Question bank — Cumulative frequency — ogive, median from graph

True — last class har ek observation ko accumulate karta hai, isliye uska running total poori population $N$ hota hai. Ye kisi bhi CF table par standard sanity check hai.

False — cumulative frequency non-negative counts ka running total hai, isliye woh sirf flat reh sakti hai ya badhti hai. Neeche jaane waala segment ka matlab hoga negative frequency, jo impossible hai.

True — yeh count karta hai "kitne is boundary par ya usse upar hain." Jaise boundary daayein jaati hai, waise kam observations upar rehte hain, isliye count $N$ se ghatkar $0$ tak aa jaata hai.

True — $C{F}_i$ class $i$ ki upper boundary ke neeche ki har cheez count karta hai, isliye honest point hai (upper boundary, $C{F}_i$). Midpoint use karne se poori curve left shift ho jaayegi.

True — kisi class ka more-than $CF$ uski lower boundary se aage ki sab cheez count karta hai, isliye honest point hai (lower boundary, more-than $CF$). Ye less-than rule ka mirror image hai jo upper boundaries use karta hai.

False — woh formula ordered ungrouped list ke liye hai. Grouped ogive ko smooth curve ki tarah treat kiya jaata hai, isliye hum use height $N/2$ par cut karte hain (aadha upar), $(N+1)/2$ nahi.

False — woh class hoti hai jahan cumulative frequency pehli baar $N/2$ cross karti hai. Sabse lamba bar us crossing se door bhi ho sakta hai; dekhein mode jahaan sabse lamba bar actually matter karta hai.

True — us point par "neeche ka number" (less-than curve) aur "upar ka number" (more-than curve) barabar hote hain, aur dono $N/2$ ke barabar sirf median par hote hain. Perpendicular giraakar median ki x-value milti hai (figure s02 dekhein).

False — median simply us boundary par baith jaata hai. Interpolation term zero ho jaata hai, isliye $\text{Median}=L$ exactly hota hai (jaisa Example 1 mein hua, jahan woh 30 par aaya).

True — pehli class ki lower boundary se pehle, koi bhi observation count nahi hua hota, isliye curve us lower boundary par $y=0$ se shuru hota hai.

False — ogive ek cumulative count curve hai, density nahi. Iske neeche ke area mein kuch meaningful nahi hota; useful information har x par height hai.

Galat — median class woh hai jiske andar CF $N/2$ cross karti hai. CF 13 (20 se neeche) se 25 tak 20–30 ke andar badhi, isliye median class 20–30 hai, agle class nahi.

Galat — ye discontinuous classes hain. Pehle continuous boundaries mein convert karo (9.5–19.5, 19.5–29.5), isliye $L=19.5$ hai. Ye step skip karne par har boundary 0.5 se shift ho jaayegi.

Ogive ke liye galat — CF sab kuch upper boundary ke neeche count karta hai, isliye points upper boundaries par hone chahiye. Midpoints frequency polygon ke liye hain, cumulative curve ke liye nahi.

Galat — $C{F}_b$ median class se pehle waali class ki CF hai (sab kuch jo $L$ par aate waqt already count ho chuka hai). Median class ki apni CF use karne se double-counting hogi aur result zyada nikalega.

Galat — $y=N$ curve ka bilkul top hai (neeche saara data). Median data ko aadha-aadha baantta hai, isliye sirf $y=N/2$ tak charhne ke baad across slide karo (figure s01).

Galat — woh height $N/2$ par cross karte hain, jahan "neeche" aur "upar" dono counts barabar hote hain. $N$ par less-than curve top out ho chuki hai lekin more-than curve zero par aa gayi hai.

Galat — $f$ sirf median class ki frequency hai. Ye woh count hai jo width $h$ par spread hai, jo interpolation mein use hone waala $f/h$ per-unit slope deta hai.

Galat — $f=0$ se formula zero se divide karta hai aur ogive wahan flat hai, isliye us class ke andar koi crossing nahi hoti. $f=0$ waali class kabhi median class nahi ho sakti; crossing wahan honi chahiye jahan curve actually rise kare.

Generally galat — unbounded class ki koi asli lower boundary nahi hoti. Common class width use karke ek reasonable finite boundary assign karo (e.g. '<20' ko 10–20 treat karo agar width 10 hai), warna interpolation base $L$ arbitrary hai aur result unreliable.

Kyunki $C{F}_i$ class $i$ ki upper boundary ke strictly neeche ki har cheez ka tally hai. Point (upper boundary, $C{F}_i$) tab ek sachcha "is x ke neeche itni values hain" represent karta hai.

$C{F}_b$ observations $L$ tak pahunchte waqt already count ho chuke hain; hume halfway mark tak pahunchne ke liye abhi bhi $N/2-C{F}_b$ aur chahiye. Wahi bacha hua count hai jo hum median class ke across interpolate karte hain (figure s03).

Median class mein $f$ observations apni width $h$ par uniformly spread hain. Isliye class ka woh hissa jitna hume enter karna hai woh hai (count still needed)$/f$, jo us class ke andar position deta hai — ek linear-interpolation step.

Less-than curve deta hai "x ke neeche kitne" aur more-than deta hai "x ke upar kitne." Ye tabhi barabar hote hain jab dono $N/2$ ke barabar hon, jo definition ke hisaab se median hai.

Hume sirf har class ka total count pata hota hai, na ki har value kahaan baithi hai. Sabse fair, least-biased guess hai even spacing, jo ogive ko class ke across straight line banata hai aur interpolate karne deta hai. Ye mirror karta hai ki median grouped data ke liye kaise define hoti hai.

Ogive grouped data ko continuous quantity ki tarah model karta hai, isliye uska output us smooth curve par ek position hai — ek representative middle value, zaruri nahi ki observed data point ho.

Ek quartile bas alag height par hoti hai: $Q_1$ ke liye $N/4$, $Q_3$ ke liye $3N/4$. Tum $N/2$ ki jagah us height par across slide karte ho; median simply $N/2$ (ya $Q_2$) case hai.

Kyunki less-than $CF$ sirf wahan $N/2$ cross kar sakti hai jahan woh actually rise karti hai, aur woh sirf wahan rise karti hai jahan class frequency positive ho. Ek flat ($f=0$) stretch skip ho jaata hai, usme kabhi land nahi hota.

Ogive ek straight line ban jaata hai $0$ se $N$ tak, isliye uska median data range ka exactly midpoint hota hai — interpolation phir bhi chalti hai, bas centrally land karti hai.

Tab $N/2=C{F}_b+f$ ya $N/2=C{F}_b$, jisse interpolation term $0$ ya poora $h$ ho jaata hai, isliye median ek boundary par land karta hai — interpolation ki zarurat nahi, jaisa Example 1 mein tha.

Pehli class khud. Tab $C{F}_b=0$ hoga aur $L$ pehli class ki lower boundary hogi; formula un substitutions ke saath phir bhi apply hoga.

Haan — frequency $0$ waali class running total mein kuch add nahi karti, isliye curve us class ki width ke across horizontal rehta hai phir dobara rise karta hai. Aisi class kabhi median class nahi ho sakti.

Ise shared class width use karke ek finite boundary do — "<20" ko ek class-width wide treat karo (e.g. 10–20) aur ">80" ko 80–90 — plot karne ya $L$ aur $h$ read karne se pehle. Defined boundary ke bina curve ka koi honest starting/ending x nahi hai.

Zyaadatar haan — median normally data ke andar kaafi gehraai mein hoti hai, isliye unbounded top class rarely $N/2$ ko touch karti hai. Tumhe sirf us class ke liye finite boundary chahiye jo actually crossing contain karti hai.

Sirf ek hi $N/2$ crossing contain kar sakta hai. Dhundo kahaan CF $N/2$ se neeche se at or above ho jaati hai; wahi akeli class median class hai, aur tie upar wale boundary rule se resolve hota hai.

Ye generally alag hote hain. Symmetric distribution mein teeno ek saath milte hain; skewed mein ogive median mean aur mode ke beech hoti hai, extreme values se resistant hoti hai.