Mode
3.2 = 30 − Mode
Mode = 30 − 3.2 = 26.8

Finding Median (using SKp = 3(Mean−Median)/SD):
0.4 = 3(30 − Median)/8
3.2 = 3(30 − Median)
1.0667 = 30 − Median
Median = 30 − 1.0667 ≈ 28.93
💡 Exam Tip: SKp is usually between −3 and +3 (though in theory unbounded) — a value of 0 means perfectly symmetric, positive means right-skewed, negative means left-skewed. Always double-check your Mean vs Mode/Median relationship matches the sign of your answer (e.g., positive SKp should give Mean > Mode).

4. Bowley's Coefficient of Skewness

📖 Bowley's Coefficient: A skewness measure based on QUARTILES (Q₁, Median, Q₃) rather than mean/mode — useful when the mean/mode are unreliable (e.g. with extreme outliers) or when the distribution is open-ended.
SKb = (Q₃ + Q₁ − 2×Median) / (Q₃ − Q₁)
💡 Worked Example: A distribution has Q₁=44.1, Q₃=56.6. If skewness is −0.8, find the Median.

SKb = (Q₃+Q₁−2×Median)/(Q₃−Q₁)
−0.8 = (56.6+44.1−2×Median)/(56.6−44.1)
−0.8 = (100.7 − 2×Median)/12.5
−0.8 × 12.5 = 100.7 − 2×Median
−10 = 100.7 − 2×Median
2×Median = 100.7 + 10 = 110.7
Median = 110.7/2 = 55.35
FeatureKarl Pearson'sBowley's
Based onMean, Mode/Median, SDQ₁, Median, Q₃ (quartiles only)
Affected by outliers?Yes (mean is sensitive)No (quartiles are robust)
RangeTheoretically −3 to +3Always −1 to +1
💡 Exam Tip: Bowley's coefficient ALWAYS lies between −1 and +1 — if your calculated answer falls outside this range, you've made an error. This built-in bound is a handy self-check, unlike Karl Pearson's which has no such fixed limit.

5. Concept of Kurtosis — Mesokurtic, Leptokurtic & Platykurtic

📖 Kurtosis: A measure of the "PEAKEDNESS" or "TAILEDNESS" of a distribution — how sharply it peaks and how heavy/light its tails are, compared to a Normal distribution.
Coefficient of Kurtosis (β₂) = μ₄ / μ₂²
Excess Kurtosis (γ₂) = β₂ − 3
Leptokurtic (sharp peak) Mesokurtic (normal) Platykurtic (flat) ▲ ▲ ▲ /│\ / \ / \ / │ \ / \ / \ __/ │ \__ _/ \_ __/ \__ (heavy tails) (β2 = 3) (light tails, β2 < 3) (β2 > 3)
Fig: Three Types of Kurtosis
Typeβ₂ valueShape
Mesokurticβ₂ = 3Normal/moderate peak (like the Normal distribution)
Leptokurticβ₂ > 3Sharp, tall peak with heavy/fat tails
Platykurticβ₂ < 3Flat, short peak with light/thin tails
💡 Worked Example: The first four moments of a distribution about the mean are 0, 20, 40, 50 (μ₁=0, μ₂=20, μ₃=40, μ₄=50). Is the distribution platykurtic?

β₂ = μ₄/μ₂² = 50/(20)² = 50/400 = 0.125

Since β₂ = 0.125 < 3, YES, the distribution IS platykurtic (flatter than normal).
💡 Exam Tip: Memory trick — "Lepto" sounds like "leap" (jumps UP sharply = peaked); "Platy" sounds like "flat" (flat top). Meso = middle/normal. Always compute β₂ = μ₄/μ₂² first, then compare to 3 to classify.
🎲 Chapter 1.3 — Probability Distributions
🎲 CHAPTER 1.3 — PROBABILITY DISTRIBUTIONS MIND MAP
Bayes' Theorem → P(Ai|B) = [P(Ai)×P(B|Ai)] / Σ[P(Aj)×P(B|Aj)] — reverses conditional probability
PMF/PDF/CDF → PMF=discrete P(X=x), PDF=continuous density, CDF=P(X≤x); ΣP(x)=1 always
Binomial → P(X=r)=nCr·p^r·q^(n-r); Mean=np, Variance=npq; "at least 1"=1−q^n
Poisson → P(X=r)=e^(−λ)λ^r/r!; Mean=Variance=λ (UNIQUE); recurrence P(r+1)=[λ/(r+1)]P(r)
Normal → Z=(X−μ)/σ; use Z-table for areas; total area=1, each half=0.5

1. Conditional Probability & Bayes' Theorem

📖 Conditional Probability: The probability of event A occurring, GIVEN that event B has already occurred.
P(A|B) = P(A∩B) / P(B)  (read: "probability of A GIVEN B")
📖 Bayes' Theorem: Lets us REVERSE a conditional probability — if we know P(B|A), we can find P(A|B). Used to update our belief about a CAUSE after observing an EFFECT.
P(Aᵢ|B) = [P(Aᵢ) × P(B|Aᵢ)] / Σⱼ[P(Aⱼ) × P(B|Aⱼ)]
💡 Worked Example: Bag X has 2 white + 3 red balls. Bag Y has 4 white + 5 red balls. A ball is drawn at random from ONE of the bags and found to be RED. Find the probability it was drawn from Bag Y.

Step 1 — Prior probabilities (equally likely to pick either bag):
P(X) = 1/2,   P(Y) = 1/2

Step 2 — Likelihood (probability of RED given each bag):
P(Red|X) = 3/5  (3 red out of 5 total in Bag X)
P(Red|Y) = 5/9  (5 red out of 9 total in Bag Y)

Step 3 — Apply Bayes' Theorem:
P(Y|Red) = [P(Y)×P(Red|Y)] / [P(X)×P(Red|X) + P(Y)×P(Red|Y)]

Numerator = (1/2)×(5/9) = 5/18 ≈ 0.2778
P(X)×P(Red|X) = (1/2)×(3/5) = 3/10 = 0.3
Denominator = 0.3 + 0.2778 = 0.5778

P(Y|Red) = 0.2778/0.5778 = 25/52 ≈ 0.481
💡 Exam Tip: Bayes' theorem questions ALWAYS follow this pattern: (1) Identify the "causes" (which bag/machine/source) and their PRIOR probabilities, (2) Find the LIKELIHOOD of the observed event under each cause, (3) Plug into the formula — numerator is just ONE term, denominator is the SUM of all such terms.

2. Random Variables — PMF, PDF & CDF

📖 Random Variable (X): A variable whose value is a numerical outcome of a random phenomenon.

📖 Discrete Random Variable: Takes only SPECIFIC, countable values (e.g. number of heads in 3 coin tosses: 0,1,2,3).

📖 Continuous Random Variable: Can take ANY value within a range (e.g. height, weight, time).
FunctionApplies toMeaning
PMF — Probability Mass FunctionDiscrete RVP(X=x) — probability of EXACTLY value x
PDF — Probability Density FunctionContinuous RVf(x) — describes relative likelihood; P(a≤X≤b) = ∫f(x)dx from a to b
CDF — Cumulative Distribution FunctionBothF(x) = P(X≤x) — probability of value ≤ x
Worked Example — Finding k, Mean, Variance from a PMF:
X-3-2-10123
P(X)k0.1k0.22k0.42k
💡 Step 1 — Find k (using ΣP(X) = 1, the fundamental rule of any probability distribution):
k + 0.1 + k + 0.2 + 2k + 0.4 + 2k = 1
6k + 0.7 = 1
6k = 0.3
k = 0.05

Step 2 — Substitute k to get all probabilities:
P(X): 0.05, 0.1, 0.05, 0.2, 0.1, 0.4, 0.1

Step 3 — Mean = ΣxP(x):
(−3×0.05)+(−2×0.1)+(−1×0.05)+(0×0.2)+(1×0.1)+(2×0.4)+(3×0.1)
= −0.15−0.2−0.05+0+0.1+0.8+0.3 = 0.8
Key Rules for ANY valid probability distribution:
1. Σ P(x) = 1   (Discrete)   OR   ∫f(x)dx = 1 over all x   (Continuous)
2. Every P(x) ≥ 0   (probabilities can never be negative)
3. Mean = ΣxP(x)   (Discrete)   OR   ∫xf(x)dx   (Continuous)
4. Variance = Σx²P(x) − (Mean)²   (Discrete)
💡 Exam Tip: "Σ P(x) = 1" is THE most important rule — almost every "find k" question uses this single equation. Once k is found, substitute it back everywhere before calculating mean/variance.

3. Binomial Distribution

📖 Binomial Distribution: Models the number of SUCCESSES in a fixed number (n) of INDEPENDENT trials, each with the SAME probability of success (p) — e.g. number of heads in 10 coin tosses.
P(X=r) = ⁿCᵣ × pʳ × qⁿ⁻ʳ
n = number of trials, r = number of successes, p = probability of success, q = 1−p (probability of failure)

Mean = np    Variance = npq
💡 Worked Example (Reverse Calculation): A shooter hits a target with probability 3/4. Find the MINIMUM number of shots needed so that the probability of hitting the target AT LEAST ONCE is more than 0.99.

p = 3/4 (hit), q = 1/4 (miss)

P(at least 1 hit in n shots) = 1 − P(no hits at all) = 1 − qⁿ
We need: 1 − qⁿ > 0.99
qⁿ < 0.01
(0.25)ⁿ < 0.01

Testing values of n:
n=3: (0.25)³ = 0.015625  → NOT less than 0.01 ✗
n=4: (0.25)⁴ = 0.00390625  → IS less than 0.01 ✓

Minimum n = 4 shots
Basic PMF Application:
💡 Find the Mean of Binomial Distribution B(4, 1/3):
n=4, p=1/3
Mean = np = 4 × 1/3 = 4/3 ≈ 1.33
💡 Exam Tip: "At least 1" success problems ALWAYS use the trick: P(at least 1) = 1 − P(none). Computing "P(none)" (all failures, q^n) is far easier than adding up P(1)+P(2)+...+P(n) directly.

4. Poisson Distribution

📖 Poisson Distribution: Models the number of RARE EVENTS occurring in a FIXED interval of time/space — e.g. number of calls received per hour, number of typos per page. Used when n is LARGE and p is SMALL (rare events).
P(X=r) = (e⁻λ × λʳ) / r!
λ (lambda) = mean number of occurrences

Mean = λ    Variance = λ (Mean = Variance is a UNIQUE property of Poisson!)
Recurrence Relation (Faster Calculation Method):
P(X=r+1) = [λ/(r+1)] × P(X=r)
(once P(0) is known, each next probability is found from the PREVIOUS one — no need to recompute factorials each time)
💡 Worked Example: Using the recurrence relation, find P(X=0) through P(X=5) if Mean of the distribution is 3 (λ=3).

P(0) = e⁻³ ≈ 0.0498

Using recurrence P(r+1) = (λ/(r+1))×P(r):
P(1) = (3/1) × 0.0498 = 0.1494
P(2) = (3/2) × 0.1494 = 0.2241
P(3) = (3/3) × 0.2241 = 0.2241
P(4) = (3/4) × 0.2241 = 0.1681
P(5) = (3/5) × 0.1681 = 0.1008
💡 Exam Tip: The single most-tested UNIQUE fact about Poisson: Mean = Variance = λ. If a question gives you BOTH mean and variance and they're different, it's NOT a valid Poisson scenario — this is a great sanity check for word problems.

5. Normal Distribution

📖 Normal Distribution: The famous "bell-shaped curve" — symmetric around the mean, describes MANY natural phenomena (heights, weights, test scores). Fully described by just 2 parameters: Mean (μ) and Standard Deviation (σ).
Standardization (Z-score):
Z = (X − μ) / σ

Converts ANY normal distribution into the STANDARD Normal Distribution (mean=0, SD=1), so we can use standard Z-tables to find probabilities/areas.
Z valueArea from 0 to Z (approx.)
Z = 1.000.3413
Z = 1.330.4082
Z = 1.6450.4500
Z = 1.960.4750
Z = 2.000.4772
💡 Worked Example: Masses of 300 students are normally distributed with Mean=68 kg, SD=3 kg. How many students have mass GREATER than 72 kg?

Step 1 — Standardize:
Z = (72−68)/3 = 4/3 ≈ 1.33

Step 2 — Find area from Z-table:
Area from 0 to Z=1.33 is 0.4082 (from standard table)

Step 3 — Find P(X>72):
Since total area on ONE side of the mean = 0.5:
P(Z>1.33) = 0.5 − 0.4082 = 0.0918

Step 4 — Convert proportion to actual student count:
Number of students = 0.0918 × 300 ≈ 27.5 ≈ 28 students
┌─── 68% ───┐ ┌──────── 95% ────────┐ ┌────────── 99.7% ──────────┐ ───┼────┼────┼────┼────┼────┼────┼─── -3σ -2σ -1σ μ +1σ +2σ +3σ
Fig: Empirical Rule (68-95-99.7 Rule)
💡 Exam Tip: Always draw the bell curve and shade the region asked for — this instantly tells you whether to ADD or SUBTRACT areas from 0.5. Remember: total area under the ENTIRE curve = 1, and each HALF (left or right of mean) = 0.5.
⚡ Ready for Exam? Sab padh liya? Ab Quick Revision karo — formulas aur worked examples ek jagah! Quick Revision Karo →
⚡ Quick Revision — Last Minute Exam Prep!
📌 How to Use: Read this 5-10 minutes before exam. Contains all important points in condensed form. Focus on tables, comparisons, and key formulas!

📊 Chapter 1.1 — Central Tendency & Dispersion

📖 Mean Formulas:
Direct: Σfx/Σf | Short-Cut: A+(Σfd/Σf) | Step-Dev: A+(Σfd'/Σf)×h — all give SAME answer
🔑 Median & Mode (Grouped):
Median = L + [(N/2−cf)/f]×h
Mode = L + [(f₁−f₀)/(2f₁−f₀−f₂)]×h
Empirical: Mode = 3Median − 2Mean
GM & HM:
GM = ⁿ√(x₁×x₂×...×xₙ) | HM = n/Σ(1/x)
Always: AM ≥ GM ≥ HM
✅ Dispersion Quick Recall:
Range = Max−Min
QD = (Q₃−Q₁)/2, Q₁ pos=(n+1)/4, Q₃ pos=3(n+1)/4
MD = Σf|x−x̄|/Σf (always absolute value!)
SD = √[Σ(x−x̄)²/n], CV = (SD/Mean)×100
TopicKey FormulaTrick
Mean (3 methods)All equal same answerStep-Dev fastest for constant class width
Median ClassFirst cf ≥ N/2Different from Modal Class!
Modal ClassHighest frequency (not cumulative)Don't confuse with Median Class
CVUse to compare 2 datasetsSmaller CV = more consistent

📈 Chapter 1.2 — Moments, Skewness & Kurtosis

📖 Moments:
mᵣ=Σ(x−A)ʳ/n (about A) | μᵣ=Σ(x−x̄)ʳ/n (about mean) | μ₁ ALWAYS = 0
🔑 Conversion (A→Mean):
Mean = A + m₁
μ₂ = m₂ − m₁²
μ₃ = m₃ − 3m₁m₂ + 2m₁³
SKEWNESS:
Karl Pearson: SKp = (Mean−Mode)/SD = 3(Mean−Median)/SD
Bowley: SKb = (Q₃+Q₁−2Median)/(Q₃−Q₁), range −1 to +1

Positive: Mean>Median>Mode | Negative: Mean<Median<Mode
✅ KURTOSIS:
β₂ = μ₄/μ₂²
β₂=3 → Mesokurtic (normal)
β₂>3 → Leptokurtic (peaked)
β₂<3 → Platykurtic (flat)
TopicKey FormulaTrick
Momentsμ₁=0 alwaysSelf-check for arithmetic errors
Karl Pearson's(Mean−Mode)/SDNo fixed range (theoretically ±3)
Bowley'sQuartile-basedALWAYS between −1 and +1
Kurtosisβ₂=μ₄/μ₂²Lepto=peaked, Platy=flat, Meso=normal

🎲 Chapter 1.3 — Probability Distributions

📖 Bayes' Theorem:
P(Aᵢ|B) = [P(Aᵢ)×P(B|Aᵢ)] / Σ[P(Aⱼ)×P(B|Aⱼ)]
Steps: Prior probabilities → Likelihoods → Plug into formula
🔑 PMF/PDF/CDF:
ΣP(x)=1 always (most "find k" questions use this)
Mean=ΣxP(x), Variance=Σx²P(x)−(Mean)²
3 DISTRIBUTIONS:
Binomial: P(X=r)=ⁿCᵣpʳqⁿ⁻ʳ; Mean=np, Var=npq
Poisson: P(X=r)=e⁻λλʳ/r!; Mean=Var=λ (UNIQUE!)
Normal: Z=(X−μ)/σ; use Z-table
✅ Quick Tricks:
"At least 1" (Binomial) = 1 − q^n
Poisson recurrence: P(r+1) = [λ/(r+1)]×P(r)
Normal: total area=1, each half=0.5
TopicKey FormulaTrick
BayesReverses conditional probabilityDenominator = sum of ALL causes
BinomialMean=np, Var=npq"at least 1" trick saves time
PoissonMean=Variance=λUnique identifying property
NormalZ=(X−μ)/σDraw bell curve, shade region asked

⚠️ Common Exam Mistakes

❌ Confusing Median Class (uses cumulative frequency ≥ N/2) with Modal Class (uses highest single frequency)
❌ Forgetting Mean Deviation ALWAYS uses absolute values |x−x̄| — without it, deviations cancel to zero
❌ Mixing up moments "about A" (mᵣ) with "about mean" (μᵣ) — always convert using Mean=A+m₁ first
❌ Forgetting μ₁ is ALWAYS 0 — a non-zero answer means an arithmetic mistake somewhere
❌ Using Karl Pearson's formula with Mode when Mode is unclear — switch to the Mean-Median version instead
❌ Forgetting Bowley's coefficient must ALWAYS fall between −1 and +1 — a good self-check
❌ Confusing Binomial (fixed n trials, np=mean) with Poisson (rare events, mean=variance=λ)
❌ Forgetting "at least 1" probability = 1 − P(none), not adding up P(1)+P(2)+...
❌ Not standardizing to Z-score before using the Normal distribution table

✅ Pre-Exam Checklist

☑ Mean — Direct, Short-Cut, Step-Deviation methods (all 3!)
☑ Median — individual + grouped data formula
☑ Mode — individual + grouped data + empirical relation
☑ GM, HM formulas + AM≥GM≥HM ordering
☑ Range, Quartile Deviation (Q₁, Q₃ calculation)
☑ Mean Deviation — full worked table method
☑ Variance, SD, CV — full worked calculation
☑ Moments about A vs about Mean + conversion formulas
☑ Karl Pearson's Coefficient of Skewness (both forms)
☑ Bowley's Coefficient of Skewness
☑ Kurtosis — β₂ formula + Meso/Lepto/Platy classification
☑ Bayes' Theorem — full worked example
☑ PMF/PDF/CDF — finding k, mean, variance
☑ Binomial Distribution — formula + mean/variance + numericals
☑ Poisson Distribution — formula + recurrence relation
☑ Normal Distribution — Z-score standardization + table usage

🎯 Exam Strategy

General Approach for ALL Numericals:
• Write the FORMULA first, before substituting any numbers.
• Show EVERY calculation step — for grouped data, always make a proper table (x, f, fx, d, fd, etc.)
• Double-check your final answer against known bounds (e.g. Bowley's between −1 and +1, μ₁ always 0).
• Time: allocate based on marks — 2M questions ~3-4 min, 5M questions ~8-10 min (statistics numericals often need more time than other subjects due to tables).

Marks-saving tip:
Even if the final numerical answer is slightly off due to a small arithmetic slip, writing the CORRECT formula and showing the CORRECT method/steps earns most of the marks — examiners give significant partial credit for correct methodology.
🌟 All the Best!
Statistics is 100% practice-based — solve as many numericals by hand as you can, following the exact step-by-step methods shown here. Don't just read the formulas, WRITE them out and solve! Tu definitely kar sakta hai! 💪📊
📄 Previous Year Questions

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