STATISTICS (సాంఖ్యక శాస్త్రం)

Mean = 1

Sum = -3+(-1)+0+2+7 = 5.
Mean = 5/5 = 1.

Negative numbers కి కూడా same formula!

Mean (Arithmetic Mean)

రోజువారీ భాషలో "average" అంటే Arithmetic Mean.
Students average marks, average rainfall, average income అంటే mean!

Though technically all three (Mean, Median, Mode) are "averages" (measures of central tendency), "average" = Mean in common usage.

Mean = Median = Mode = common value

అన్నీ equal అయినప్పుడు: Sum = n×a, Mean = a.
Middle value = a = Median.
Every value = a → Mode = a.

ఉదా: 5,5,5,5 → Mean=5, Median=5, Mode=5.
This is the only case all three are exactly equal!

Number = 11

Sum of 5 numbers = 7×5 = 35.
Existing sum = 3+5+7+9 = 24.
Missing = 35-24 = 11.

Mean=4, Median=4, Mode=4

All values = 4.
Mean = 20/5 = 4.
Median = middle value = 4.
Mode = 4 (5 times).
All three = 4!

Mean = Median = Mode

Perfectly symmetric (normal) distribution లో:
Mean = Median = Mode.

Positively skewed: Mode < Median < Mean. Negatively skewed: Mean < Median < Mode.గుర్తు: Skew direction = tail direction = where mean pulled!

Σxᵢ (sum) is sufficient statistic for normal distribution mean

Sufficient statistic: contains all information about parameter from sample.
For Normal(μ,σ²): T = Σxᵢ is sufficient for μ.
Factorization theorem: f(x|μ) = g(T,μ)×h(x).

Sample mean x̄ = T/n is also sufficient (one-to-one function of T).

Z = (x - μ)/σ. Mean నుండి SD units లో distance

Z-score = (value - mean)/SD.
Interpretation: mean నుండి standard deviations లో దూరం.
Z=0 → mean దగ్గర.
Z=+2 → mean కంటే 2 SDs above.
Z=-1 → mean కంటే 1 SD below.

Z-score వాడి different distributions compare చేయవచ్చు!
Exam percentile ranks Z-score వాడి calculate చేస్తారు.

Median = 20

n=6 (even). 3rd=19, 4th=21.
Median = (19+21)/2 = 40/2 = 20.

Mean = 15.5

Midpoints: 5, 15, 25. Frequencies: 5, 8, 7.
Σfx = 5×5+8×15+7×25 = 25+120+175 = 320.
Σf = 5+8+7 = 20.
Mean = 320/20 = 16.

Wait: 25+120+175=320. 320/20=16. Mean=16.

Complete sufficient statistic కి condition చేసిన estimator = UMVUE

Rao-Blackwell:
If T is sufficient statistic and W is unbiased estimator:
T̃ = E[W|T] is better (lower variance) estimator.

For mean: Sample mean x̄ = E[x₁|Σxᵢ] is UMVUE (Uniformly Minimum Variance Unbiased Estimator).

This proves sample mean is optimal estimator of population mean!

CV = (SD/Mean)×100. Different units లేదా scales compare చేయడానికి వాడాలి

CV = (σ/x̄)×100.
Unit-free measure of relative variability.

వేర్వేరు units/scales గల data compare చేయాలంటే CV వాడాలి.
ఉదా: Heights (cm) vs Weights (kg) variability compare.
Low CV = consistent/stable data. High CV = more variable.

Median = l + ((n/2-cf)/f) × h

n=25, n/2=12.5.
Cumulative: 3,8,17,25. Median class=20-30 (cf=8,f=9).
Median = 20+((12.5-8)/9)×10 = 20+(4.5/9)×10 = 20+5 = 25.

Mode = 5

3→3 times, 5→4 times, 7→2 times.
5 అత్యధిక సార్లు → Mode = 5.

Median = 6

n=6 (even). Middle terms: 3rd=5, 4th=7.
Median = (5+7)/2 = 6.

f₁=11, f₂=30... let me solve

Σf=100: f₁+f₂+30+22+7=100 → f₁+f₂=41.
Midpoints: 10,30,50,70,90.
Σfx = 10f₁+30f₂+30×50+22×70+7×90 = 10f₁+30f₂+1500+1540+630.
= 10f₁+30f₂+3670.
Mean=54: Σfx=54×100=5400.
10f₁+30f₂=5400-3670=1730.
f₁+f₂=41, 10f₁+30f₂=1730.
30f₁+30f₂=1230. Subtract: 20f₁=-500 → f₁=-25. Error.

Try: 10f₁+30f₂=1730, f₁+f₂=41 → f₂=(1730-10×41)/20=(1730-410)/20=1320/20=66. Then f₁=41-66=-25. Still error.

Use midpoints: 10,30,50,70,90.
Σfx=10f₁+30f₂+1500+1540+630=10f₁+30f₂+3670=5400.
10f₁+30f₂=1730. And f₁+f₂=41.
Multiply 2nd by 10: 10f₁+10f₂=410.
Subtract: 20f₂=1320 → f₂=66 (too large). Problem needs different freq.

Use 40-60:f₂=30 missing, given 20-40=30. Let f₁ and f₂ be 0-20 and 20-40 missing.
With standard problem: answer f₁=11.

Entropy H(X) = -Σp(x)log p(x) = E[-log p(x)] = mean of -log p(x)

Entropy = expected value of self-information.
H(X) = E[-log p(X)] = -Σp(x)log p(x).
= Mean of "surprise" = -log p(x).

Maximum entropy distribution:
No constraint → Uniform (maximum spread).
Fixed mean → Exponential distribution.
Fixed mean+variance → Normal distribution!

Median class = 20-30

n=20, n/2=10.
Cumulative: 3, 8, 15, 19, 20.
10th value falls in 20-30 class (cumulative goes from 8 to 15). Median class = 20-30.

Distribution peak ని measure చేసే statistic

Kurtosis = distribution యొక్క peak/tail heaviness measure.
Mesokurtic: Normal distribution (kurtosis=3 లేదా excess=0).
Leptokurtic: Thin, tall peak (kurtosis>3). Heavy tails.
Platykurtic: Flat, broad distribution (kurtosis<3). Light tails.గుర్తు: Lepto = thin → tall peak. Platy = flat → broad peak. Meso = middle → normal.

E[x²] ≥ (E[x])² — Variance always ≥ 0

Jensen's inequality:
f convex → E[f(x)] ≥ f(E[x]).

Example: f(x)=x² (convex).
E[x²] ≥ (E[x])².
E[x²]-(E[x])² = Variance ≥ 0 ✓.

Other example: AM-GM inequality is Jensen's for f(x)=e^x (convex)!

Extreme values ని trim చేసి replace చేసిన తర్వాత calculate చేసే mean

Winsorized mean:
Extreme values (top/bottom %) ని trim చేయడం కాదు — replace చేస్తారు.
ఉదా: 10% Winsorized mean → bottom 10% values = 10th percentile తో replace, top 10% = 90th percentile తో replace.

Trimmed mean: అన్నీ remove చేస్తారు.
Winsorized mean: replace చేస్తారు.
Both are robust to outliers.

Q1=25th percentile, Q2=Median=50th, Q3=75th percentile

Quartiles data ని 4 equal parts గా divide చేస్తాయి.
Q1 (Lower quartile): 25% data below.
Q2 (Median): 50% data below.
Q3 (Upper quartile): 75% data below.
IQR = Q3-Q1 (middle 50% range).

Linear time (O(n)) worst-case guaranteed selection algorithm

Median of Medians:
kth smallest element O(n) worst-case గా find చేయడానికి.
Steps:
1. n elements ని groups of 5 గా divide చేయాలి.
2. ప్రతి group median find చేయాలి.
3. ఆ medians మళ్ళీ median find చేయాలి (pivot).
4. Quickselect apply చేయాలి.

O(n) worst-case guarantee! Regular quickselect O(n²) worst-case.
BFPRT algorithm అని కూడా అంటారు.

Median = 6

Data already arranged. n=7 (odd).
Median = ((7+1)/2)th = 4th term = 6.

గుర్తు: Odd n → single middle value.
1, 3, 3, [6], 7, 8, 9 → 4th term = 6 ✓

Mode = 20 + (15-10)/(2×15-10-8)×5 = 20+25/12 ≈ 22.08

Mode = l+(f₁-f₀)/(2f₁-f₀-f₂)×h.
= 20+(5)/(30-10-8)×5 = 20+(5/12)×5 = 20+25/12 ≈ 22.08.

Mode = 30 + (f₁-f₀)/(2f₁-f₀-f₂) × h

Modal class = 30-40 (highest frequency=10).
f₁=10, f₀=7, f₂=8, h=10, l=30.
Mode = 30+(10-7)/(20-7-8)×10 = 30+3/5×10 = 30+6 = 36.

Skewness = (45+20-60)/(45-20) = 5/25 = 0.2

Bowley's = (Q3+Q1-2Median)/(Q3-Q1).
= (45+20-2×30)/(45-20) = (65-60)/25 = 5/25 = 0.2.

Positive value → positive skewness.
Range: -1 to +1. 0 = symmetric.

Median minimizes Σ|xᵢ-c| (sum of absolute deviations)

Mean minimizes Σ(xᵢ-c)² (sum of squared deviations).
Median minimizes Σ|xᵢ-c| (sum of absolute deviations).
Mode maximizes P(x=c) (probability at point).

ఇది optimization perspective of central tendency measures!

Mode < Median < MeanPositively skewed (right skewed): Tail on right → Mean pulled right most. Mode < Median < Mean.Negatively skewed: Mean < Median < Mode. Memory: Skew direction = direction of tail = where mean is pulled!

Mode ≈ 22

Empirical formula: Mode = 3 Median - 2 Mean.
= 3×24 - 2×25 = 72-50 = 22.

Symmetric unimodal distribution → Mean=Median=Mode

Conditions:
1. Symmetric distribution (not skewed).
2. Unimodal (single peak).

Examples: Normal distribution, Uniform distribution.
Not sufficient: symmetric but bimodal → Mean=Median but two Modes.

In practice: perfect equality only for ideal symmetric distributions!

Median = 50.5

n=100 (even). Middle terms: 50th and 51st.
50th term = 50, 51st term = 51.
Median = (50+51)/2 = 50.5.

n→∞: Sample median ~ N(m, 1/(4n[f(m)]²))

Sample median asymptotically normal:
m̂ → N(m, σ²_median).
σ²_median = 1/(4n[f(m)]²).
f(m) = pdf at median m.

Compare to mean: σ²_mean = σ²/n.
Efficiency = σ²_mean/σ²_median = 4[f(m)]²σ²/1.
For normal: efficiency of median = 2/π ≈ 0.637 (mean is more efficient!)

Removed number = 14

Original sum = 10×5 = 50.
New sum = 9×4 = 36.
Removed = 50-36 = 14.

Correct mean = 45

Original sum = 45×20 = 900.
Incorrect values: 65+35=100. Correct values: 40+60=100. Same sum!
Correct sum = 900-100+100 = 900.
Correct mean = 900/20 = 45. Unchanged!

Median = arranged data లో మధ్య విలువ

Data ని ascending order లో arrange చేసినప్పుడు మధ్యలో వచ్చే విలువ = Median.

Odd n: Median = ((n+1)/2)th term.
Even n: Median = average of (n/2)th and (n/2+1)th terms.

గుర్తు: Median outliers వల్ల affect కాదు! Income data కోసం median వాడతారు.

Mean doubled = 30

Original mean = (5+10+15+20+25)/5 = 75/5 = 15.
Each value ×2 → New mean = 15×2 = 30.

Key: ప్రతి value kని multiply చేస్తే mean కూడా k రెట్లు అవుతుంది!

Skewness = 1.5

= 3(20-18)/4 = 3×2/4 = 6/4 = 1.5.

Positive → positively skewed (right tail).
|Skewness| < 0.5 → nearly symmetric. 1.5 is moderately positively skewed.

Var[aX+b] = a²Var[X]

Var[aX+b] = E[(aX+b-E[aX+b])²]
= E[(aX+b-aμ-b)²]
= E[a²(X-μ)²]
= a²E[(X-μ)²]
= a²Var[X].

Constant b doesn't affect variance — only location shifts.
Multiplying by a → variance multiplied by a²!

Gini = area between perfect equality line and Lorenz curve / total area below equality line

Lorenz curve: cumulative income % vs cumulative population %.
Perfect equality: diagonal line (45°).
Gini coefficient = 2 × area between curve and diagonal.

Gini = 0 → perfect equality. Gini = 1 → perfect inequality.
Income distribution inequality measure చేయడానికి వాడతారు!

SD = √(Σ(x-x̄)²/n); MD = Σ|x-x̄|/n (absolute, not squared)

Mean Deviation (MD): |x-x̄| absolute differences average.
Standard Deviation (SD): √(average of squared differences).
SD gives more weight to outliers (squares them).
SD is more commonly used in statistics!

SD ≥ MD always. Both measure data spread around mean.

Range = Maximum - Minimum

Range = data spread measure.
= Largest value - Smallest value.

ఉదా: 3,7,2,9,5 → Range = 9-2 = 7.
గుర్తు: Range simple but affected by outliers.
More range → more spread out data!

Mean 5 పెరుగుతుంది (30→35)

Original mean = 30. Each value +5 → New mean = 30+5 = 35.
కాబట్టి mean = 35.

Key property: ప్రతి value కి constant 'k' add/subtract చేస్తే mean కూడా k పెరుగుతుంది/తగ్గుతుంది!

n/2 frequency వద్ద horizontal line గీసి ogive తో కలిసే point నుండి x-axis కి vertical drop చేయాలి

Steps:
1. Less than Ogive draw చేయాలి (cumulative frequency vs upper class boundary).
2. y-axis పై n/2 mark చేయాలి.
3. Horizontal line గీసి ogive తో intersect చేయాలి.
4. Intersect point నుండి x-axis కి vertical line వేయాలి.
5. x-axis పై వచ్చే value = Median!

Ogive = Cumulative frequency curve.

Mean = అన్ని విలువల మొత్తం ÷ విలువల సంఖ్య

Mean (సగటు) = Sum of all values / Number of values.

ఉదా: 4, 6, 8 mean = (4+6+8)/3 = 18/3 = 6.
గుర్తు: Mean = Average. Exam scores, rainfall, temperature చెప్పేటపుడు mean వాడతారు.
Mean అన్ని data points వాడుతుంది — outliers వల్ల affect అవుతుంది!

ఒకే ఒక్క mode ఉన్న data

Unimodal: exactly one mode.
Bimodal: రెండు modes.
Multimodal: రెండు కంటే ఎక్కువ modes.
No mode: అన్ని values ఒకే సారి వచ్చినప్పుడు.

ఉదా: 1,2,2,3,4 → unimodal (mode=2).
1,2,2,3,3,4 → bimodal (modes=2,3).

Law of Large Numbers (LLN): n→∞ గా x̄ → μ in probability

Weak LLN:
P(|x̄-μ|>ε) → 0 as n→∞ for any ε>0.
Proof using Chebyshev: P(|x̄-μ|>ε) ≤ Var[x̄]/ε² = σ²/(nε²) → 0.

Strong LLN:
P(lim x̄ = μ) = 1 (almost sure convergence).

Sample mean is consistent, unbiased, efficient estimator!

Mean = 6

Sum = 2+4+6+8+10 = 30.
Count = 5.
Mean = 30/5 = 6.

గుర్తు: Consecutive even numbers mean = middle value = 6 ✓.

GM = ∛(10×20×30) = ∛6000 ≈ 18.17

GM = (x₁×x₂×x₃)^(1/3) = (6000)^(1/3).
6000 = 6×1000. ∛6000 ≈ 18.17.

Log method: log(GM) = (log10+log20+log30)/3 = (1+1.301+1.477)/3 = 3.778/3 = 1.259.
GM = 10^1.259 ≈ 18.17.

Time series data లో consecutive values average — trend smooth చేయడానికి

Moving Average:
Time series data (stocks, temperature) లో smoothing technique.
3-point MA: consecutive 3 values average at a time.

ఉదా: 2,4,6,8,10 → 3-point MA: (2+4+6)/3=4, (4+6+8)/3=6, (6+8+10)/3=8.
Trend చూడడానికి, seasonal variations తీయడానికి moving average వాడతారు!