Box-and-whisker plots — quartiles, IQR: Visual walkthrough

2.7.4 · D2 · Maths › Statistics & Probability — Intermediate › Box-and-whisker plots — quartiles, IQR

Hum poore walkthrough mein ek hi running dataset use karenge taaki pictures connect hon:

$8, 2, 15, 5, 6, 3, 9, 7, 30, 4$

Das numbers, jaanbujhkar jumbled, ek suspiciously badi value ( $30$ ) ke saath. End tak tumhe pata hoga ke har ek exactly kahan land karta hai.

Step 1 — Numbers ko line mein lagao (sort karo)

KYA. Hum list ko smallest se largest tak rearrange karte hain:

$2, 3, 4, 5, 6, 7, 8, 9, 15, 30$

KYU. Ek quartile ek position hai — "list mein ek-quarter aage ki value." Yeh phrase ek scrambled pile mein meaningless hai. Jis waqt "ek-quarter aage" ka kuch matlab hona ho, data zaroor order mein hona chahiye. Socho log queue mein height ke hisaab se khade hain: "line mein 1/4 neeche wala person" tabhi sense karta hai jab woh line up ho jaayein.

PICTURE. Figure dekho: jumbled numbers (upar ki row) apne sorted slots mein gir jaate hain (neeche ki row). Har slot ke neeche ek position index $1, 2, \dots, 10$ likha hai — woh indices woh ruler hai jisse hum quartiles measure karte hain.

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

Step 2 — Median $Q_{2}$ dhundho (central cut)

KYA. $n = 10$ ke saath (ek even count), koi single middle value nahi hai — middle positions $5$ aur $6$ ke beech mein padti hai. Hum un dono ko average karte hain:

$Q_{2} = \frac{6 position 5 + 7 position 6}{2} = 6.5$

$Q_{2}$ — "Q-two" padho, median: woh value jiske neeche aadha data hai, aadha upar.
The $6$ and $7$ — dono taraf se close karne par bache hue do innermost values (sorted list ki positions $5$ aur $6$ ).
$2$ se divide karna — unke beech exact midpoint leta hai, kyunki koi data point wahan baithta nahi.

Do ko average kyun karein? Even count ke saath queue mein koi akela centre person nahi hota; fair "middle" do central logon ke beech ki gap hai, isliye hum difference split karte hain. (Agar $n$ odd hota toh hum simply single middle value lete — Step 6 dekho.)

PICTURE. Do arrows dono ends se inward march karte hain aur $6$ – $7$ gap mein milte hain. Dashed pink line $Q_{2} = 6.5$ ko gap mein baithte hua mark karti hai, kisi bhi data point ko touch kiye bina.

Step 3 — Do halves mein split karo

KYA. Median cut sorted list ko ek lower half aur ek upper half mein divide karta hai:

$lower half (5 values) 2, 3, 4, 5, 6 upper half (5 values) 7, 8, 9, 15, 30$

KYU. Plan hai "har half ka median," toh pehle humare paas do clean halves hone chahiye. Kyunki $n = 10$ even hai, cut $6$ (position $5$ ) aur $7$ (position $6$ ) ke beech land karta hai, aur har half ko exactly $n /2 = 5$ values milti hain — koi value orphan nahi hoti. (Jab $n$ odd hota hai toh single median value kisi bhi half ki nahi hoti; Step 6 woh case dikhata hai.)

PICTURE. Ek vertical pink bar $6$ – $7$ gap ke through girta hai. Uske left mein sab kuch blue lower half hai; right mein sab kuch yellow upper half hai. Har ek ke neeche printed counts dekho: $5$ aur $5$ , balanced.

Step 4 — Har half ka median $Q_{1}$ aur $Q_{3}$ deta hai

KYA. "Middle dhundho" wali trick har half ke andar repeat karo. Har half mein $5$ values hain (odd), isliye har ek ka ek genuine single middle hai:

$Q_{1} = median of {2, 3, middle 4, 5, 6} = 4$ $Q_{3} = median of {7, 8, middle 9, 15, 30} = 9$

$Q_{1}$ — "Q-one", lower quartile: lower half ka median, isliye saare data ka roughly ek quarter iske neeche hai.
$Q_{3}$ — "Q-three", upper quartile: upper half ka median, isliye roughly three-quarters iske neeche hai.

Half-ka-median rule kyun (naa ki "position $n /4$ ")? Jaise parent note ki steel-man warning thi: raw index counting points ke beech land karti hai aur textbooks mein agree nahi karti. Har half ka median lena unambiguous hai — yeh wahi reliable operation hai jis par hum already trust karte hain, bas do baar apply kiya gaya hai.

PICTURE. Step 2 ke inward-arrows ke do mini-versions, ek har half mein. Blue arrows $4$ par milte hain; yellow arrows $9$ par milte hain. Ab teen cuts $Q_{1} = 4$ , $Q_{2} = 6.5$ , $Q_{3} = 9$ data ko four equal-count blocks mein kaatte hain.

Step 5 — IQR aur 1.5×IQR fences

KYA. Interquartile range $Q_{1}$ se $Q_{3}$ tak ki width hai:

$IQR = 9 Q_{3} - 4 Q_{1} = 5$

Phir hum fences banate hain — "normal" ki boundary:

$Lower fence = 4 Q_{1} - yardsticks 1.5 \times 5 IQR = 4 - 7.5 = - 3.5$ $Upper fence = 9 Q_{3} + 1.5 \times 5 IQR = 9 + 7.5 = 16.5$

IQR — middle-50% box ki width; humara natural "step size."
$1.5$ — kitne box-widths hum bahar step karte hain kisi point ko weird kehne se pehle (ek convention jo normal data ka ≈0.7% flag karta hai).
Subtraction/addition — $Q_{1}$ ke left aur $Q_{3}$ ke right mein us distance se walk karo.

IQR ko yardstick kyun use karein? Kyunki IQR robust hai: yeh extreme outer quarters ko ignore karta hai, isliye fence usi outlier se bahar nahi phool sakti jise woh pakadne ki koshish kar rahi hai. Plain range use karna giant $30$ ko apna khud ka "normal" define karne deta.

PICTURE. Box ( $Q_{1}$ se $Q_{3}$ tak) draw ki jaati hai, phir $- 3.5$ aur $16.5$ par do fence-flags $1.5$ box-widths bahar step karke lagaaye jaate hain. Value $30$ upper flag ke saaf past baithti hai.

Step 6 — Edge case: odd $n$ (median kahan jaati hai?)

KYA. $30$ drop karo toh $n = 9$ (odd): $2, 3, 4, 5, 6, 7, 8, 9, 15$ . Ab middle ek single value hai position $\frac{n + 1}{2} = 5$ par, yaani $Q_{2} = 6$ . Khaas baat, woh $6$ halving se pehle remove ki jaati hai:

$lower half 2, 3, 4, 5 [6] upper half 7, 8, 9, 15$

$Q_{1} = \frac{3 + 4}{2} = 3.5, Q_{3} = \frac{8 + 9}{2} = 8.5$

Median ko exclude kyun karein? Odd $n$ ke saath middle value kisi bhi side ki nahi hoti — woh khud pivot hai. Agar hum use ek half se chipka dein, us half mein ek extra member hoga aur chaar blocks equal-count nahi rahenge. Use bahar rakhne se dono halves $4$ values par rehti hain.

PICTURE. Single median value $6$ boxed hai aur greyed out hai, arrows dikhate hain ke woh step aside kar rahi hai taaki dono equal halves size $4$ ki dono taraf ban sakein.

Step 7 — Finished plot assemble karo

KYA. Original $n = 10$ data ke liye sab kuch ek saath rakkho. Whisker fence tak nahi pahunchti — woh us last real data point par rukti hai jo abhi bhi usके andar hai:

Left whisker → lower fence $= 2$ ke $\geq$ wali smallest value.
Box: left edge $Q_{1} = 4$ , line $Q_{2} = 6.5$ , right edge $Q_{3} = 9$ .
Right whisker → upper fence $= 16.5$ ke $\leq$ wali largest value. Woh $15$ hai (kyunki $16.5 \geq 15$ ), naa ki $30$ .
$30 > 16.5$ → ek akele dot ke roop mein draw kiya: ek outlier.

Whisker $15$ par kyun rukti hai, $16.5$ par nahi. Ek whisker ek real observation mark karti hai; fence ek imaginary threshold hai. Last observed value jo fence clear karti hai woh $15$ hai, isliye arm exactly wahan pahunchti hai aur $30$ ek dot mein exile ho jaata hai.

PICTURE. Complete box-and-whisker: whisker $\to$ box $\to$ whisker, faint fence lines uske peeche aur $30$ right whisker ke aage ek pink dot ki tarah hovering karta hua.

Ek-picture summary

Har pehla step, ek single ladder mein stack kiya hua: sort karo $\to$ $Q_{2}$ par kato $\to$ har half ko $Q_{1}, Q_{3}$ par kato $\to$ IQR measure karo $\to$ $1.5 \times$ IQR step karo $\to$ box, whiskers, outlier draw karo.

Recall Feynman retelling — poora walkthrough simple words mein

Pehle maine sabko ek line mein khada karaya, shortest se tallest, kyunki "ek-quarter aage" ka matlab tabhi hota hai jab woh line up ho jaayein. Maine exactly beech wale person ko dhundha — even crowd ke saath woh do middle logon ke beech ki gap hai, isliye maine un dono ke beech wala point choose kiya: wahi $Q_{2}$ hai. Us middle cut ne line ko ek shorter half aur ek taller half mein split kar diya. Maine wahi middle-dhundhne wali trick har half ke andar ki, mujhe do aur cuts mile, $Q_{1}$ aur $Q_{3}$ — ab line chaar equal groups mein hai. Un do inner cuts ke beech ki distance IQR hai, "middle bunch" ki width. Weirdos pakadne ke liye, maine un inner cuts ke past un widths ka dedh guna step kiya aur ek fence lagaai. Jo bhi fence ke past tha woh suspicious loner tha (woh $30$ tha). Finally maine $Q_{1}$ se $Q_{3}$ tak ek box draw ki beech mein ek line ke saath, dono taraf last normal person tak arms failaaye, aur loner ko ek lonely dot ke roop mein draw kiya. Jab crowd odd ho, toh single middle person step aside karta hai aur kisi bhi half mein nahi aata, isliye halves equal rehti hain.

Connections

Box-and-whisker plots — quartiles, IQR — woh parent jise yeh walkthrough build karta hai.
Median and Measures of Central Tendency — $Q_{2}$ median hai, yahan teen baar apply kiya gaya.
Range and Spread — Step 5 mein IQR vs plain range.
Outliers and Robust Statistics — $1.5 \times$ IQR fence logic.
Percentiles and Quantiles — quartiles 25/50/75 cut points ke roop mein.
Skewness — off-centre median line skew hint karti hai.
Normal Distribution — kyun $1.5$ ≈0.7% flag karta hai.