2.7.3 · D3 · Maths › Statistics & Probability — Intermediate › Measures of dispersion — variance, standard deviation
Intuition Yeh page kis liye hai
Parent note ne formulas banaye. Yahan hum unhe har us situation mein chalate hain jo ek exam ya real world throw kar sakta hai — chhota data, degenerate data, shifting aur scaling, samples vs populations, aur word problems. Agar koi case exist karta hai, toh uska ek worked example neeche hai, aur jahan geometry help kare wahan picture bhi hai.
Hum sirf woh do engines use karte hain jo parent ne di hain. Main unhe simple words mein restate karta hoon taaki yahan koi bhi symbol mystery na rahe.
Neeche "case classes" ki poori list hai. Har cell mein kam se kam ek example hai — har [!example] mein tag batata hai ki woh kaunsi cell hit karta hai.
Case class
Kya tricky hai
Covered by
A. Definition, plain data
bas steps follow karo
Ex 1
B. Shortcut on same data
A se exactly match karna chahiye
Ex 2
C. Degenerate — all equal
σ should be 0
Ex 3
D. Shift invariance (x + c )
variance nahi hilna chahiye
Ex 4
E. Scaling (c x , incl. negative c )
variance × c 2 , SD $\times
c
F. Sample vs population (n vs n − 1 )
kaunsa divisor?
Ex 6
G. Frequency / grouped data
weights, raw list nahi
Ex 7
H. Word problem, real units
SD ko context mein interpret karo
Ex 8
I. Exam twist — combine sets
summary stats se derive karo
Ex 9
J. Limiting behaviour (n → ∞ , one outlier grows)
σ kaise react karta hai
Ex 10
Upar wali figure mental map hai: har example ek dart-scatter picture hai, aur σ measure karta hai ki scatter kitna wide hai.
Worked example Example 1 — Case A (definition engine)
Data (ek population ): 3 , 5 , 5 , 7 , 10 .
Forecast: padhne se pehle guess karo ki σ 1 , 2.5 , ya 5 ke zyada kareeb hai.
Mean. x ˉ = 5 3 + 5 + 5 + 7 + 10 = 5 30 = 6 .
Yeh step kyun? Spread centre se measure hoti hai, isliye pehle centre chahiye.
Deviations d i = x i − x ˉ : − 3 , − 1 , − 1 , 1 , 4 .
Kyun? Middle se har point ki raw signed distance.
Square karo: 9 , 1 , 1 , 1 , 16 .
Kyun? Squaring signs khatam karta hai (warna woh 0 pe cancel ho jaate hain) aur door ke points ko zyada punish karta hai.
Sum & average: sum = 28 , toh σ 2 = 28/5 = 5.6 .
Kyun? Variance average squared distance hai.
Square-root: σ = 5.6 ≈ 2.37 .
Kyun? Hume data ke apne units mein wapas le jaata hai.
Verify: deviations ka sum zero hona chahiye: − 3 − 1 − 1 + 1 + 4 = 0 ✓. Answer 2.37 "closer to 2.5 " wale forecast se match karta hai.
Worked example Example 2 — Case B (shortcut engine, same data)
Same 3 , 5 , 5 , 7 , 10 , ab σ 2 = x 2 − ( x ˉ ) 2 se.
Forecast: iska 5.6 hi aana chahiye — parent note mein algebra ne prove kiya ki dono engines identical hain.
Data ke squares: 9 , 25 , 25 , 49 , 100 ; unka sum = 208 .
Kyun? Hume x 2 = squares ka mean chahiye.
x 2 = 208/5 = 41.6 .
( x ˉ ) 2 = 6 2 = 36 .
Kyun? Square-of-mean, doosra term.
σ 2 = 41.6 − 36 = 5.6 .
Verify: Example 1 se exactly match karta hai ✓. Ordering trap ka sanity check: x 2 = 41.6 ≥ ( x ˉ ) 2 = 36 , jaisa hona chahiye, kyunki unka difference ek variance hai aur variances kabhi negative nahi hote.
Worked example Example 3 — Case C (zero spread)
Data: 8 , 8 , 8 , 8 .
Forecast: sab ek hi jagah par hain. Spread kya honi chahiye ?
Mean = 32/4 = 8 .
Deviations sab = 8 − 8 = 0 .
Yeh kyun important hai? Har point hi centre hai.
Squares sab 0 , sum 0 , σ 2 = 0/4 = 0 , σ = 0 .
Verify (shortcut): x 2 = 64 , ( x ˉ ) 2 = 64 , difference = 0 ✓. σ 2 = 0 hone ka yahi ek tarika hai — parent ki non-negativity property: variance zero hoti hai tabhi jab sab values identical hon .
Yeh dono exam questions ke workhorse hain: yeh allow karte hain ki jo variance tumne already compute ki ho, use reuse karo.
Intuition Pehle figure padho
Poore scatter ko sideways slide karna (add c ) darts ke beech ke gaps kabhi nahi badalta — isliye variance nahi hilti. Scatter ko stretch karna (multiply by c ) har gap ko c se multiply karta hai; squared gaps c 2 se badhte hain.
Worked example Example 4 — Case D (shift invariance)
Example 1 ka data lo aur har value mein c = 100 add karo: 103 , 105 , 105 , 107 , 110 .
Forecast: kya σ 2 5.6 rahega ya kuch bahut bada ho jaayega?
Naya mean: old mean + 100 = 106 .
Kyun? Har point mein c add karne se balance point c se shift hota hai.
Naye deviations = x i − 106 : − 3 , − 1 , − 1 , 1 , 4 — pehle jaisi bilkul .
Kyun? + 100 dono x i aur x ˉ mein appear karta hai, isliye x i − x ˉ ke andar cancel ho jaata hai.
Isliye σ 2 = 5.6 unchanged.
Verify: Var ( x + c ) = Var ( x ) — Example 1 ke 5.6 se match karta hai ✓.
Worked example Example 5 — Case E (scaling, negative factor bhi)
Example 1 ka data c = − 2 se multiply karo: − 6 , − 10 , − 10 , − 14 , − 20 .
Forecast: property kehti hai variance × c 2 aur SD × ∣ c ∣ . c = − 2 ke saath, woh hai variance × 4 , SD × 2 . Numbers guess karo.
Naya mean: − 2 × 6 = − 12 .
Kyun? Har value ko c se multiply karne se balance point c se multiply hota hai.
Naye deviations: x i − x ˉ har ek − 2 × (old deviation) ban jaata hai: 6 , 2 , 2 , − 2 , − 8 .
Kyun? c x i − c x ˉ = c ( x i − x ˉ ) .
Squares: 36 , 4 , 4 , 4 , 64 , sum = 112 , σ 2 = 112/5 = 22.4 .
σ = 22.4 ≈ 4.73 .
Verify: old σ 2 = 5.6 ; 5.6 × c 2 = 5.6 × 4 = 22.4 ✓. Old σ = 2.37 ; 2.37 × ∣ c ∣ = 2.37 × 2 = 4.73 ✓. c ka negative sign gayab ho gaya kyunki variance c 2 use karta hai aur SD ∣ c ∣ — spread ki koi direction nahi hoti.
Worked example Example 6 — Case F (kaunsa divisor?)
Teen exam times (minutes) ek poore cohort ko estimate karne ke liye sample ke roop mein liye gaye: 10 , 12 , 14 .
Forecast: sample variance n − 1 use karta hai, isliye woh population version se bada hona chahiye. Guess karo kitna bada.
Mean = 36/3 = 12 .
Squared deviations: ( − 2 ) 2 , 0 2 , 2 2 = 4 , 0 , 4 ; sum = 8 .
Agar population: σ 2 = 8/3 ≈ 2.67 .
Sample ki tarah: s 2 = 8/ ( 3 − 1 ) = 8/2 = 4 .
n − 1 = 2 se kyun divide karein? Bessel's correction: sample mean precisely inhi points ke liye tune hua tha, jisse deviations thodi chhoti lag rahi hain, toh hum chhote divisor use karke inflate karte hain.
s = 4 = 2 minutes.
Verify: s 2 = 4 > σ 2 ≈ 2.67 ✓ (sample version bada hai, jaise forecast tha). Ratio 4/ ( 8/3 ) = 1.5 = n − 1 n = 2 3 ✓ — exactly correction factor.
Definition Engines ka frequency version
Jab ek value x i f i baar aati hai, toh har ek ko f i copies maano. Total count N = ∑ f i ke saath:
x ˉ = N 1 ∑ f i x i , σ 2 = N 1 ∑ f i ( x i − x ˉ ) 2 = N ∑ f i x i 2 − ( x ˉ ) 2 .
Same idea — har squared distance ko bas weight milta hai ki woh kitni baar hota hai.
Worked example Example 7 — Case G (weighted data)
Marks aur kitne students ko mile:
mark x i
2
4
6
frequency f i
1
3
2
Forecast: zyaatar students 4 par hain; guess karo ki σ 2 se neeche hai ya upar.
Total count N = 1 + 3 + 2 = 6 .
Weighted mean: ∑ f i x i = 1 ⋅ 2 + 3 ⋅ 4 + 2 ⋅ 6 = 2 + 12 + 12 = 26 , toh x ˉ = 26/6 ≈ 4.33 .
Weight kyun? Ek mark jo 3 baar appear karta hai, centre ko teen guna zyada pull karta hai.
Squares ka weighted mean: ∑ f i x i 2 = 1 ⋅ 4 + 3 ⋅ 16 + 2 ⋅ 36 = 4 + 48 + 72 = 124 ; N se divide karo: 124/6 ≈ 20.67 .
Shortcut: σ 2 = 20.67 − ( 4.33 ) 2 = 20.67 − 18.78 ≈ 1.89 .
σ ≈ 1.37 .
Verify: exact fractions — x ˉ = 13/3 , x 2 = 62/3 , toh σ 2 = 3 62 − 9 169 = 9 186 − 169 = 9 17 ≈ 1.89 ✓. 2 se neeche, jaise forecast tha, kyunki zyaatar students 4 ke paas cluster karte hain.
Worked example Example 8 — Case H (units interpret karo)
Ek machine bottles bharta hai. Paanch bottles measure ki gayi (ml), is batch ki population maano: 498 , 502 , 500 , 501 , 499 .
Forecast: yeh 500 ke paas tightly bunched hain; σ ml mein guess karo.
Mean = 5 498 + 502 + 500 + 501 + 499 = 5 2500 = 500 ml.
Deviations: − 2 , 2 , 0 , 1 , − 1 ml.
Squares: 4 , 4 , 0 , 1 , 1 ; sum = 10 .
σ 2 = 10/5 = 2 ml² .
"ml²" kyun? Deviations ko square karne se units bhi square ho gayi.
σ = 2 ≈ 1.41 ml .
Square-root kyun? Spread real millilitres mein report karne ke liye, ml² mein nahi.
Verify: deviations sum − 2 + 2 + 0 + 1 − 1 = 0 ✓. Interpretation: typical bottle 500 ml target se lagbhag 1.41 ml off hai — machine precise hai. Square-root liye bina "2 " report karna classic forgot-to-root mistake hogi, aur woh ml² mein hogi, jo ek bottle ke liye bekar unit hai .
Kabhi kabhi tumhe raw numbers nahi diye jaate, sirf summary sums. Shortcut engine yahan shine karta hai.
Worked example Example 9 — Case I (
∑ x aur ∑ x 2 use karke merge karo)
Group P: n P = 3 values with ∑ x = 12 , ∑ x 2 = 56 .
Group Q: n Q = 2 values with ∑ x = 18 , ∑ x 2 = 164 .
Combined population ki variance nikalo.
Forecast: low-mean group ko high-mean group se combine karna aksar spread badhata hai. Dhyan rakhna.
Combined count N = 3 + 2 = 5 .
Combined mean: x ˉ = 5 12 + 18 = 5 30 = 6 .
∑ x kyun add karein? Total sum over total count overall balance point hai.
Squares ka combined mean: x 2 = 5 56 + 164 = 5 220 = 44 .
∑ x 2 kyun add karein? ∑ x 2 disjoint groups mein additive hai, bilkul ∑ x ki tarah.
Shortcut: σ 2 = 44 − 6 2 = 44 − 36 = 8 .
σ = 8 ≈ 2.83 .
Verify: consistent example — Group P ho sakta hai 2 , 4 , 6 (∑ = 12 , ∑ x 2 = 4 + 16 + 36 = 56 ✓) aur Group Q 8 , 10 (∑ = 18 , ∑ x 2 = 64 + 100 = 164 ✓). Merged list 2 , 4 , 6 , 8 , 10 : mean 6 , squared devs 16 , 4 , 0 , 4 , 16 sum 40 , σ 2 = 40/5 = 8 ✓.
Worked example Example 10 — Case J (ek value bhaag jaati hai)
0 , 0 , 0 , 0 (population) se shuru karo. Ek entry x tak badhti hai: data ban jaata hai x , 0 , 0 , 0 . x → ∞ hone par σ 2 kaise behave karta hai?
Forecast: ek wild dart spread ko unlimited blow up kar dena chahiye. Kya woh x ki tarah badhta hai ya x 2 ki tarah?
Mean: x ˉ = x /4 .
Squares ka mean: x 2 = 4 x 2 + 0 + 0 + 0 = 4 x 2 .
Shortcut: σ 2 = 4 x 2 − ( 4 x ) 2 = 4 x 2 − 16 x 2 = 16 3 x 2 .
Limit: x → ∞ hone par, σ 2 = 16 3 x 2 → ∞ , x 2 ki tarah badhta hai (SD x ki tarah badhti hai).
x 2 kyun, x nahi? Deviations ko square karne ka matlab hai ki ek door ka point apni distance squared contribute karta hai — outliers dominate karte hain. Yahi woh reason hai jis par parent note ne warn kiya tha ki squaring "penalises big deviations more."
Verify: x = 4 plug karo: data 4 , 0 , 0 , 0 , mean 1 , squared devs 9 , 1 , 1 , 1 sum 12 , σ 2 = 12/4 = 3 ; formula 16 3 ⋅ 16 = 3 ✓.
Recall Self-test: case batao, phir solve karo
Har mini-prompt ke liye, pehle batao kaunsa matrix cell (A–J) hai.
Data 5 , 5 , 5 population variance ::: Case C — degenerate, σ 2 = 0 .
Tumne already σ 2 = 9 find ki; ab har value 3 se multiply hoti hai. Naya variance? ::: Case E — × c 2 = 9 × 9 = 81 .
Same data, lekin ab har value mein 7 add karo. Naya variance? ::: Case D — unchanged, still 9 .
n = 3 sample, sum of squared deviations = 18 . Sample variance? ::: Case F — 18/ ( 3 − 1 ) = 9 .
Merge: N = 4 , ∑ x = 8 , ∑ x 2 = 24 . Variance? ::: Case I — 24/4 − ( 8/4 ) 2 = 6 − 4 = 2 .
Mnemonic Sab survive karne ke liye ek line
Centre, subtract, square, average, root — aur sirf do dials ise change karte hain: shift kuch nahi karta, stretch square kar deta hai.