Cumulative frequency — ogive, median from graph: Visual walkthrough

2.7.2 · D2 · Maths › Statistics & Probability — Intermediate › Cumulative frequency — ogive, median from graph

Hum sirf teen ideas pe build karte hain jo tumhare paas pehle se hain: ek frequency table (ek list — "kitne values har bin mein gire"), median (woh value jiske neeche half data ho), aur seedhi-line interpolation (ek seedhi ramp pe chalna).

Step 1 — "Kitne IN hain" ko "kitne BELOW hain" mein badlo

KYA. Hum ek plain frequency table se shuru karte hain: har class (values ki ek range, jaise $20$ – $30$ ) ka ek frequency $f$ hota hai — us mein girne wale observations ki count. Hum ise ek running total mein convert karte hain.

KYUN. Median define hota hai "kitne values kisi value ke neeche hain" se, na ki "kitne ek bin mein hain" se. Toh humein ek aisi quantity chahiye jo neeche-waala sawaal seedha answer kare. Woh quantity hai cumulative frequency — har bin ko left se jaate hue jodते raho.

PICTURE. Dekho kaise bars ek utha hua staircase banate hain. Har naye bar ki height pehle se saare ke upar stack hoti hai.

Figure — Cumulative frequency — ogive, median from graph

$C F_{i} = every bin up to and including bin i f_{1} + f_{2} + \dots + f_{i} = \sum_{j = 1}^{i} f_{j}$

$f_{j}$ — bin number $j$ mein count.
Sum $j = 1$ se $i$ tak chalta hai — "class $i$ se pehle ya class $i$ tak sab kuch."
$C F_{i}$ — class $i$ ki upper edge ke neeche baitha hua total.

Sabse aakhri value hoti hai $C F_{k} = N$ , observations ki total sankhya — hamaara sanity check.

Step 2 — Har total ko SAHI jagah anchor karo

KYA. Hum har cumulative total ko graph pe ek dot ki tarah rakhte hain. Dot ki height hai $C F_{i}$ . Horizontal position hai class $i$ ki upper boundary.

KYUN upper boundary, midpoint nahi? $C F_{i}$ class ki upper edge se neeche sab kuch count karta hai. Toh us height ke liye ek hi honest x-position hai: upper edge khud — dot tab padhta hai "itne values ( $C F_{i}$ ) is x ke neeche hain." Midpoint pe rakho toh claim karte ho ki count bin ke aadhe mein hi complete ho gaya — yeh jhooth hai jo poori curve ko left shift kar deta hai.

PICTURE. Red dot bin ki right wall pe baitha hai, running total ki height par.

Step 3 — Dots jodo: ogive paida hota hai

KYA. Dots ko ek rising curve se connect karo. Ise floor se shuru karo: pehli class ki lower boundary par, abhi tak kuch bhi neeche nahi, toh $C F = 0$ . Yeh right mein $N$ tak climb karta hai.

KYUN jodo? Plotted upper boundaries ke beech humhare paas data nahi hai — toh hum honest sabse simple assumption karte hain: har class ke andar observations uniformly spread hain. Uniform spread matlab count bin ke across constant rate se badhta hai, jo ek straight line hai. Dots ko segment-by-segment jodna exactly wahi draw karta hai.

PICTURE. Ogive — seedhe ramps ka ek staircase jo kabhi neeche nahi jaata.

Step 4 — Graph PE median ka sawaal poochho

KYA. Median woh value hai jiske neeche half data ho. "Half data" hamare y-axis par ek height hai: $N /2$ . Toh hum $y = N /2$ mark karte hain, curve tak slide karte hain, aur seedha neeche drop karte hain.

KYUN $N /2$ aur $(N + 1) /2$ nahi? $(N + 1) /2$ rule items ki ek discrete ordered list count karne ke liye hai. Yahan ogive data ko ek smooth, continuous flow ki tarah treat karta hai — hum woh height chahte hain jo total rise ko do barabar halves mein split kare, aur rise $N$ ka aadha exactly $N /2$ hai.

PICTURE. Height $N /2$ par ek red horizontal line, curve se milti hui, phir median par drop hoti ek red vertical line.

Step 5 — Pata karo HUM KAUNSE ramp par land karte hain (median class)

KYA. Horizontal line $y = N /2$ ek particular ramp se milti hai. Us ramp ka bin hai median class: woh class jiske andar running total $N /2$ se guzarta hai.

KYUN care karein? Kyunki interpolation usi ek ramp par hoti hai. Hum ise identify karte hain yeh dhundhke ki $C F$ kahan $N /2$ se neeche se at-or-above $N /2$ ho jaata hai.

PICTURE. Ek single ramp par zoom karo. Iska bottom-left corner hai $(L, C F_{b})$ ; top-right $(L + h, C F_{b} + f)$ par.

Corners ke naam (yahi woh chaar letters hain jo final formula mein hain):

$L$ — median class ki lower boundary (x jahan yeh ramp shuru hota hai).
$C F_{b}$ — pehle waali class ki cumulative frequency, yaani ramp ki starting height.
$f$ — median class ki frequency, toh ramp bin ke across $f$ se rise karta hai.
$h$ — class width, toh ramp horizontally $h$ se run karta hai.

Step 6 — Ramp ek straight line hai: iska slope nikaalo

KYA. Is single ramp par count uniformly badhta hai, isliye yeh ek straight line hai. Rise over run iska slope deta hai.

KYUN slope? Slope batata hai "x ki har unit pe kitne extra counts." Jab hum jaante hain ki $N /2$ tak pahunchne ke liye abhi kitne extra counts chahiye, slope us count ko x mein ek distance mein convert karta hai.

PICTURE. Ramp ke neeche right triangle: vertical leg $= f$ (rise), horizontal leg $= h$ (run).

$slope = \frac{rise}{run} = \frac{f}{h} (\frac{counts}{unit of x})$

$f$ — poora bin cross karne par total extra count.
$h$ — bin ke across total x-distance.
$f / h$ — x ki har unit par add hone wale counts. Yeh "count" aur "distance" ke beech exchange rate hai.

Step 7 — Missing count ko ek distance mein convert karo

KYA. Hum ramp ko height $C F_{b}$ par shuru karte hain aur $N /2$ tak climb karna hai. Zaroorat extra count hai $(\frac{N}{2} - C F_{b})$ . Slope (counts-per-x) se divide karo x-distance $Δ x$ paane ke liye jitna hum right walk karein.

KYUN divide? Slope x ki har unit par counts hai. Count ko x-distance mein badalne ke liye, rate undo karo — isse divide karo. ("7 extra counts hain, $12/10$ counts per step milte hain $\Rightarrow$ $7 \div \frac{12}{10}$ steps chalo.")

PICTURE. Chhota red sub-triangle: vertical leg $(\frac{N}{2} - C F_{b})$ , horizontal leg $Δ x$ — bade triangle jaisa wahi slope.

$Δ x = \frac{\frac{N}{2} - C F _{b}}{f / h} = (\frac{\frac{N}{2} - C F _{b}}{f}) \times h$

$\frac{N}{2} - C F_{b}$ — bin mein enter karte waqt abhi bhi missing counts.
$f / h$ se divide karna — ramp ki rate use karke counts ko x-distance mein trade karna.
fraction $\frac{\frac{N}{2} - C F _{b}}{f}$ — "bin ke kitne fraction raste par hain," $0$ aur $1$ ke beech ek number.
$\times h$ — us fraction ko real x-units mein scale karo.

Phir median starting x plus woh walk hai:

Step 8 — Do edge cases jinpar kabhi mat fisar

KYA. Do boundary situations jo confusing lagti hain lekin same rule cleanly handle karta hai.

Case A — $N /2$ exactly ek corner par land karta hai. Missing count $\frac{N}{2} - C F_{b}$ $f$ ke barabar nikalta hai (ya $0$ ). Tab fraction $1$ (ya $0$ ) hai aur median exactly ek class boundary par baitha hai. Koi special handling nahi — formula corner return kar deta hai.

Case B — twin-ogive crossing. More-than ogive bhi banao (ek running total right se count karke, toh yeh x badhne ke saath girta hai). Less-than curve padhta hai "x ke neeche count"; more-than padhta hai "x ke upar count." Woh equal sirf tab hote hain jab dono $N /2$ ke barabar hon — unka crossing. Wahan se ek perpendicular drop karo: same median, koi arithmetic nahi.

PICTURE. Dono curves; red dot unka intersection height $N /2$ par mark karta hai.

Worked check — derivation action mein

Ek-picture summary

Upar sab kuch ek ramp par baitha ek triangle hai: $(L, C F_{b})$ par enter karo, $\frac{N}{2} - C F_{b}$ aur counts chahiye, unhe slope $f / h$ ke through distance mein trade karo, $Δ x$ right chalo, median par pahuncho.

Recall Feynman: poora walkthrough retell karo

Ek frequency table kehta hai kitne bacche har height band mein aate hain. Main bands ko ek running total mein stack karta hoon — "is line se chhote kitne bacche hain" — aur woh total kabhi nahi ghatta: draw karo aur ogive milta hai. Beech waala baccha woh hai jiske neeche half bacche hain, toh $N$ ka aadha hai $N /2$ ; main height axis par $N /2$ tak jaata hoon aur curve tak walk karta hoon. Main ek ramp par land karta hoon — median class, $(L, C F_{b})$ se $(L + h, C F_{b} + f)$ tak. Woh ramp straight hai kyunki maine assume kiya ki andar ke bacche evenly spread hain, toh yeh $h$ ke run par $f$ se rise karta hai: slope $f / h$ counts per step. Ramp mein enter karte waqt mujhe $\frac{N}{2} - C F_{b}$ counts ki kami hai, toh main slope se divide karta hoon us shortfall ko ek sideways walk $Δ x$ mein badalne ke liye, aur median hai $L + Δ x$ . Agar main right-se-aane waali curve bhi draw karoon, jahan woh cross karti hain wahan "below" equals "above" equals $N /2$ — koi sum kiye bina same answer.

Recall Self-test

Upper boundary par plot kyun karte hain? ::: CF upper edge se neeche sab count karta hai, isliye woh iska honest x hai. Slope $f / h$ physically kya matlab rakhta hai? ::: median class ke andar x ki har unit par add hone wale extra counts. Missing count ko distance mein kaise badlate ho? ::: missing count ko slope $f / h$ se divide karo. Twin-ogive crossing ko median kyun force karta hai? ::: wahan below = above, toh dono $N /2$ ke barabar hote hain.