Measures of central tendency — mean (grouped - ungrouped), median (grouped), mode (grouped): Exercises

2.7.1 · D4 · Maths › Statistics & Probability — Intermediate › Measures of central tendency — mean (grouped - ungrouped), m

Shuru karne se pehle, yeh hai symbol dictionary jo is page ke har grouped-data problem mein use hoti hai — pehle picture mein har ek se milo, phir words mein:

Figure — Measures of central tendency — mean (grouped - ungrouped), median (grouped), mode (grouped)

Figure 1 — Table $T$ ko histogram ki tarah draw kiya gaya hai. Har coloured bar ek class hai; uski height frequency $f$ hai (upar print ki gayi hai: 4, 9, 15, 11, 6). Coral bar (20–30) char working symbols label karne ke liye use hota hai: coral arrow $l$ mark karta hai, jo lower boundary hai (left edge, $= 20$ ); base ke saath double-headed arrow $h$ mark karta hai, jo class width hai ( $= 10$ ); top arrow $f$ mark karta hai, jo bar ki height hai ( $= 15$ ); aur $n$ saari heights ka sum hai ( $= 45$ ). Horizontal axis class boundaries dikhata hai; vertical axis frequency dikhata hai. Jab bhi koi symbol neeche abstract lage, yeh picture padho.

Definition Chhe grouped-data symbols (figure ke saath padho)

Ek class ek bar hai (values ka ek range, jaise $20$ – $30$ ). Uske do vertical edges class boundaries hain.

$x_{i}$ = class midpoint — ek bar ke centre par value, jo $\frac{lower boundary + upper boundary}{2}$ se milti hai (jaise 20–30 ke liye yeh $\frac{20 + 30}{2} = 25$ hai). Hum class ke har item ko aise treat karte hain jaise woh exactly is midpoint par baitha ho, isliye $x_{i}$ woh single number hai jo mean compute karte waqt poori class ki jagah kaam karta hai.
$l$ = us class ki lower class boundary jisme hum kaam kar rahe hain — us bar ka left edge.
$h$ = class width — ek bar kitna wide hai (right edge minus left edge, jaise $30 - 20 = 10$ ).
$f$ = us ek class ki frequency — us single bar ki height (kitne items woh hold karta hai).
$n = \sum f_{i}$ = items ki total number — har bar ki height add karo.
$c f$ = cumulative frequency — left to right sweep karta hua running total: kisi given boundary tak pahunchte waqt aapne kitne items pass kar liye hain (dekho Cumulative Frequency & Ogives). Median formula mein, $c f$ ka matlab hai median class mein enter karne se just before ka running total.

Bars khud ek histogram hain (dekho Histograms & Frequency Distributions); unki heights frequencies hain. Yeh figure mind mein rakho — neeche har formula bas yahi hai: "answer ek bar ke andar kitni door baithta hai?"

$f_{1}, f_{0}, f_{2}$ (sirf mode ke liye use hote hain) teen neighbouring bar heights hain: $f_{1}$ sabse tall (modal) bar, $f_{0}$ uske just left wala bar, $f_{2}$ just right wala bar. Mode tak pahunchne par hum inhe figure par phir se banayenge.

Level 1 — Recognition

Recall Solution

KYA poochha gaya hai: woh value jo data ko do equal halves mein split kare → woh median hai (mean nahi, mode nahi). Formula: $Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$ jahaan (upar wali symbol dictionary se) $l$ = median class ki lower boundary, $n$ = total count, $c f$ = median class se pehle wali class ki cumulative frequency, $f$ = median class ki frequency, $h$ = class width. YEH WORDS IS PAR KYU MAP KARTE HAIN: "50% below" literally middle position $\frac{n}{2}$ ki definition hai, aur formula find karta hai ki woh middle person ek class ke andar kahan land karta hai.

Recall Solution

KYA: modal class woh class hai jisme sabse badi individual frequency ho — kabhi cumulative nahi. Sabse bada $f$ hai $15$ → modal class = 20–30.

$f_{1} = 15$ (modal class khud)
$f_{0} = 9$ (class pehle, 10–20)
$f_{2} = 11$ (class baad mein, 30–40) KYU: mode formula measure karta hai ki peak apne dono neighbours se kitni sharply upar uthti hai, isliye humein dono sides ki frequencies chahiye.

Level 2 — Application

Running table $T$ (already continuous hai, isliye limits = boundaries) — yeh wahi table hai jo figure mein draw ki gayi hai:

Class	$f_{i}$	Midpoint $x_{i}$
0–10	4	5
10–20	9	15
20–30	15	25
30–40	11	35
40–50	6	45

$\sum f_{i} = n = 45$ .

Recall Solution

Rule: pehle sort karo, phir median middle position ki value hai.

Agar $n$ odd hai, toh ek hi middle hai: $(\frac{n + 1}{2})$ th value.
Agar $n$ even hai, toh koi single middle nahi hai, isliye hum do central values ka average karte hain: $\frac{n}{2}$ th aur $(\frac{n}{2} + 1)$ th.

(a) Sort karo: $3, 4, 7, 8, 9$ . Yahaan $n = 5$ (odd), middle position $= \frac{5 + 1}{2} = 3$ rd value $= 7$ . Median $= 7$ . (b) Sort karo: $2, 3, 4, 7, 8, 9$ . Yahaan $n = 6$ (even), do central positions hain 3rd aur 4th values: $4$ aur $7$ . Average $= \frac{4 + 7}{2} = 5.5$ . Median $= 5.5$ . Even $n$ ke liye average KYU: even count ke saath data cleanly do neighbours ke beech split hota hai, aur unke beech ki koi bhi value ka 50% neeche aur 50% upar hota hai; convention ke hisaab se hum unka midpoint lete hain.

Recall Solution

$u_{i}$ KYA hai? Yeh class ka midpoint hai jo apne guess $a$ se $h$ ki steps mein measure kiya gaya hai: $u_{i} = \frac{x _{i} - a}{h}$ . Yeh midpoints ko chhote integers mein convert kar deta hai. $u_{i} = - 2, - 1, 0, 1, 2$ YEH KYU KARTE HAIN? Frequency se weighted $- 2, - 1, 0, 1, 2$ add karna $5, 15, 25, 35, 45$ add karne se kahin zyada aasaan hai — aur answer identical hai kyunki $x_{i} = a + h u_{i}$ . $\sum f_{i} u_{i} = 4 (- 2) + 9 (- 1) + 15 (0) + 11 (1) + 6 (2) = - 8 - 9 + 0 + 11 + 12 = 6$ $\overset{x}{ˉ} = a + h (\frac{\sum f _{i} u _{i}}{\sum f _{i}}) = 25 + 10 \cdot \frac{6}{45} = 25 + 1.333 \dots = 26.33$ Answer: $\overset{x}{ˉ} \approx 26.33$ .

Recall Solution

Step 1 — cumulative frequency (cf): left to right sweep karo, running total rakho. $4, 13, 28, 39, 45$ Step 2 — median class dhundho. $\frac{n}{2} = \frac{45}{2} = 22.5$ . Pehla cf jo $22.5$ tak pahunche ya pass kare woh $28$ hai → median class = 20–30. Step 3 — pieces padho: $l = 20$ , $c f = 13$ (median class se pehle wala cf), $f = 15$ , $h = 10$ . Step 4 — interpolate karo. Is class se pehle hum $13$ items use kar chuke hain; humein $22.5 - 13 = 9.5$ aur chahiye, out of $15$ items jo width $10$ par spread hain: $Median = 20 + (\frac{22.5 - 13}{15}) \times 10 = 20 + \frac{9.5}{15} \cdot 10 = 20 + 6.333 = 26.33$ Answer: Median $\approx 26.33$ .

Recall Solution

Modal class = 20–30 (sabse bada $f = 15$ ). Padho: $l = 20$ , $f_{1} = 15$ , $f_{0} = 9$ , $f_{2} = 11$ , $h = 10$ . Fraction ka KYA matlab hai: left neighbour se upar rise, $f_{1} - f_{0} = 6$ , divide by dono neighbours se total excess, $2 f_{1} - f_{0} - f_{2} = 30 - 9 - 11 = 10$ . $Mode = 20 + (\frac{15 - 9}{2 ( 15 ) - 9 - 11}) \times 10 = 20 + \frac{6}{10} \cdot 10 = 20 + 6 = 26$ Answer: Mode $= 26$ . YEH WAHAN KYU LAND KARTA HAI: kyunki $f_{0} = 9 < f_{2} = 11$ hai, class ke baad wala heavier hai, isliye peak class centre ( $25$ ) se right ki taraf lean karti hai. Aur sach mein $26 > 25$ . ✓

Level 3 — Analysis

Recall Solution

Step 1 — total se $f$ recover karo. $4 + f + 15 + 11 + 6 = 45 \Rightarrow f = 9$ . Theek hai — lekin problem chahti hai ki hum verify karein ki yeh median $= 25$ ke saath consistent hai, toh doosri frequency ko unknown $f$ maanke median condition se solve karte hain, phir confirm karte hain.

Doosri frequency ko $f$ maano. Phir $n = 36 + f$ aur $\frac{n}{2} = \frac{36 + f}{2}$ . Median class 20–30 ke saath: $l = 20$ , $c f = 4 + f$ , median class ki frequency $= 15$ , $h = 10$ . $25 = 20 + (\frac{\frac{36 + f}{2} - ( 4 + f )}{15}) \times 10$ Step 2 — numerator simplify karo. $\frac{36 + f}{2} - (4 + f) = 18 + \frac{f}{2} - 4 - f = 14 - \frac{f}{2}$ . $25 - 20 = \frac{14 - \frac{f}{2}}{15} \times 10 \Rightarrow 5 = \frac{10 ( 14 - \frac{f}{2} )}{15} = \frac{2 ( 14 - \frac{f}{2} )}{3}$ Step 3 — solve karo. $5 \cdot 3 = 2 (14 - \frac{f}{2}) \Rightarrow 15 = 28 - f \Rightarrow f = 13.$

Lekin total check karo: $f = 13$ ke saath, $n = 49$ , na ki $45$ . Dono conditions ( $n = 45$ aur median $= 25$ ) inconsistent hain — dono ek saath nahi ho saktiñ. Yahi analysis ka lesson hai. Answer: Agar $n = 45$ fixed hai, toh $f = 9$ aur resulting median $26.33$ hai (L2·2 se), $25$ nahi. Agar median $= 25$ in doosri frequencies ke saath fixed hai, toh $f = 13$ aur $n = 49$ . Jaisi di gayi data over-constrained hai. Contradiction report karo — galat number force mat karo.

Recall Solution

TRAP KYA HAI YAHAAN: inclusive classes mein gaps hote hain (10 aur 11 ke beech kuch missing hai), isliye printed limit true lower boundary nahi hai. Step 1 — continuous boundaries mein convert karo. Gap $= 11 - 10 = 1$ ; har lower limit se $\frac{1}{2} = 0.5$ ghataao aur har upper limit mein $0.5$ jodo: $0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5, 30.5 - 40.5$ Step 2 — cf: $6, 16, 30, 40$ . Yahaan $n = 40$ , $\frac{n}{2} = 20$ . Step 3 — median class: pehla cf $\geq 20$ hai $30$ → class 20.5–30.5. Padho: $l = 20.5$ (boundary, na ki $21$ ), $c f = 16$ , $f = 14$ , $h = 10$ . $Median = 20.5 + (\frac{20 - 16}{14}) \times 10 = 20.5 + \frac{4}{14} \cdot 10 = 20.5 + 2.857 = 23.357$ Answer: Median $\approx 23.36$ . $l = 21$ use karna $23.86$ deta — exactly $0.5$ boundary shift ka fark.

Recall Solution

Median: already sorted hai, $n = 7$ (odd) → $\frac{n + 1}{2} = 4$ th value $= 5$ . Mean: $\frac{2 + 4 + 4 + 5 + 5 + 6 + 40}{7} = \frac{66}{7} = 9.43$ . YEH KYU DISAGREE KARTE HAIN: single value $40$ ek outlier hai. Mean ek balance point hai — ek bhaari value usse kaafi right mein drag karti hai ( $9.43$ tak). Median sirf position ki chinta karta hai, isliye outlier sirf far end par baithta hai aur middle ko move nahi karta. Answer: Median $= 5$ yahaan better "typical" value hai; mean $9.43$ outlier se inflated hai. Isliye hum teen measures rakhte hain (dekho Skewness).

Level 4 — Synthesis

Recall Solution

Empirical estimate: $3 Median - 2 Mean = 3 (26.33) - 2 (26.33) = 79.00 - 52.67 = 26.33$ True Mode $= 26$ se compare karo: $0.33$ ke andar agreement. ✓ Interpretation — YEH ITNE CLOSE KYU HAIN: mean $\approx$ median $\approx$ mode ka matlab hai distribution almost symmetric hai (thodi Skewness). Jab mean $=$ median exactly ho, toh relation predict karta hai Mode $=$ Median bhi, aur tiny gap ( $26$ vs $26.33$ ) sirf mode formula ka peak-leaning correction reflect karta hai.

Recall Solution

Step 1 — class midpoints. $x_{i} = \frac{lower + upper}{2}$ use karke: midpoints hain $10, 30, 50$ . Step 2 — direct grouped-mean equation set up karo. Mean total count se divide kiya gaya $f_{i} x_{i}$ ka total hai: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{8 ( 10 ) + 12 ( 30 ) + x ( 50 )}{8 + 12 + x} = 34$ Step 3 — known parts compute karo. Numerator $= 80 + 360 + 50 x = 440 + 50 x$ ; denominator $= 20 + x$ . $\frac{440 + 50 x}{20 + x} = 34 \Rightarrow 440 + 50 x = 34 (20 + x) = 680 + 34 x$ Step 4 — $x$ ke liye solve karo. $50 x - 34 x = 680 - 440 \Rightarrow 16 x = 240 \Rightarrow x = 15$ Answer: $x = 15$ students. Check karo: mean $= \frac{440 + 750}{35} = \frac{1190}{35} = 34$ . ✓

Level 5 — Mastery

Recall Solution

Kya $T$ ke liye 20–30 correct hai? Haan — lekin sahi karan se, student ke karan se nahi. Sahi test cumulative frequency use karta hai: cf $= 4, 13, 28, 39, 45$ ; $\frac{n}{2} = 22.5$ ; pehla cf $\geq 22.5$ hai $28$ , jo class 20–30 mein baitha hai. Yeh modal class ke saath coincide karta hai, lekin yeh ek coincidence hai, koi rule nahi.

Ek clean counterexample. 0–10, 10–20, 20–30, 30–40, 40–50 frequencies ke saath paanch classes lo: $1, 40, 2, 3, 40.$

Modal classes: sabse badi frequency $40$ hai, aur yeh do baar appear karta hai — 10–20 aur 40–50 mein. Toh data bimodal hai; koi ek class "the" mode nahi hai.
Median class (cf se): cf $= 1, 41, 43, 46, 86$ . Total $n = 86$ , toh $\frac{n}{2} = 43$ . Pehla cf $\geq 43$ hai $43$ , jo class 20–30 se belong karta hai.
Mismatch: median class 20–30 hai, jo koi bhi modal class (10–20 ya 40–50) nahi hai.

Lesson: median class hamesha cumulative frequency column se $\frac{n}{2}$ tak pahunchne par aana chahiye, kabhi "kaunsa bar sabse tall hai" se nahi. Symmetric single-peaked data ke liye dono often coincide karte hain, lekin yeh example prove karta hai ki zaroori nahi.

Recall Solution

Target ordering: mean sabse right, mode sabse left → ek right (positive) tail. Ek kaam karne wala dataset: $2, 2, 3, 5, 18$ .

Mode $= 2$ (do baar appear hota hai; baaki har value ek baar — sabse crowded value).
Median $=$ $5$ sorted values $2, 2, 3, 5, 18$ ke middle → $\frac{5 + 1}{2} = 3$ rd value $= 3$ .
Mean $= \frac{2 + 2 + 3 + 5 + 18}{5} = \frac{30}{5} = 6$ . Ordering check karo: $6 > 3 > 2$ , toh $Mean > Median > Mode$ . ✓ YEH POSITIVE SKEW KYU SIGNAL KARTA HAI: akela bada value $18$ ek right tail banata hai. Mode crowded left par baitha hai, median positional middle par, aur mean tail ki taraf rightward drag hota hai — toh yeh right to left mein line up hote hain: mean → median → mode. Yeh positive skew ka fingerprint hai (dekho Skewness). Wahi door wala value spread bhi inflate karta hai, jo Measures of Dispersion — Variance & Standard Deviation se connect hota hai.

Recap

Recall Kaun sa column median class dhundta hai?

Cumulative frequency — woh pehli class jiska cf $\frac{n}{2}$ tak pahunche. ::: Cumulative frequency; pehla cf $\geq n /2$ .

Recall Kaun si frequency modal class dhundti hai?

Sabse bada individual frequency $f_{i}$ , kabhi cumulative nahi. ::: Sabse bada individual $f_{i}$ .

Recall Even

$n$ ke liye, ungrouped median kaise nikala jaata hai? Sort karo, phir do central values ka average nikalo ( $\frac{n}{2}$ th aur $(\frac{n}{2} + 1)$ th). ::: Do middle values ka average nikalo.

Recall Inclusive classes ke saath grouped median nikalne se pehle, pehle kya fix karna hoga?

Printed limits ko continuous boundaries mein convert karo (half gap se shift karo). ::: Continuous class boundaries mein convert karo.