Measures of dispersion — variance, standard deviation: Question bank

2.7.3 · D5 · Maths › Statistics & Probability — Intermediate › Measures of dispersion — variance, standard deviation

Ye traps jinpe rely karte hain, unki vocabulary

Ek bhi trap se pehle, har symbol ko pin down karte hain taaki kuch surprise na kare.

Figure s01 mein yellow dashed line mean $\overset{x}{ˉ}$ (balance point) ko mark karti hai, aur har coloured stick ek deviation $x_{i} - \overset{x}{ˉ}$ hai: left ki taraf point karne wali sticks negative hain, right ki taraf point karne wali sticks positive. Apni aankhon se notice karo ki left-lengths aur right-lengths balance ho jaati hain — yahi " $\sum (x_{i} - \overset{x}{ˉ}) = 0$ " ke peeche ka picture hai, aur isi liye average karne se pehle hume square karna padta hai.

$σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ kyun hota hai — sirf quote nahi, derive kiya hua

Ye shortcut magic nahi hai. Definition se shuru karo aur andar ka square expand karo: $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{n} \sum (x_{i}^{2} - 2 x_{i} \overset{x}{ˉ} + \overset{x}{ˉ}^{2}) .$ Ek sum ko teen mein split karo (summation $+$ par distribute ho jaati hai): $= \overline{x^{2}} \frac{1}{n} \sum x_{i}^{2} - 2 \overset{x}{ˉ} \cdot \overset{x}{ˉ} \frac{1}{n} \sum x_{i} + \frac{1}{n} n \overset{x}{ˉ}^{2} \sum \overset{x}{ˉ}^{2} .$ Beech wala piece $2 \overset{x}{ˉ} \cdot \overset{x}{ˉ} = 2 (\overset{x}{ˉ})^{2}$ hai aur aakhri $(\overset{x}{ˉ})^{2}$ hai, isliye $σ^{2} = \overline{x^{2}} - 2 (\overset{x}{ˉ})^{2} + (\overset{x}{ˉ})^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2} .$

Traps mein le jaane ke liye teen quick anchors:

Variance $σ^{2}$ — centre se average squared distance, kabhi negative nahi.
Standard deviation $σ = σ^{2}$ — wahi idea data ke apne units mein.
Shortcut $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ — "mean of squares minus square of mean".

True or false — justify karo

Har line ek claim hai. Reveal karne se pehle true ya false decide karo aur ek-sentence reason do. Sirf "true"/"false" ka koi matlab nahi — reason hi point hai.

Variance negative ho sakta hai agar data mein mean se neeche wali values zyada hon mean se upar wali values se.

False. Har term

(x_{i} - \overset{x}{ˉ})^{2}

ek square hai, isliye

\geq 0

hai; non-negative numbers ka sum

n > 0

se divide karne par kabhi negative nahi ho sakta, chahe data mean ke aas-paas kaise bhi ho.

Agar do datasets ka mean same ho, toh unka variance bhi same hona chahiye.

False. Mean sirf centre fix karta hai; parent note ka poora point yahi hai ki do classes averaging 60% tightly bunched ya wildly scattered ho sakti hain — same mean, different spread.

Standard deviation hamesha variance se chhoti ya barabar hoti hai.

False. Sirf tab jab

σ^{2} \geq 1

. Agar

σ^{2} = 0.25

toh

σ = 0.5 > 0.25

; 1 se chhote number ka square root lena use bada kar deta hai.

Har exam mark mein 10 add karne se standard deviation 10 badh jaati hai.

False. Har value ko shift karna poore cloud ko sideways slide karta hai bina points ke beech koi gap badlaye, isliye variance aur SD dono unchanged rehte hain — variance shift-invariant hota hai.

Har data value ko double karne se variance double ho jaata hai.

False.

c

se scale karne par variance

c^{2}

se multiply hota hai, isliye double karne (

c = 2

) par variance

4

se multiply hota hai; ye SD ko double karta hai, variance ko nahi. Stretch picture ke liye Figure s03 dekho.

\overline{x^{2}}

aur

(\overset{x}{ˉ})^{2}

ek hi number ke do naam hain.

False.

\overline{x^{2}}

har value ko square karke average karta hai;

(\overset{x}{ˉ})^{2}

pehle average karta hai phir square karta hai. Unka difference exactly variance hai, isliye ye tab barabar hote hain jab variance

0

ho (saari values identical hon).

Agar variance

0

hai, toh data ka mean, median aur mode sab equal hone chahiye.

True. Variance

0

har value ko identical hone par force karta hai, isliye ek hi repeated number simultaneously mean, median aur mode hota hai.

Same data ke liye sample variance

s^{2}

hamesha population variance

σ^{2}

se bada hota hai.

True (

n > 1

aur non-constant data ke liye). Dono mein squared deviations ka same sum share hota hai, lekin

s^{2}

chhote

n - 1

se divide karta hai, isliye bada hota hai; agar saari values equal hon toh dono

0

hain.

Figure s03 dikhata hai ki scaling rule $c$ ko kyun square karta hai: number line ko $c$ se stretch karo aur har deviation stick $c$ se stretch hoti hai, isliye har squared stick — aur isliye unka average, variance — $c^{2}$ se stretch hota hai, jabki SD (phir se ek length) sirf $c$ se stretch hoti hai.

Spot the error

Har line ek calculation ya statement describe karti hai jo kisine actually ki. Mistake dhundho aur correct principle batao.

"Mujhe variance

= 5.6

marks mila, toh spread lagbhag

5.6

marks hai."

Error units mein hai. Variance marks² mein hai, marks mein nahi; interpret karne layak spread

σ = 5.6 \approx 2.37

marks hai. Real units mein spread ke liye SD report karo, variance nahi.

"Maine deviations

(x_{i} - \overset{x}{ˉ})

average kiye aur

0

mila, toh is data mein koi spread nahi."

Error squaring skip karne mein hai.

\sum (x_{i} - \overset{x}{ˉ}) = 0

har dataset ke liye hota hai kyunki positives aur negatives cancel ho jaate hain (Figure s01 dekho); woh zero spread ke baare mein kuch measure nahi karta. Pehle square karna zaroori hai.

2, 4, 4, 6, 9

ke liye maine

\overline{x^{2}} = (\overset{x}{ˉ})^{2} = 25

use kiya, isliye variance

= 25 - 25 = 0

Error

\overline{x^{2}}

aur

(\overset{x}{ˉ})^{2}

ko confuse karne mein hai. Yahan

\overline{x^{2}} = 153/5 = 30.6

hai jabki

(\overset{x}{ˉ})^{2} = 25

hai, isliye variance

= 30.6 - 25 = 5.6

hai,

0

nahi.

"Mera data 5 students ka poora population hai, isliye main sum of squares ko

n - 1 = 4

se divide karta hoon."

Error Bessel's correction ko full population par apply karne mein hai.

n - 1

se sirf tab divide karo jab data ek bade population ko estimate karne wala sample ho; poore population ke liye

n

use karo.

"Maine mean

5

compute kiya, phir deviations find karte waqt alag rounded mean

5.2

use kiya."

Error inconsistent centre use karne mein hai. Har deviation ko usi same

\overset{x}{ˉ}

se measure karna chahiye jo tumne actually compute kiya, warna deviations cancel nahi honge aur variance corrupt ho jaayega.

"Variance

- 3

aaya, jo mujhe batata hai ki data downward trend kar raha hai."

Error ye hai ki variance negative banana impossible hai — negative result matlab arithmetic mein koi slip hai (likely

(\overset{x}{ˉ})^{2} > \overline{x^{2}}

, jo sachchi mein kabhi nahi ho sakta). Variance mein koi direction/trend information bilkul nahi hoti.

"Do SDs

4

cm aur

9

cm hain, toh milake combined SD

4 + 9 = 13

cm hai."

Error SDs ko directly add karne mein hai. Standard deviations add nahi hote; variances bhi sirf independence ke under add hote hain, aur spread ko SDs ke simple addition se kabhi combine nahi karte.

Why questions

Pehle apne words mein reason batao, phir check karo.

Hum absolute values lene ki jagah deviations ko square kyun karte hain?

Squaring smooth/differentiable hota hai (calculus aur least-squares ke liye acha), large deviations ko zyada penalise karta hai, aur geometry aur normal distribution se cleanly link karta hai — ye advantages

∣ x ∣

ki kink mein nahi hain.

\overline{x^{2}}

hamesha

(\overset{x}{ˉ})^{2}

se kam se kam utna bada kyun hota hai?

Kyunki unka difference hi variance hai (Figure s02 ka green gap), aur variance squares ka sum

n

se divided hai, isliye

\geq 0

; toh

\overline{x^{2}} - (\overset{x}{ˉ})^{2} \geq 0

hamesha.

n - 1

se divide karna (

n

ki jagah) sample variance ko bada kyun karta hai?

Sample mean

\overset{x}{ˉ}

squared deviations ke sum ko minimize karne ke liye choose hota hai, isliye woh deviations average par thodi si chhoti hoti hain; chhote

n - 1

se divide karna result ko inflate karta hai taaki woh downward bias remove ho sake.

Variance constant add karne ko ignore kyun karta hai lekin multiply karne ka effect square kyun karta hai?

Har point ko

+ c

se slide karna saare pairwise distances unchanged chodta hai, isliye spread untouched rehta hai;

c

se multiply karna har distance ko

c

se stretch karta hai, aur kyunki variance squared distances use karta hai woh

c^{2}

se stretch hote hain (Figure s03).

Jab variance already spread measure karta hai toh SD se kyun pareshaan hon?

Variance squared units (marks², cm²) mein rehta hai jo interpret karna mushkil hai; square root lene se SD original units mein aa jaati hai isliye "2.37 marks ka spread" sense banata hai.

Range spread ka poor measure kyun ho sakta hai variance se compare karte hue?

Range sirf do extreme points use karta hai aur beech ki saari cheezein ignore karta hai, isliye ek akela outlier isko control karta hai; variance har value ki distance centre se use karta hai.

Akele

σ

ki jagah

σ / \overset{x}{ˉ}

(coefficient of variation) kabhi kabhi kyun prefer kiya jaata hai?

SD of

5

cm ek beetle ke liye bada hai lekin ek building ke liye tiny; mean se divide karne par ek unitless relative spread milta hai — Coefficient of Variation dekho — jo tumhe scales ko fairly compare karne deta hai.

Edge cases

Boundary aur degenerate scenarios. Agar tumne ye kabhi nahi dekhe, toh definition poori tarah se tumhari nahi hai.

Ek single data point

x_{1} = 7

(population ke roop mein) ka variance kya hai?

Zero. Mean

7

ke barabar hai, ek hi deviation

0

hai, isliye

σ^{2} = 0/1 = 0

; ek point mein koi spread nahi hoti.

Ek single data point ka sample variance kya hai?

Undefined. Formula

n - 1 = 0

se divide karta hai, aur ek single observation population ke aas-paas spread ke baare mein koi information nahi deta — tum isko estimate nahi kar sakte.

Data hai

5, 5, 5, 5

. Mean, variance aur SD kya hain?

Mean

= 5

, variance

= 0

, SD

= 0

. Saari values identical matlab har deviation

0

hai; ye woh unique case hai jahan variance apna minimum hit karta hai.

Data hai

0, 0, 0, 10

. Kya variance zero hai kyunki zyaatar values zero hain?

Nahi. Mean

2.5

hai, isliye har point isse deviate karta hai;

- 2.5

ke teen deviations aur

+ 7.5

ki ek deviation positive variance deti hai. "Zyaatar zeros" ka matlab "sab identical" nahi hai.

Agar har value negative ho, jaise

- 2, - 4, - 6

, kya variance phir bhi positive ho sakta hai?

Haan. Variance spread measure karta hai, sign nahi; mean

- 4

se deviations

2, 0, - 2

hain, jinke squares positive variance dete hain. Negative data kabhi negative ya zero variance force nahi karta.

Do datasets:

{10, 10, 10}

aur

{9, 10, 11}

. Kiska SD bada hai aur kyun?

Doosre ka. Pehle mein saari values identical hain isliye SD

= 0

; doosre mein mean

10

ke aas-paas genuine spread hai, jo positive SD deta hai.

Ek fixed population ke liye

n \to \infty

hone par,

n

-divisor aur

(n - 1)

-divisor variance ke beech ka gap matter karta hai?

Nahi — dono estimates converge ho jaate hain, kyunki

\frac{n}{n - 1} \to 1

; Bessel's correction sirf chhote samples ke liye noticeably matter karta hai.

Agar tum current mean ke exactly barabar ek point append karo toh SD ka kya hoga?

Mean unchanged rehta hai lekin squared deviations ka sum unchanged rehta hai jabki

n

badhta hai, isliye variance aur SD thode se shrink ho jaate hain — centre par hi ek point add karna average spread ko dilute karta hai.

Recall Jane se pehle ek-line self-test

Zor se bolo: variance kabhi negative nahi hota, squared units mein measure hota hai, jab saara data shift karo toh unchanged rehta hai, aur $c$ se scale karne par == $c^{2}$ == se multiply hota hai. Agar in chaar mein se koi bhi surprising laga, toh woh trap dobara padho.

Connections

Measures of dispersion — variance, standard deviation — woh parent jise ye bank stress-test karta hai.
Mean, Median, Mode — measures of central tendency — woh centre jisse har deviation measure hoti hai.
Coefficient of Variation — upar wale relative-spread "why" question ke liye zaroori.
Normal Distribution — jahan SD 68-95-99.7 rule ke zariye apni interpretation earn karta hai.
Covariance and Correlation — inhi traps ka two-variable generalisation.