Box-and-whisker plots — quartiles, IQR: Exercises

2.7.4 · D4 · Maths › Statistics & Probability — Intermediate › Box-and-whisker plots — quartiles, IQR

Shuru karne se pehle, ek shared reminder — poore toolkit ka, bilkul ground up se — koi bhi symbol use karne se pehle uska meaning fix kar lete hain.

Definition Wo paanch numbers jo tum hamesha compute karte ho

Sort = data ko sabse chhote se sabse bade tak likho. Quartiles positions hain, isliye unsorted list ka koi matlab nahi.
== $Q_{2}$ == = median = woh value jiske neeche aadha data ho.
== $Q_{1}$ == = lower half ka median.
== $Q_{3}$ == = upper half ka median.
== $IQR$ == = $Q_{3} - Q_{1}$ = beech ke 50% ki width.

"Lower half" / "upper half" exactly kya hota hai? $n$ values ki sorted list ke baad:

Agar $n$ odd hai: median ek single value hoti hai; lower half mein har woh value hai jo usse pehle aati hai, upper half mein har woh value hai jo uske baad aati hai. Median khud kisi bhi half mein nahi hoti.
Agar $n$ even hai: koi single middle value nahi hoti, toh hum sorted list ko seedha beech se split karte hain — lower half pehli $n /2$ values hain, upper half aakhri $n /2$ values hain. (Woh do middle values, jinka average $Q_{2}$ hai, apne-apne halves mein rehti hain.)

Har ek idea ki poori build-up ke liye Box-and-whisker plots — quartiles, IQR dekho jo yahan use ki gayi hai.

Level 1 — Recognition

L1·Q1

Neeche diya gaya box plot ek number line pe drawn hai. Picture se seedha padho: median, $Q_{1}$ , $Q_{3}$ , dono whisker ends, aur koi bhi outlier.

Figure — Box-and-whisker plots — quartiles, IQR

Recall Solution

Axis pe marked positions padhne par:

Left whisker end (sabse chhota non-outlier) $= 4$
Box ka left edge $= Q_{1} = 8$
Box ke andar ki line $= Q_{2} = 12$
Box ka right edge $= Q_{3} = 16$
Right whisker end (sabse bada non-outlier) $= 22$
Right whisker ke baad akela dot $= 30 \Rightarrow$ outlier

Yahi L1 ki puri skill hai: ek box plot ek map hai, aur har landmark ka ek hi matlab hai. Inse hum $IQR = 16 - 8 = 8$ bhi padh sakte hain.

L1·Q2

Sorted list $3, 5, 8, 11, 14$ ( $n = 5$ , odd) ke liye, batao woh value kaun si hai jo $Q_{2}$ hai aur kaho ki lower half mein konsi values hain.

Recall Solution

$n = 5$ odd hai, toh $Q_{2}$ position $\frac{n + 1}{2} = \frac{6}{2} = 3$ par single middle value hai, jo 3rd value $= 8$ deti hai. Lower half = median position ke neeche ki values = $3, 5$ (median $8$ kisi bhi half mein nahi aata kyunki $n$ odd hai).

Level 2 — Application

L2·Q1

$12, 4, 7, 15, 9, 2, 20, 6, 11$ ( $n = 9$ ) ke liye $Q_{1}, Q_{2}, Q_{3}$ aur IQR compute karo.

Recall Solution

Sort: $2, 4, 6, 7, 9, 11, 12, 15, 20$ . Median: position $\frac{9 + 1}{2} = 5$ value $9 = Q_{2}$ deta hai. Lower half ( $n$ odd, pehli 4, median excluded): $2, 4, 6, 7 \Rightarrow$ uska median $= \frac{4 + 6}{2} = 5 = Q_{1}$ . Upper half (aakhri 4): $11, 12, 15, 20 \Rightarrow$ median $= \frac{12 + 15}{2} = 13.5 = Q_{3}$ . IQR $= 13.5 - 5 = 8.5$ .

L2·Q2

Data (sorted): $2, 4, 4, 5, 7, 8, 10, 12$ ( $n = 8$ ). Five-number summary (min, $Q_{1}$ , $Q_{2}$ , $Q_{3}$ , max) aur IQR nikalo.

Recall Solution

$n = 8$ even hai, toh hum list ko pehli $n /2 = 4$ values aur aakhri $4$ values mein split karte hain. Median: even $n$ , 4th aur 5th values ka average $= \frac{5 + 7}{2} = 6 = Q_{2}$ . Lower half (pehli 4): $2, 4, 4, 5 \Rightarrow Q_{1} = \frac{4 + 4}{2} = 4$ . Upper half (aakhri 4): $7, 8, 10, 12 \Rightarrow Q_{3} = \frac{8 + 10}{2} = 9$ . Five-number summary: min $= 2$ , $Q_{1} = 4$ , $Q_{2} = 6$ , $Q_{3} = 9$ , max $= 12$ . IQR $= 9 - 4 = 5$ .

L2·Q3

L2·Q2 ke data ke liye outlier fences nikalo, phir batao ki har whisker kahan end hoti hai.

Recall Solution

$IQR = 5$ . Lower fence $= Q_{1} - 1.5 IQR = 4 - 1.5 (5) = 4 - 7.5 = - 3.5$ . Upper fence $= Q_{3} + 1.5 IQR = 9 + 7.5 = 16.5$ . Har value $- 3.5$ aur $16.5$ ke beech hai, toh koi outlier nahi. Whiskers extreme data points tak pahunchti hain: left whisker $2$ tak, right whisker $12$ tak (fences $- 3.5$ ya $16.5$ tak nahi).

Level 3 — Analysis

L3·Q1

Data (sorted): $5, 6, 7, 8, 9, 10, 11, 40$ ( $n = 8$ ). Numerically dikhao ki agar $40$ delete karo toh IQR barely change hota hai, lekin range collapse ho jaata hai. Explain karo ki yeh kya reveal karta hai.

Recall Solution

$40$ ke saath: median $= \frac{8 + 9}{2} = 8.5$ ; lower half $5, 6, 7, 8 \Rightarrow Q_{1} = 6.5$ ; upper half $9, 10, 11, 40 \Rightarrow Q_{3} = \frac{10 + 11}{2} = 10.5$ . IQR $= 10.5 - 6.5 = 4$ . Range $= 40 - 5 = 35$ . $40$ delete karo ( $n = 7$ ): $5, 6, 7, 8, 9, 10, 11$ ; median $= 8$ ; lower half $5, 6, 7 \Rightarrow Q_{1} = 6$ ; upper half $9, 10, 11 \Rightarrow Q_{3} = 10$ . IQR $= 10 - 6 = 4$ . Range $= 11 - 5 = 6$ . Reveal: IQR $4$ par raha (unchanged), jabki range $35$ se crash karke $6$ ho gayi. Range do most extreme points se bani hoti hai, toh ek wild value usse dominate karti hai; IQR sirf inner quartiles use karta hai, isliye woh robust hai. Yahi Outliers and Robust Statistics aur Range and Spread ka point hai.

L3·Q2

$2, 4, 6, 7, 9, 11, 12, 15, 20$ (L2·Q1: $Q_{1} = 5, Q_{2} = 9, Q_{3} = 13.5$ ) ke liye, median ke neeche ka gap ( $Q_{2} - Q_{1}$ ) aur upar ka gap ( $Q_{3} - Q_{2}$ ) compare karo. Yeh asymmetry data ki shape ke baare mein kya kehti hai?

Recall Solution

Lower gap $= Q_{2} - Q_{1} = 9 - 5 = 4$ . Upper gap $= Q_{3} - Q_{2} = 13.5 - 9 = 4.5$ . Upper gap bada hai, toh box ka right half left half se wider hai. High side pe zyada stretch ka matlab hai ki values median ke upar zyada spread hain, toh distribution right-skewed hai (large values ki taraf tail). Aise ek box plot Skewness ko bina kisi formula ke reveal karta hai.

Level 4 — Synthesis

L4·Q1

Tumhe bataya gaya hai ki $n = 10$ ke ek data set mein $Q_{1} = 20$ , $Q_{3} = 32$ hai, aur high side par exactly ek outlier hai. Uski sorted values hain $14, 20, 22, 25, 28, 30, 31, 32, 35, v$ Sabse chhota integer $v$ nikalo jo last value ko genuine upper outlier banaye, aur batao ki right whisker phir kahan end hogi.

Recall Solution

$IQR = Q_{3} - Q_{1} = 32 - 20 = 12$ . Upper fence $= Q_{3} + 1.5 IQR = 32 + 1.5 (12) = 32 + 18 = 50$ . Ek point outlier hota hai jab woh fence se strictly greater ho: $v > 50$ . Sabse chhota integer $v = 51$ hai. $v = 51$ ke saath, fence ke andar sabse badi value $35$ hai, toh right whisker $35$ par end hoti hai aur $51$ ek akele dot ki tarah draw hota hai. (Check: hume $Q_{1}, Q_{3}$ as given rakhne hain. Sirf last value change karne se $Q_{3}$ nahi hilta: upper half $30, 31, 32, 35, v$ hai, jiska median middle value $32$ hai chahe $v$ kitna bhi bada ho — consistent hai.)

L4·Q2

Ek 7-value data set (integers) banao jiska median $10$ ho, IQR $6$ ho, aur jisme koi outlier na ho. Phir apna set verify karo.

Recall Solution

Design plan: $n = 7$ odd matlab halves se median drop karo. Chahiye $Q_{2} = 10$ , aur $Q_{3} - Q_{1} = 6$ . $Q_{1} = 8, Q_{3} = 14$ pick karo (difference $6$ ). Lower half (3 values) ka median $8$ hona chahiye; upper half (3 values) ka median $14$ . Ek valid set: $6, 8, 9, 10, 12, 14, 16$ . Verify: sorted ✓. Median position $4$ deta hai $10 = Q_{2}$ ✓. Lower half $6, 8, 9$ ka median $8 = Q_{1}$ ✓. Upper half $12, 14, 16$ ka median $14 = Q_{3}$ ✓. IQR $= 14 - 8 = 6$ ✓. Fences: lower $= 8 - 1.5 (6) = 8 - 9 = - 1$ ; upper $= 14 + 9 = 23$ . Saara data $[- 1, 23]$ mein hai, toh koi outlier nahi ✓. (Bahut saare answers possible hain; koi bhi set jo teeno targets match kare, sahi hai.)

Level 5 — Mastery

L5·Q1 (degenerate / edge case)

Data $7, 7, 7, 7, 7, 7$ ( $n = 6$ , sab identical) consider karo. Five-number summary, IQR, aur fences compute karo. Box plot kaisa dikhega?

Recall Solution

Sorted (already): sab $7$ . Median $= \frac{7 + 7}{2} = 7 = Q_{2}$ . Lower half $7, 7, 7 \Rightarrow Q_{1} = 7$ . Upper half $7, 7, 7 \Rightarrow Q_{3} = 7$ . $IQR = 7 - 7 = 0$ . Fences: lower $= 7 - 1.5 (0) = 7$ ; upper $= 7 + 1.5 (0) = 7$ . Dono fences exactly $7$ par hain. Plot: box ki zero width hai (ek single vertical line at $7$ ), median line uske upar, whiskers ki length zero. Har point $7$ ke barabar hai, jo fence par hai (usse bahar nahi), toh kuch bhi flag nahi hota. Zero IQR "bilkul koi spread nahi" ki honest picture hai.

L5·Q2 (four-value halves & tie-breaking)

Data (sorted): $1, 1, 2, 2, 2, 8, 9, 9, 10, 10$ ( $n = 10$ ). Five-number summary, IQR, fences nikalo, koi bhi outliers list karo, aur dono whisker ends batao.

Recall Solution

Median: even $n$ , 5th aur 6th ka average $= \frac{2 + 8}{2} = 5 = Q_{2}$ . Lower half (pehli 5): $1, 1, 2, 2, 2 \Rightarrow$ median (3rd) $= 2 = Q_{1}$ . Upper half (aakhri 5): $8, 9, 9, 10, 10 \Rightarrow$ median (3rd) $= 9 = Q_{3}$ . IQR $= 9 - 2 = 7$ . Fences: lower $= 2 - 1.5 (7) = 2 - 10.5 = - 8.5$ ; upper $= 9 + 1.5 (7) = 9 + 10.5 = 19.5$ . Saari values $[- 8.5, 19.5]$ ke andar hain, toh koi outlier nahi. Whiskers: left se $1$ tak (min), right se $10$ tak (max). Dhyan do ki median $5$ ek aisi value hai jo data mein appear hi nahi karti — yeh theek hai, median ek position-based cut point hai, observation hona zaroori nahi. Yeh distribution clearly bimodal hai (ek low clump aur ek high clump), phir bhi summary ise cleanly handle kar leti hai. Dekho Percentiles and Quantiles.

L5·Q3 (reverse-engineering from a summary)

Ek report kehti hai: min $= 3$ , $Q_{1} = 10$ , median $= 15$ , $Q_{3} = 22$ , max $= 60$ , aur kam se kam ek outlier hai. Prove karo ki ek outlier zaroor exist karta hai aur identify karo ki kaun sa extreme offender hai.

Recall Solution

$IQR = Q_{3} - Q_{1} = 22 - 10 = 12$ . Upper fence $= 22 + 1.5 (12) = 22 + 18 = 40$ . Lower fence $= 10 - 1.5 (12) = 10 - 18 = - 8$ . Max $= 60 > 40$ , toh woh upper fence se bahar hai aur outlier hai (offender high side par hai). Min $= 3$ satisfy karta hai $3 > - 8$ , toh low end theek hai. Isliye ek outlier zaroor exist karta hai, aur woh $60$ ke paas wali badi value hai. Hum exact data recover nahi kar sakte, lekin summary akele hi yeh conclusion force karti hai — five-number summary ki power ek compact fingerprint ke roop mein.

Recall checklist

Connections

Box-and-whisker plots — quartiles, IQR — parent recipe jise yeh drills karte hain.
Median and Measures of Central Tendency — har quartile ek median hai.
Range and Spread — IQR se contrast L3·Q1 mein.
Outliers and Robust Statistics — L2–L5 mein poora fence rule.
Percentiles and Quantiles — quartiles as the 25/50/75 cut points.
Skewness — L3·Q2 mein half-widths se padha jaata hai.
Normal Distribution — $1.5$ multiplier ki origin.