Measures of dispersion — variance, standard deviation: Exercises

2.7.3 · D4 · Maths › Statistics & Probability — Intermediate › Measures of dispersion — variance, standard deviation

Shuru karne se pehle, ek shared reminder un chaar symbols ka jo hum use karte hain, taaki kuch bhi unexplained na rahe:

Level 1 — Recognition

L1.1 Ek dataset ka variance $σ^{2} = 49 cm^{2}$ hai. Standard deviation aur uski units batao.

Recall Solution L1.1

SD, variance ka square root hota hai: $σ = 49 = 7$ . Variance $cm^{2}$ mein tha, toh root lene se hum data ki apni units par wapas aa jaate hain: $σ = 7 cm$ .

L1.2 Data $5, 5, 5, 5$ ke liye, variance kya hai? Bina koi sum calculate kiye jawab do.

Recall Solution L1.2

Har value identical hai, toh har deviation $d_{i} = 5 - 5 = 0$ hai. Zeros ka average zero hota hai: $σ^{2} = 0$ . (Ye ek maatra case hai jahan variance exactly 0 hota hai.)

L1.3 Variance shortcut kaun sa expression hai: (a) $(\overset{x}{ˉ})^{2} - \overline{x^{2}}$ , ya (b) $\overline{x^{2}} - (\overset{x}{ˉ})^{2}$ ?

Recall Solution L1.3

(b) $\overline{x^{2}} - (\overset{x}{ˉ})^{2}$ . Naam padho: squares ka mean minus mean ka square. Ye $\geq 0$ hona chahiye; option (a) negative hoga, jo koi spread kabhi nahi ho sakta.

Level 2 — Application

L2.1 Data (population): $3, 7, 7, 19$ . Pehle $\overset{x}{ˉ}$ nikalo, phir definition use karke $σ^{2}$ , phir $σ$ .

Recall Solution L2.1

Mean: $\overset{x}{ˉ} = \frac{3 + 7 + 7 + 19}{4} = \frac{36}{4} = 9$ . Deviations: $3 - 9 = - 6$ , $7 - 9 = - 2$ , $7 - 9 = - 2$ , $19 - 9 = 10$ . Squares: $36, 4, 4, 100$ . Sum $= 144$ . Variance: $σ^{2} = \frac{144}{4} = 36$ . SD: $σ = 36 = 6$ .

L2.2 Same data $3, 7, 7, 19$ , ab shortcut $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ se.

Recall Solution L2.2

$\sum x_{i}^{2} = 9 + 49 + 49 + 361 = 468$ , toh $\overline{x^{2}} = \frac{468}{4} = 117$ . $(\overset{x}{ˉ})^{2} = 9^{2} = 81$ . $σ^{2} = 117 - 81 = 36$ ✓ — L2.1 se identical, bilkul waisa jैसा algebra ne promise kiya tha.

L2.3 Data (population): $1, 2, 3, 4, 5$ . $σ^{2}$ aur $σ$ nikalo.

Recall Solution L2.3

$\overset{x}{ˉ} = \frac{15}{5} = 3$ . Deviations: $- 2, - 1, 0, 1, 2$ ; squares $4, 1, 0, 1, 4$ ; sum $= 10$ . $σ^{2} = \frac{10}{5} = 2$ . $σ = 2 \approx 1.414$ .

Level 3 — Analysis

L3.1 3 volunteers ke reaction times (seconds), saare logon ke sample ke roop mein: $0.20, 0.30, 0.40$ . Sample variance $s^{2}$ aur sample SD $s$ nikalo.

Recall Solution L3.1

$\overset{x}{ˉ} = \frac{0.20 + 0.30 + 0.40}{3} = 0.30$ . Deviations: $- 0.10, 0, 0.10$ ; squares $0.01, 0, 0.01$ ; sum $= 0.02$ . Sample → $n - 1 = 2$ se divide karo: $s^{2} = \frac{0.02}{2} = 0.01$ . $s = 0.01 = 0.10$ s.

L3.2 Yahi teen numbers $0.20, 0.30, 0.40$ ab ek chhote experiment ki poori population hain. $σ^{2}$ nikalo. L3.1 se compare karo aur difference ki direction explain karo.

Recall Solution L3.2

Population → $n = 3$ se divide karo: $σ^{2} = \frac{0.02}{3} \approx 0.00667$ . Ye sample $s^{2} = 0.01$ se chhota hai. Same sum of squares ( $0.02$ ) ko bade number ( $3$ vs $2$ ) se divide karne par chhoti value milti hai. Bessel ka $n - 1$ jaanbujhkar sample estimate ko inflate karta hai taaki woh downward bias correct ho jo $\overset{x}{ˉ}$ use karne se aata hai (jo ki us sum ko hi minimize karta hai).

L3.3 Ek student " $s^{2} = 0.02/3$ " report karta hai 3 ke sample ke liye. Kya divisor sahi hai? Sahi value do.

Recall Solution L3.3

Nahi. Sample ke liye, $n = 3$ se nahi, $n - 1 = 2$ se divide karo. Sahi sample variance hai $s^{2} = 0.02/2 = 0.01$ . $3$ se divide karne par population variance $\approx 0.00667$ milta, jo broader population ki spread ko underestimate karta.

Level 4 — Synthesis

L4.1 (Shift invariance). Dataset $A = {10, 20, 30}$ ka population variance $σ_{A}^{2}$ hai. Ek naya set har value mein $c = 100$ add karta hai: $B = {110, 120, 130}$ . $σ_{A}^{2}$ aur $σ_{B}^{2}$ nikalo aur property $Var (x + c) = Var (x)$ confirm karo.

Recall Solution L4.1

A: $\overset{x}{ˉ}_{A} = 20$ ; deviations $- 10, 0, 10$ ; squares $100, 0, 100$ ; sum $200$ ; $σ_{A}^{2} = 200/3 \approx 66.67$ . B: $\overset{x}{ˉ}_{B} = 120$ ; deviations $- 10, 0, 10$ (bilkul same, kyunki dono values aur mean $+ 100$ se shift hue); squares identical; $σ_{B}^{2} = 200/3 \approx 66.67$ . Equal ✓. Ek constant add karna har point aur mean ko same amount se slide kar deta hai, toh deviations — aur is tarah spread — unchanged rehti hai.

L4.2 (Scaling). $A = {10, 20, 30}$ lo jiska $σ_{A}^{2} = 200/3$ hai. Har value ko $c = 3$ se multiply karo: $C = {30, 60, 90}$ . Scaling rule se $σ_{C}^{2}$ aur $σ_{C}$ predict karo, phir verify karo.

Recall Solution L4.2

Predict: $Var (c x) = c^{2} Var (x)$ , toh $σ_{C}^{2} = 3^{2} \cdot \frac{200}{3} = 9 \cdot \frac{200}{3} = 600$ . Aur $σ_{C} = ∣ c ∣ σ_{A} = 3 200/3 = 3 \cdot 66.67 \approx 24.49$ . Directly verify karo: $\overset{x}{ˉ}_{C} = 60$ ; deviations $- 30, 0, 30$ ; squares $900, 0, 900$ ; sum $1800$ ; $σ_{C}^{2} = 1800/3 = 600$ ✓. $3$ se stretch karne par har distance triple hua; squared distances $3^{2} = 9$ se badhe.

L4.3 (Dono combine karo). $A$ se $D = 3 x + 100 = {130, 160, 190}$ banao. Deviations scratch se recompute kiye bina, dono properties use karke $σ_{D}^{2}$ nikalo.

Recall Solution L4.3

$+ 100$ shift variance par kuch nahi karta (shift invariance). $\times 3$ variance ko $3^{2} = 9$ se multiply karta hai. $σ_{D}^{2} = 9 \cdot σ_{A}^{2} = 9 \cdot \frac{200}{3} = 600$ . ( $C$ ke jaisa hi, kyunki constant $+ 100$ spread ke liye irrelevant hai.) SD: $σ_{D} = 3 200/3 \approx 24.49$ .

Figure — Measures of dispersion — variance, standard deviation

Upar red band dekho: $A$ aur $A + c$ ke liye ye same width hai (pure slide) lekin $3 A$ ke liye factor $3$ se stretch hoti hai — ye L4.1–L4.3 ke peeche ki picture hai.

Level 5 — Mastery

L5.1 ( $\overline{x^{2}} \geq (\overset{x}{ˉ})^{2}$ prove karo). Dikhao ki squares ka mean kabhi mean ke square se neeche nahi hota, sirf shortcut aur variance ki definition use karke.

Recall Solution L5.1

Definition se, $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}$ . Ye squares ka average hai, aur har square $(x_{i} - \overset{x}{ˉ})^{2} \geq 0$ hai, toh unka average $\geq 0$ hai; is tarah $σ^{2} \geq 0$ . Shortcut se, $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ . Combine karo: $\overline{x^{2}} - (\overset{x}{ˉ})^{2} \geq 0 \Rightarrow \overline{x^{2}} \geq (\overset{x}{ˉ})^{2}$ . Equality tab hoti hai iff har $(x_{i} - \overset{x}{ˉ})^{2} = 0$ , yaani saari values identical hों. $■$

L5.2 (Data reverse-engineer karo). Ek two-point population ${a, b}$ jahan $a < b$ ka mean $\overset{x}{ˉ} = 5$ aur SD $σ = 3$ hai. $a$ aur $b$ nikalo.

Recall Solution L5.2

Do points ke liye, dono deviations ki magnitude equal hoti hai $\frac{b - a}{2}$ (har ek midpoint mean se exactly aadhe gap par hota hai). $σ^{2} = \frac{1}{2} [(\frac{b - a}{2})^{2} + (\frac{b - a}{2})^{2}] = (\frac{b - a}{2})^{2}$ . Toh $σ = \frac{b - a}{2} = 3 \Rightarrow b - a = 6$ . Midpoint $\overset{x}{ˉ} = 5$ ke saath: $a = 5 - 3 = 2$ , $b = 5 + 3 = 8$ . Check: ${2, 8}$ , mean $5$ , deviations $\pm 3$ , squares $9, 9$ , $σ^{2} = 18/2 = 9$ , $σ = 3$ ✓.

L5.3 (Kaun zyada spread hai?). Do classes: Class P ke scores ${40, 50, 60}$ ; Class Q ke scores ${20, 50, 80}$ . Dono ka same mean hai. Bina poori computation ke argue karo ki kiski variance badi hai, phir numerically confirm karo.

Recall Solution L5.3

Dono ka $\overset{x}{ˉ} = 50$ hai. Q ki values $50$ se zyada door baithi hain ( $\pm 30$ ) P ki tulna mein ( $\pm 10$ ), toh Q ki variance badi honi chahiye. P: deviations $- 10, 0, 10$ ; squares $100, 0, 100$ ; sum $200$ ; $σ_{P}^{2} = 200/3 \approx 66.67$ . Q: deviations $- 30, 0, 30$ ; squares $900, 0, 900$ ; sum $1800$ ; $σ_{Q}^{2} = 1800/3 = 600$ . $σ_{Q}^{2} = 600 > 66.67 = σ_{P}^{2}$ ✓. Actually Q ki spread P ki $3 \times$ hai, toh uski variance $3^{2} = 9 \times$ badi hai — scaling rule chhupe hue roop mein.

Connections

Mean, Median, Mode — measures of central tendency — har exercise centre locate karne se shuru hoti hai.
Coefficient of Variation — L5.3 ke liye, $σ$ ko $\overset{x}{ˉ}$ se divide karna spread ko fairly compare karta hai jab means alag hon.
Normal Distribution — jo SD tum yahan compute karte ho, woh 68-95-99.7 rule ka ruler hai.
Covariance and Correlation — L4 ki scaling generalize hoti hai ki paired variables kaise co-vary karti hain.
Least Squares Regression — L5.1 ka " $\overset{x}{ˉ}$ sum of squares minimize karta hai" regression ka seed hai.

Exercises — Measures of dispersion — variance, standard deviation