Cumulative frequency — ogive, median from graph: Exercises

2.7.2 · D4 · Maths › Statistics & Probability — Intermediate › Cumulative frequency — ogive, median from graph

Shuru karne se pehle, teen shared reminders seedhe alfazon mein — taaki koi symbol aur koi idea bina samjhe na aaye.

Intuition Formula aisa kyun dikhta hai (derive karo, memorise mat karo)

Median class ke left edge $L$ se shuru karo. Bilkul us point pe exactly $C F_{b}$ observations hamare neeche baith jaate hain — lekin hum wahan khade hona chahte hain jahan $N /2$ observations hamare neeche hon. Toh hume abhi bhi $aur counts chahiye jo abhi baki hain \frac{N}{2} - C F_{b}$ aur observations ko cross karna hai. Ab key assumption: median class ke andar $f$ observations ko width $h$ ke across evenly spread maana jaata hai (neeche uniform-spread reminder dekho). Even spreading ka matlab hai ki x ka har ek unit same number of counts carry karta hai, yaani $f / h$ counts per unit of x. $(\frac{N}{2} - C F_{b})$ counts gain karne ke liye hume ek horizontal distance $Δ x = \frac{chahiye counts}{counts per unit x} = \frac{\frac{N}{2} - C F _{b}}{f / h} = (\frac{\frac{N}{2} - C F _{b}}{f}) \times h$ walk karni padegi. Is $Δ x$ ko apne starting point $L$ mein add karne se milta hai $Median = L + Δ x$ — exactly wahi boxed formula. Ab har letter earned hai, blindly accept nahi kiya.

Figure — Cumulative frequency — ogive, median from graph

Level 1 — Recognition

(Kya tum pieces ko naam de sakte ho aur graph padh sakte ho?)

Exercise 1.1. Ek table mein classes $0 - 10, 10 - 20, 20 - 30$ hain frequencies $3, 7, 5$ ke saath. Less-than cumulative frequency table likho aur $N$ state karo.

Recall Solution 1.1

Left se right accumulate karo — har CF pichli CF plus current frequency hai.

Less than	CF
10	3
20	$3 + 7 = 10$
30	$10 + 5 = 15$

Final CF $15$ hai, toh $N = 15$ . Woh last value sabhi frequencies ke sum ke barabar honi chahiye ( $3 + 7 + 5 = 15$ ) — yahi tumhara sanity check hai. ✓

Exercise 1.2. Less-than ogive ke liye, class $20 - 30$ ki CF ko hum kis x-value ke against plot karte hain: midpoint $25$ ya upper boundary $30$ ? Ek sentence mein explain karo.

Recall Solution 1.2

Upper boundary $30$ . Us class ki CF $30$ se neeche sab kuch count karti hai, toh honest point $(30, C F)$ hai. Midpoint pe plot karne se poori curve left shift ho jaati aur galat median milti. parent note ka "midpoint mistake" dekho.

Exercise 1.3. Ek ogive point $(40, 32)$ se guzarti hai. Plain words mein, woh dot kya mean karta hai?

Recall Solution 1.3

"== $32$ observations $40$ se kam hain==." Less-than ogive ki y-value $x = 40$ pe literally $40$ se neeche data ka count hai.

Level 2 — Application

(Median formula mein sahi plug-in karo.)

Exercise 2.1. $40$ students ke marks:

Marks	$f$
0–10	4
10–20	6
20–30	10
30–40	12
40–50	8

Median nikalo.

Recall Solution 2.1

Step 1 — CF table. $4, 10, 20, 32, 40$ . Toh $N = 40$ aur $N /2 = 20$ . Step 2 — median class. Hume dekhna hai kahan CF pehli baar $20$ tak pahonchti hai. CF $20$ ke class ke neeche $10$ hai, phir $20 - 30$ class ke andar $20$ tak pohonchti hai → median class $20 - 30$ hai. Step 3 — letters padho. $L = 20, C F_{b} = 10, f = 10, h = 10$ . $Median = 20 + (\frac{20 - 10}{10}) \times 10 = 20 + 1 \times 10 = 30.$ Yeh exactly wahi slide-across hai jo Figure s01 mein draw ki gayi hai: $y = 20$ , right to the curve, down to $x = 30$ . ✓

Exercise 2.2. Usi class limits ke liye frequencies $5, 8, 12, 10, 5$ ho jaati hain (total $40$ ). Median nikalo.

Recall Solution 2.2

CF: $5, 13, 25, 35, 40$ . $N = 40, N /2 = 20$ . Pehla CF $\geq 20$ hai $25$ ("less than $30$ " pe), aur CF $13$ (below $20$ ) se $25$ tak $20 - 30$ ke andar jaata hai → median class $20 - 30$ . $L = 20, C F_{b} = 13, f = 12, h = 10$ : $Median = 20 + (\frac{20 - 13}{12}) \times 10 = 20 + \frac{7}{12} \times 10 = 20 + 5.8 \overline{3} = 25.83.$ Yahan median mid-class land karta hai, toh Linear interpolation sach mein kaam aata hai. Figure s02 dekho, jo sirf $20 - 30$ class ko zoom in karta hai: teal segment seedhi $C F_{b} = 13$ (at $x = 20$ ) se $25$ (at $x = 30$ ) tak chadhti hai — seedhi kyunki uniform-spread assumption hai. Hume $20$ tak pohonchne ke liye $13$ ke aage $7$ aur counts chahiye the, aur $12$ counts width ko evenly fill karte hain, toh plum marker $7/12$ of the way across baith jaata hai, $x \approx 25.83$ pe. ✓

Level 3 — Analysis

(Median class carefully dhundho; boundaries aur unequal widths handle karo.)

Exercise 3.1. Classes $0 - 10, \dots, 40 - 50$ (width $10$ , $N = 50$ ) ke liye CF values $8, 20, 20, 35, 50$ hain, median nikalo. Repeated $20$ pe dhyan do.

Recall Solution 3.1

$N /2 = 25$ . CFs scan karo: $8, 20, 20, 35, 50$ . Pehla CF $\geq 25$ hai $35$ , $30 - 40$ class mein. Isse pehle wali class ka $C F_{b} = 20$ hai. CF se median class ka $f$ recover karo: $35 - 20 = 15$ . $L = 30, C F_{b} = 20, f = 15, h = 10$ : $Median = 30 + (\frac{25 - 20}{15}) \times 10 = 30 + \frac{5}{15} \times 10 = 30 + 3.3 \overline{3} = 33.33.$ Flat stretch (CF $20 \to 20$ ) ek zero frequency wali class hai — ogive wahan horizontal hai, koi observations nahi. Yeh kabhi $25$ tak nahi pohonchti, toh yeh median class nahi ho sakti. ✓

Exercise 3.2. Discontinuous classes $1 - 10, 11 - 20, 21 - 30$ jinka frequencies $6, 14, 10$ hain ( $N = 30$ ). Continuous boundaries mein convert karne ke baad median nikalo.

Recall Solution 3.2

Pehle boundaries convert karo: har lower limit se $0.5$ ghataao, har upper limit mein $0.5$ jodao → $0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5$ . Width $h = 10$ . CF: $6, 20, 30$ . $N /2 = 15$ . Pehla CF $\geq 15$ hai $20$ (class $10.5 - 20.5$ ); $C F_{b} = 6$ . $L = 10.5, C F_{b} = 6, f = 14, h = 10$ : $Median = 10.5 + (\frac{15 - 6}{14}) \times 10 = 10.5 + \frac{9}{14} \times 10 = 10.5 + 6.4286 = 16.93.$ $0.5$ shift skip karne se $\approx 16.43$ aata — exactly $0.5$ se galat. ✓

Exercise 3.3 (unequal class widths). $50$ logon ki ages deliberately uneven classes mein grouped hain:

Age	$f$	width $h$
0–10	8	10
10–30	22	20
30–35	12	5
35–50	8	15

Median nikalo, median class ki apni width use karne ka dhyan rakho.

Recall Solution 3.3

CF: $8, 30, 42, 50$ . $N = 50, N /2 = 25$ . Pehla CF $\geq 25$ hai $30$ , toh median class $10 - 30$ hai; $C F_{b} = 8$ . Yahan is exercise ka poora point hai: median class ki width $h = 30 - 10 = 20$ hai, na ki pehli row ka $10$ . Median class ki width usi class se lo. $L = 10, C F_{b} = 8, f = 22, h = 20$ : $Median = 10 + (\frac{25 - 8}{22}) \times 20 = 10 + \frac{17}{22} \times 20 = 10 + 15.4545 = 25.45.$ Uniform-spread assumption abhi bhi apply hoti hai, lekin counts-per-unit slope ab $f / h = 22/20$ hai, width-10 class se zyada gentle — exactly isliye sahi $h$ matter karta hai. ✓

Level 4 — Synthesis

(Ogive graph, twin-ogive idea, aur doosre statistics ko combine karo.)

Exercise 4.1 se pehle hume doosri curve chahiye, jis par exercise rely karti hai:

Exercise 4.1. Twin-ogive reasoning. $N = 60$ ke liye, less-than ogive aur more-than ogive $x = 27$ pe intersect karti hain. (a) Intersection pe y-value kya hai? (b) Median kya hai?

Recall Solution 4.1

(a) Crossing pe, "number below" $=$ "number above" $= N /2 = 60/2 = 30$ . Toh y-value $30$ hai — yeh seedha Figure s03 se padho, jahan dono curves halfway height pe milti hain. (b) Crossing se x-axis pe perpendicular giraao → median $x = 27$ hai. Kyun kaam karta hai: less-than curve "count below" deti hai, more-than "count above" deti hai; yeh sirf $N /2$ pe equal hote hain, jo exactly median position hai. Isi slide-across trick ko $N /4$ aur $3 N /4$ pe apply karne ke liye Quartiles and percentiles from ogive dekho.

Exercise 4.2. Ek distribution mein width $5$ ki classes hain: $0 - 5, 5 - 10, 10 - 15, 15 - 20$ jinka frequencies $7, 13, 20, 10$ hain ( $N = 50$ ). (a) Median, aur (b) lower quartile $Q_{1}$ nikalo ( $N /2$ ki jagah height $N /4$ pe slide-across karke milta hai).

Recall Solution 4.2

CF: $7, 20, 40, 50$ . (a) $N /2 = 25$ . Pehla CF $\geq 25$ hai $40$ (class $10 - 15$ ), $C F_{b} = 20$ , $f = 20$ , $L = 10$ , $h = 5$ : $Median = 10 + (\frac{25 - 20}{20}) \times 5 = 10 + \frac{5}{20} \times 5 = 10 + 1.25 = 11.25.$ (b) $N /4 = 12.5$ . Pehla CF $\geq 12.5$ hai $20$ (class $5 - 10$ ), $C F_{b} = 7$ , $f = 13$ , $L = 5$ , $h = 5$ : $Q_{1} = 5 + (\frac{12.5 - 7}{13}) \times 5 = 5 + \frac{5.5}{13} \times 5 = 5 + 2.1154 = 7.12.$ Same interpolation machine, sirf alag target height. ✓

Level 5 — Mastery

(Machinery reverse karo: unknowns, missing frequencies.)

Exercise 5.1. Ek grouped table (width $10$ ) mein frequencies $5, x, 20, 15, y$ hain classes $0 - 10, \dots, 40 - 50$ ke liye. Total $N = 70$ hai aur median $28.5$ known hai. Median class $20 - 30$ hai. $x$ aur $y$ nikalo.

Recall Solution 5.1

Equation 1 (total): $5 + x + 20 + 15 + y = 70 \Rightarrow x + y = 30$ . Equation 2 (median): median class $20 - 30$ deta hai $L = 20, f = 20, h = 10$ , aur $C F_{b} = 5 + x$ (median class se pehle CF). $N /2 = 35$ . $28.5 = 20 + (\frac{35 - ( 5 + x )}{20}) \times 10.$ Solve karo: $8.5 = \frac{30 - x}{20} \times 10 = \frac{30 - x}{2}$ , toh $30 - x = 17 \Rightarrow x = 13$ . Phir $y = 30 - 13 = 17$ . Median class assumption check karo: CF hai $5, 18, 38, \dots$ — CF $35$ ko $20 - 30$ ke andar cross karti hai ( $18 \to 38$ jaati hai). Consistent. ✓

Exercise 5.2. 5.1 ki completed table ( $f = 5, 13, 20, 15, 17$ ) ke liye modal class bhi state karo aur confirm karo ki woh median class se alag nahi hai — yeh dikhate hue ki "biggest bar $\neq =$ median class."

Recall Solution 5.2

Modal class woh class hai jiska frequency sabse zyada hai: $20 - 30$ ka $f = 20$ hai, maximum. Yahan modal aur median classes ittifaq se dono $20 - 30$ mein land karti hain — toh yeh table ek counter-counterexample hai: woh agree kar sakte hain. General lesson phir bhi valid hai: median class fix hoti hai waahan se jahan CF $N /2$ ko cross kare, bar height se nahi. Aur gehri contrast ke liye Mean and mode of grouped data dekho. (Agar hum $f$ shift karte $5, 25, 20, 15, 5$ mein, toh modal class $10 - 20$ hoti jabki CF $= 5, 30, 50, \dots$ phir bhi median class ko $20 - 30$ pe rakhta — alag classes.)

Active Recall

Recall One-line self-tests

Exercise 3.1 mein flat CF stretch kyun ignore ki gayi? ::: Yeh zero-frequency class hai; ogive wahan horizontal hai aur kabhi $N /2$ tak nahi pohonchti.
Exercise 3.3 mein kaunsa $h$ use karte ho? ::: Median class ki apni width, $20$ (na ki $10$ ).
$Q_{1}$ kaunsi target height se milti hai? ::: $N /4$ .
5.1 mein $x$ aur $y$ kya the? ::: $x = 13, y = 17$ .
Ek class ke across straight-line interpolation valid kyun hai? ::: Uniform-spread assumption ogive ko wahan ek straight segment banati hai.