Box-and-whisker plots — quartiles, IQR: Worked examples

2.7.4 · D3 · Maths › Statistics & Probability — Intermediate › Box-and-whisker plots — quartiles, IQR

The scenario matrix

Kuch bhi work karne se pehle, aao hum har case class list karte hain jo is topic mein aa sakti hai. Isse ek "shapes of data" ki checklist ki tarah socho — har row ek aisi situation hai jahan procedure branches hoti hai (kuch alag karta hai).

#	Case class	Isme tricky kya hai	Example jo cover karta hai
A	$n$ odd	Median ek real data point hai; dono halves se usse drop karo	Ex 1
B	$n$ even	Median do points ka average hai; halves cleanly split hoti hain	Ex 2
C	Half-size odd ( $n$ leaves 3-per-half etc.)	$Q_{1}, Q_{3}$ khud single middle values hain	Ex 1, Ex 3
D	Half-size even ( $Q_{1}, Q_{3}$ averages hain)	$Q_{1}, Q_{3}$ ko apna averaging step chahiye	Ex 3
E	Repeated values	Repeats phir bhi alag positions ke roop mein count hote hain	Ex 4
F	Outlier high side par	Upper fence use pakadta hai; whisker short ruk jaata hai	Ex 2, Ex 5
G	Outlier low side par	Lower fence use pakadta hai (log is side ko bhool jaate hain!)	Ex 5
H	Sab values identical (degenerate)	$IQR = 0$ ; fences value par collapse ho jaati hain	Ex 6
I	Sabse chota possible set ( $n = 4$ )	Halving rule ke liye boundary case	Ex 3
J	Real-world word problem (units, meaning)	Sirf compute nahi, interpret bhi karna hoga	Ex 7
K	Exam twist — $Q_{1}$ diya hai, missing value dhundho	Rule ko ulta use karo	Ex 8

Upar ka har cell neeche ke kam se kam ek worked example se cover hota hai. Chalo shuru karte hain.

Ex 1 — Case A + C: odd $n$ , single-value quartiles

Step 1 — Ascending sort karo. $6, 8, 9, 11, 14, 15, 22$ . Yeh step kyun? Quartiles position se define hote hain ek sorted queue mein; position tab tak meaningless hain jab tak sort na ho.

Step 2 — Median $Q_{2}$ . Position $\frac{n + 1}{2} = \frac{7 + 1}{2} = 4$ → 4th value $= 11$ . Yeh step kyun? 7 items mein exactly ek item dead centre mein hota hai — teen neeche, teen upar.

Step 3 — Split karo, median drop karo. Lower half $= 6, 8, 9$ ; upper half $= 14, 15, 22$ . Yeh step kyun? $n$ odd hai, toh median $11$ kisi bhi side mein nahi aata — isse include karne par halves unequal ho jaayenge.

Step 4 — $Q_{1}, Q_{3}$ = halves ke medians. Har half mein 3 values hain → middle wala. $Q_{1} = 8$ , $Q_{3} = 15$ . Yeh step kyun? $Q_{1}$ "lower half ka median" hai — yeh parent note ka median-of-half rule hai.

Step 5 — IQR. $IQR = Q_{3} - Q_{1} = 15 - 8 = 7$ .

Verify: $Q_{1} = 8$ ke neeche sirf ${6}$ hai (1 of 7 ≈ 14%, chote $n$ ke liye 25% ke paas ✓). Median $11$ ke neeche ${6, 8, 9}$ = 3 of 7 ≈ 43% ≈ 50% ✓. Saare quartiles actual data points hain, jaise odd $n$ ke liye expect kiya tha. ✓

Ex 2 — Case B + F: even $n$ , high-side outlier

Step 1 — Pehle se sorted hai. $0, 2, 2, 3, 4, 5, 6, 40$ . Ascending confirm ✓. Yeh step kyun? Kabhi assume mat karo — ek hidden unsorted value sab kuch tod deta hai.

Step 2 — Median $Q_{2}$ . Even $n$ : positions 4 aur 5 ka average → $\frac{3 + 4}{2} = 3.5$ . Yeh step kyun? Even $n$ ke liye koi single middle item nahi hota; fair centre dono middle values ke beech mein hota hai.

Step 3 — Dono halves of 4 mein split karo. Lower $= 0, 2, 2, 3$ ; upper $= 4, 5, 6, 40$ . Yeh step kyun? Even $n$ cleanly do equal groups mein divide hota hai; median ek data point nahi hai toh kuch drop nahi hota.

Step 4 — $Q_{1}, Q_{3}$ (har half even hai → average karo). Lower middle pair $2, 2$ → $Q_{1} = \frac{2 + 2}{2} = 2$ . Upper middle pair $5, 6$ → $Q_{3} = \frac{5 + 6}{2} = 5.5$ . Yeh step kyun? Har half mein 4 items hain — even — toh uska median khud do middles ka average hai.

Step 5 — IQR aur fences. $IQR = 5.5 - 2 = 3.5$ $Upper fence = Q_{3} + 1.5 IQR = 5.5 + 1.5 (3.5) = 5.5 + 5.25 = 10.75$ $Lower fence = Q_{1} - 1.5 IQR = 2 - 5.25 = - 3.25$ Yeh step kyun? Fences "1.5 boxes out" ka yardstick hain — jo bhi isse bahar ho woh unusual hai.

Step 6 — Outliers flag karo. $40 > 10.75$ → outlier. Right whisker largest value $\leq 10.75$ par rukti hai, jo $6$ hai. Yeh step kyun? Whiskers last non-outlier tak jaati hain, fence tak nahi.

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

Verify: Saari fences consistent hain (lower < $Q_{1}$ < median < $Q_{3}$ < upper ✓). $40$ upper fence ke upar hai; kuch bhi $- 3.25$ ke neeche nahi, toh koi low outlier nahi. Right whisker end $= 6$ . ✓

Ex 3 — Case D + I: sabse chota set ( $n = 4$ ), averaged quartiles

Step 1 — Sort karo. $8, 12, 16, 20$ . Yeh step kyun? Har baar same reason — positions ko order chahiye.

Step 2 — Median $Q_{2}$ . Even $n = 4$ : 2nd aur 3rd values ka average $= \frac{12 + 16}{2} = 14$ . Yeh step kyun? Koi single middle nahi; centre $12$ aur $16$ ke beech mein hai.

Step 3 — Split karo. Lower $= 8, 12$ ; upper $= 16, 20$ . Yeh step kyun? $n$ even → clean 2/2 split, koi value drop nahi.

Step 4 — Quartiles averages ke roop mein. $Q_{1} = \frac{8 + 12}{2} = 10$ , $Q_{3} = \frac{16 + 20}{2} = 18$ . Yeh step kyun? Har half mein 2 values hain (even) → median unka average hai. Yeh Case D hai: quartiles khud averaging maangti hain.

Step 5 — IQR. $18 - 10 = 8$ .

Verify: $Q_{1} < Q_{2} < Q_{3}$ → $10 < 14 < 18$ ✓, spacing symmetric kyunki data symmetric hai ✓. Yeh boundary case hai: $n = 4$ sabse kam points hain jinke liye "har half ka median" two-element halves produce karta hai. ✓

Ex 4 — Case E: repeated values

Step 1 — Pehle se sorted, $n = 9$ (odd). Confirm ✓. Yeh step kyun? Pehle order — repeats apni jagah par rehte hain.

Step 2 — Median $Q_{2}$ . Position $\frac{9 + 1}{2} = 5$ → 5th value $= 8$ . Yeh step kyun? Nine items → 5th exact centre hai. Hum positions count karte hain, distinct values nahi — teen 7s positions 2, 3, 4 par separately occupy karte hain.

Step 3 — Split karo, median drop karo. Lower $= 6, 7, 7, 7$ ; upper $= 8, 9, 10, 10$ . Yeh step kyun? Odd $n$ → median (position 5) kisi bhi half mein nahi aata.

Step 4 — Quartiles. Lower half mein 4 values hain (even) → $Q_{1} = \frac{7 + 7}{2} = 7$ . Upper half → $Q_{3} = \frac{9 + 10}{2} = 9.5$ . Yeh step kyun? Repeats collapse nahi hote: lower half ka middle pair 2nd aur 3rd 7 hai, aur $\frac{7 + 7}{2} = 7$ .

Step 5 — IQR. $9.5 - 7 = 2.5$ .

Verify: $Q_{1} = 7$ (repeated value legitimately ek quartile ke equal ho sakta hai), $Q_{3} = 9.5$ , IQR $= 2.5$ ; ordering $7 \leq 8 \leq 9.5$ holds ✓.

Ex 5 — Case F + G: DONO sides par outliers

Step 1 — Sort karo. $- 50, 10, 11, 12, 13, 14, 15, 60$ — pehle se ascending, $n = 8$ . Yeh step kyun? Order confirm karo; negative pehle aana chahiye.

Step 2 — Median $Q_{2}$ . 4th & 5th ka average $= \frac{12 + 13}{2} = 12.5$ . Yeh step kyun? Even $n$ → averaged centre.

Step 3 — 4/4 mein split karo. Lower $= - 50, 10, 11, 12$ ; upper $= 13, 14, 15, 60$ .

Step 4 — Quartiles. $Q_{1} = \frac{10 + 11}{2} = 10.5$ , $Q_{3} = \frac{14 + 15}{2} = 14.5$ . Yeh step kyun? Har half even → middle pair ka average. Notice karo ki $- 50$ aur $60$ halves ke edges par hain, toh $Q_{1}, Q_{3}$ par unka barely effect hai — yahi robustness in action hai.

Step 5 — Fences. $IQR = 14.5 - 10.5 = 4$ $Lower fence = 10.5 - 1.5 (4) = 10.5 - 6 = 4.5$ $Upper fence = 14.5 + 1.5 (4) = 14.5 + 6 = 20.5$

Step 6 — DONO sides flag karo. $- 50 < 4.5$ → low outlier; $60 > 20.5$ → high outlier. Left whisker $10$ par rukti hai (smallest value $\geq 4.5$ ); right whisker $15$ par rukti hai. Yeh step kyun? Fence rule symmetric hai — tumhe har baar dono directions test karni chahiye.

Verify: Do outliers, ek har side; IQR $= 4$ wild edge values se unchanged (compare karo: plain range $= 60 - (- 50) = 110$ !) — yeh ek vivid demo hai ki IQR robust hai jabki range nahi. (Range and Spread, Outliers and Robust Statistics.) ✓

Ex 6 — Case H: degenerate, sab values identical

Step 1 — Sorted (trivially), $n = 6$ . Sab equal. Yeh step kyun? Identical data ko bhi same rule se treat karna hoga — koi shortcuts nahi.

Step 2 — Median. 3rd & 4th ka average $= \frac{20 + 20}{2} = 20$ .

Step 3 — Halves. Lower $= 20, 20, 20$ , upper $= 20, 20, 20$ → $Q_{1} = 20$ , $Q_{3} = 20$ .

Step 4 — IQR aur fences. $IQR = 20 - 20 = 0, fences = 20 \pm 1.5 (0) = 20.$ Yeh step kyun? Zero spread ke saath box ki zero width hoti hai — yeh 20 par ek single vertical line mein collapse ho jaata hai, whiskers vanish ho jaati hain, aur koi bhi point outlier nahi ho sakta ( $20$ se bahar kuch nahi).

Verify: IQR $= 0$ , dono fences $= 20$ , koi outlier nahi — degenerate case sensibly behave karta hai. Yahan ek box plot sirf ek line hai. ✓

Ex 7 — Case J: real-world word problem (units ke saath)

Step 1 — Sort karo. Pehle se ascending, $n = 12$ . Yeh step kyun? Confirm karo; $25$ last mein hai.

Step 2 — Median $Q_{2}$ . 6th & 7th ka average $= \frac{5 + 5}{2} = 5$ minutes. Yeh step kyun? Even $n = 12$ → centre positions 6 aur 7 ke beech mein.

Step 3 — 6/6 mein split karo. Lower $= 2, 3, 3, 4, 4, 5$ ; upper $= 5, 6, 7, 8, 9, 25$ .

Step 4 — Quartiles. Lower middle pair (3rd,4th) $= 3, 4$ → $Q_{1} = 3.5$ min. Upper middle pair (upper ka 3rd,4th $= 7, 8$ ) → $Q_{3} = 7.5$ min. Yeh step kyun? Har half mein 6 (even) → middle pair ka average.

Step 5 — IQR + fences interpret karo. $IQR = 7.5 - 3.5 = 4 min$ $Upper fence = 7.5 + 1.5 (4) = 13.5 min, Lower fence = 3.5 - 6 = - 2.5 min .$ Kyunki wait negative nahi ho sakta, lower fence effectively "koi lower outlier possible nahi" hai. $25 > 13.5$ → 25-minute wait ek outlier hai.

Step 6 — Words mein jawab. "Middle half customers ne ek 4-minute band ke andar wait kiya (3.5 se 7.5 min tak), typical wait 5 min thi. Ek customer ki 25-minute wait ek outlier thi — ise investigation ke liye flag karo." Yeh step kyun? Word problem mein units ke saath interpretation chahiye, sirf ek number nahi.

Verify: $Q_{1} = 3.5$ , $Q_{3} = 7.5$ , IQR $= 4$ min, upper fence $= 13.5$ min, $25 > 13.5$ → outlier ✓. Units poore time carry hue (minutes). ✓

Ex 8 — Case K: exam twist, rule ko BACKWARDS use karo

Step 1 — $n$ aur structure identify karo. $n = 8$ (even) → lower half pehli 4 values hain $= 4, 6, x, 10$ . Yeh step kyun? $Q_{1}$ lower half ka median hai; sirf woh 4 values matter karti hain.

Step 2 — $Q_{1}$ ko $x$ ke terms mein likho. Lower half mein 4 values hain (even) → $Q_{1} = \frac{2nd + 3rd}{2} = \frac{6 + x}{2}$ . Yeh step kyun? Hum usual computation reverse karte hain: jo quantity hum jaante hain use unknown mein ek formula ke roop mein express karo.

Step 3 — Solve karo. $\frac{6 + x}{2} = 7 \Rightarrow 6 + x = 14 \Rightarrow x = 8$ . Yeh step kyun? Relationship set up hone ke baad simple algebra.

Step 4 — Constraint check karo. $x = 8$ satisfies karta hai $6 \leq 8 \leq 10$ ✓ aur list sorted rakhta hai ( $6, 8, 10$ ) ✓. Yeh step kyun? Exam chahta hai ki tum confirm karo ki answer admissible hai, sirf algebraically valid nahi.

Verify: $x = 8$ ke saath lower half hai $4, 6, 8, 10$ , jiska median $\frac{6 + 8}{2} = 7 = Q_{1}$ ✓ hai. Dataset sorted rehta hai. ✓

Recall Har example ne kaunsa cell hit kiya?

Ex 1 → A,C · Ex 2 → B,F · Ex 3 → D,I · Ex 4 → E · Ex 5 → F,G · Ex 6 → H · Ex 7 → J · Ex 8 → K. Matrix ka har cell cover hua hai.

Connections

Box-and-whisker plots — quartiles, IQR — parent: woh rules jinhe yeh examples exercise karte hain.
Median and Measures of Central Tendency — $Q_{2}$ throughout.
Range and Spread — Ex 5 mein IQR ( $4$ ) vs range ( $110$ ) contrast hai.
Outliers and Robust Statistics — Ex 2, 5, 7 mein fences apply hoti hain.
Percentiles and Quantiles — quartiles as 25/50/75 percentiles.
Skewness — Ex 2 aur Ex 7 right-skewed hain (lamba high whisker).
Normal Distribution — 1.5 multiplier kyun ≈0.7% flag karta hai.