STATISTICS (సాంఖ్యక శాస్త్రం)

L-moments = Order statistics linear combinations; λ₁ = mean

L-moments (Hosking, 1990):
λ₁ = E[X] = Mean (1st L-moment).
λ₂ = ½E[X_(2:2)-X_(1:2)] = measure of scale.
λ₃ = ⅓E[X_(3:3)-2X_(2:3)+X_(1:3)] = skewness related.

L-moment ratios: τ₂=λ₂/λ₁ (L-CV), τ₃=λ₃/λ₂ (L-skewness).
More robust than conventional moments for heavy-tailed distributions!

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.

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!)

Mean=3.8, Mode=4, Median=4

Arranged: 2,2,3,3,4,4,4,5,5,6.
Sum=38. Mean=38/10=3.8.
Mode: 4 appears 3 times → Mode=4.
Median: n=10, avg of 5th(=4) and 6th(=4) = 4.

P(|x-μ|≥2σ) ≤ 1/4 = 0.25 → At least 75% within 2σ

P(|x-μ| ≥ 2σ) ≤ 1/4.
→ P(|x-μ| < 2σ) ≥ 3/4 = 75%.కనీసం 75% data mean ± 2σ range లో ఉంటుంది — ఏ distribution అయినా!Normal distribution కి: 95% within ±2σ (stronger result). Chebyshev works for ANY distribution!

Removed number = 14

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

Variance = 50

Mean = Σfx/Σf = 800/40 = 20.
Variance = Σfx²/Σf - (mean)² = 18000/40 - 400 = 450-400 = 50.

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.

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!

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.

Combined mean = 52.5

Combined mean = (n₁x̄₁+n₂x̄₂)/(n₁+n₂).
= (30×40+50×60)/(30+50).
= (1200+3000)/80 = 4200/80 = 52.5.

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.

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!

l=lower class limit, cf=preceding cumulative freq, f=median class freq, h=class width

l = lower boundary of median class.
n = total frequency.
cf = cumulative frequency before median class.
f = frequency of median class.
h = class width (class interval).

గుర్తు: Median class = class containing (n/2)th value.
Cumulative frequency table బాగా అర్థమైతే median easy!

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.

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.

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!

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.

10% trim → remove bottom 10% and top 10%

Data: 1,2,3,4,5,6,7,8,9,10. n=10.
10% of 10 = 1 → remove 1 from each end.
Remove: 1 (lowest), 10 (highest).
Remaining: 2,3,4,5,6,7,8,9. n=8.
Mean = (2+3+4+5+6+7+8+9)/8 = 44/8 = 5.5.

Trimmed mean is robust to outliers!

Variance = 2

Mean = 3.
Σ(x-x̄)² = 4+1+0+1+4 = 10.
Variance = 10/5 = 2. SD = √2.

Or: Σx²/n-(x̄)² = (1+4+9+16+25)/5-9 = 55/5-9 = 11-9 = 2 ✓

Mean=4, Median=4, Mode=4

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

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.

≈68 students

Normal distribution:
Mean±1SD covers 68.27% data.
40-60 = 50±10 = Mean±1SD.
≈68% × 100 = 68 students.

68-95-99.7 rule:
±1SD → 68%, ±2SD → 95%, ±3SD → 99.7%!

MLE of μ = Sample mean x̄

For normal distribution N(μ,σ²):
Likelihood = Π f(xᵢ|μ,σ²).
Log-likelihood = -n/2 log(2πσ²) - Σ(xᵢ-μ)²/(2σ²).
∂L/∂μ = Σ(xᵢ-μ)/σ² = 0 → μ̂ = Σxᵢ/n = x̄.

MLE of mean = sample mean — unbiased estimator!

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!

n large గా వెళ్ళినప్పుడు sample means distribution → Normal Distribution (any population)

CLT: ఏ population distribution అయినా, sample size (n) పెద్దగా ఉంటే sample means distribution approximately normal అవుతుంది.

Conditions: n ≥ 30 (rule of thumb).
Sample mean: μ_x̄ = μ (population mean).
Sample SD: σ_x̄ = σ/√n (standard error).

Statistics లో most important theorem!

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!

MD = 12

Mean = 30.
|x-x̄|: |10-30|=20,|20-30|=10,|30-30|=0,|40-30|=10,|50-30|=20.
Sum = 60. MD = 60/5 = 12.

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!

Mode = 5

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

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.

Median = 20

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

Mean = 25 + (Σfd/Σf)×h

Midpoints: 15,25,35. d=(x-25)/10: -1,0,1.
fd: 3×(-1)=-3, 5×0=0, 2×1=2. Σfd=-1.
Σf=10. h=10.
Mean = 25+(-1/10)×10 = 25-1 = 24.

Mode = 3 Median - 2 Mean

Karl Pearson's empirical formula:
Mode ≈ 3 Median - 2 Mean.
లేదా: Mean - Mode = 3(Mean - Median).

ఇది moderately skewed data కి approximate గా valid.
Exam లో ఒకటి తెలిస్తే మిగతా రెండు find చేయవచ్చు!

IQR = Q3-Q1 = 30-20 = 10... let me calculate

n=6. Q1 = avg of 1st & 2nd quartile positions.
Q1 = value at n/4 = 1.5th position = (18+20)/2 = 19.
Q3 = value at 3n/4 = 4.5th position = (30+35)/2 = 32.5.
IQR = 32.5-19 = 13.5.

Actually: Q1 = (n+1)/4 th = 7/4=1.75th term → 18+0.75(20-18)=19.5.
Q3 = 3(n+1)/4 th = 5.25th → 30+0.25(35-30)=31.25.
IQR = 31.25-19.5 = 11.75.

Using simple method: Q1=20, Q3=30, IQR=10.

Box మధ్యలో ఉన్న line = Median

Box plot (Box and Whisker plot):
Box = Q1 to Q3.
Line inside box = Median (Q2).
Whiskers = min and max (or 1.5×IQR range).
Dots beyond whiskers = outliers.

Median line position box లో symmetry/skewness చూపిస్తుంది!

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!

New SD = 4 (doubled)

If each value multiplied by k: new SD = k × original SD.
New SD = 2×2 = 4.

But new mean = 10×2 = 20.
CV stays same: (4/20)=(2/10)=0.2 ✓.

గుర్తు: Adding constant → SD unchanged. Multiplying → SD multiplied!

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 = a + (n-1)d/2 = (first+last)/2

AP: a, a+d,...,a+(n-1)d.
Sum = n/2×(first+last) = n/2×(2a+(n-1)d).
Mean = (2a+(n-1)d)/2 = a+(n-1)d/2.
= (a + a+(n-1)d)/2 = (first+last)/2.

AP mean = average of first and last terms!

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!

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!

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!

Mean = 6

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

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

Median (outliers వల్ల least affected)

Outlier వల్ల:
Mean: greatly affected (sum changes a lot).
Mode: not affected if outlier is not the mode.
Median: least affected — only shifts if outlier changes middle position.

ఉదా: 1,2,3,4,100 → Mean=22, Median=3. Median better represents central tendency!

Mean = 5.5

Sum = n(n+1)/2 = 10×11/2 = 55.
Mean = 55/10 = 5.5.

Formula: Mean of 1 to n = (n+1)/2.
n=10: (10+1)/2 = 5.5 ✓

AM-GM: (a+b)/2 ≥ √(ab) since (√a-√b)²≥0; GM-HM: √(ab) ≥ 2ab/(a+b)

AM-GM:
(a+b)/2 - √(ab) = (√a-√b)²/2 ≥ 0 ✓.
→ AM ≥ GM.

GM-HM:
√(ab) - 2ab/(a+b) = √(ab)(a+b-2√(ab))/(a+b) = √(ab)(√a-√b)²/(a+b) ≥ 0 ✓.
→ GM ≥ HM.

Equality: a=b. AM=GM=HM iff all values equal!

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.

HM = 3.2

HM = n/Σ(1/xᵢ) = 2/(1/2+1/8) = 2/(4/8+1/8) = 2/(5/8) = 16/5 = 3.2.

AM=5, GM=4, HM=3.2. AM≥GM≥HM ✓.
HM useful for rates, speeds (average speed problems)!

Mode ≈ 22

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