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

2.7.1 · D3 · Maths › Statistics & Probability — Intermediate › Measures of central tendency — mean (grouped - ungrouped), m

Yeh page Measures of Central Tendency ke liye "sab kuch ek saath" wali drill hai. Parent note ne teen tools (mean, median, mode) aur unke formulas build kiye. Yahan hum ensure karte hain ki koi bhi situation tumhe surprise na kare — odd counts, even counts, inclusive classes with hidden gaps, ek class jo "most crowded" ke liye tie karti hai, ek extreme outlier se pull hota data, aur ek real word problem.

Numbers touch karne se pehle, hum har tarah ke case lay out karte hain jo yeh topic tumpar throw kar sakta hai, phir har case ke liye ek example solve karte hain.

Symbol reminder (is poore page mein use hota hai)

Yahan har symbol parent note mein define kiya gaya tha; in chaar ko saamne rakhho kyunki Ex 5–7 inpar constantly lean karte hain.

Scenario matrix

#	Case class	Kya cheez tricky banati hai	Example
A	Ungrouped, odd $n$	ek clean middle value	Ex 1
B	Ungrouped, even $n$	koi single middle nahi — do ko average karo	Ex 2
C	Outlier present	mean mislead karta hai; median/mode nahi	Ex 3
D	Grouped mean, teen methods agree karte hain	SAME number dena zaroori hai	Ex 4
E	Inclusive classes (hidden gap)	limits → boundaries convert karni padegi	Ex 5
F	Grouped median + ogive picture	"middle person" ki geometry	Ex 6
G	Grouped mode, peak left ya right lean karti hai	neighbour frequencies decide karte hain	Ex 7
H	Degenerate: saari values equal / mode ke liye tie	limiting behaviour, koi unique peak nahi	Ex 8
I	Word problem + empirical relation	words → table → estimate translate karo	Ex 9

Ex 1 — Case A: ungrouped, odd count

Worked example Odd

$n$ 7 matches mein score kiye runs: $12, 45, 30, 8, 22, 30, 19$ . Mean, median, mode nikalo.

Forecast: $n = 7$ (odd) ke saath sort karne ke baad ek middle value hogi. Guess karo kaunsa number exactly centre mein aayega.

Mean. Sab add karo: $12 + 45 + 30 + 8 + 22 + 30 + 19 = 166$ . $n = 7$ se divide karo: $\overset{x}{ˉ} = 166/7 \approx 23.71$ . Yeh step kyun? Mean = fair share = total ÷ count. Kuch bhi grouped nahi hai, isliye raw values use karo.
Median — pehle sort karo. $8, 12, 19, 22, 30, 30, 45$ . Yeh step kyun? Median ordered data par defined hai; unsorted middle meaningless hoti hai.
Middle pick karo. $n = 7$ odd → position $\frac{n + 1}{2} = 4$ . 4th value $22$ hai. Yeh step kyun? Odd count mein exactly ek central slot hoti hai; 3 neeche, 3 upar.
Mode. Value $30$ do baar aata hai, baaki sab ek baar → mode $= 30$ . Yeh step kyun? Mode = sabse crowded value = highest count.

Verify: 3 values ( $8, 12, 19$ ) $22$ se neeche hain aur 3 ( $30, 30, 45$ ) upar hain → median genuinely set ko aadha-aadha split karti hai. ✓ Mean $23.71$ median se thoda upar hai kyunki bada $45$ use upar kheench raha hai.

Ex 2 — Case B: ungrouped, even count

Worked example Even

$n$ Daily temperatures (°C): $18, 21, 17, 24, 20, 19$ . Median nikalo.

Forecast: $n = 6$ even hai — koi single middle nahi hai. Kaunsi do values average karoge?

Sort karo. $17, 18, 19, 20, 21, 24$ . Yeh step kyun? Median ordered data ka middle define hoti hai — "50% below, 50% above" tabhi sense deta hai jab values increasing order mein hon.
Do central slots locate karo. $\frac{n}{2} = 3$ aur $\frac{n}{2} + 1 = 4$ → 3rd aur 4th values $19$ aur $20$ hain. Yeh step kyun? Even count centre ke beech do items ke beech straddle karta hai; dono ka equal claim hai.
Unhe average karo. Median $= \frac{19 + 20}{2} = 19.5$ °C. Yeh step kyun? Saccha "middle" do central values ke beech halfway par hota hai.

Verify: $19.5$ se neeche ${17, 18, 19}$ (3 values) hain; upar ${20, 21, 24}$ (3 values) hain. Perfect 50/50 split, aur answer mein °C bhi hai jaise data mein. ✓

Ex 3 — Case C: woh outlier jo mean ko fool karta hai

Worked example Ek extreme value

5 staff ki monthly salaries (₹ thousand): $20, 22, 25, 23, 210$ (last owner ka hai). Kaunsa measure ek "typical" salary best describe karta hai?

Forecast: ek value ( $210$ ) baaki sab se 8× zyada hai. Guess karo: kya mean $20$ – $25$ ke cluster ke andar rahega, ya bahut upar kheench jayega?

Mean. $\frac{20 + 22 + 25 + 23 + 210}{5} = \frac{300}{5} = 60$ . Yeh step kyun? Mean total deviation balance karta hai, isliye ek huge value use bahut door kheench leti hai. $60$ har ordinary worker se upar hai.
Median. Sort karo: $20, 22, 23, 25, 210$ ; $n = 5$ odd → 3rd value $= 23$ . Yeh step kyun? Median sirf position ki parwah karta hai, size ki nahi — giant $210$ bas ek end par baithta hai aur kabhi middle nahi hilata.
Mode. Saari values distinct hain → koi mode nahi (koi value repeat nahi hoti). Yeh step kyun? Mode ko genuine pile-up chahiye; yahan har salary unique hai.

Verify: $23$ (median) typical worker describe karta hai; $60$ (mean) kisi ko describe nahi karta. Yahi wajah hai ki Skewness matter karta hai — ek lamba right tail mean ko right kheenchta hai jabki median wahi rehta hai. ✓

Ex 4 — Case D: teenon methods se grouped mean (unhe agree karna chahiye)

Worked example Direct = assumed-mean = step-deviation

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

Forecast: table $25$ ke aas-paas symmetric hai. Compute karne se pehle mean guess karo — symmetry ek strong hint hai.

Direct method. $\sum f_{i} = 30$ . $\sum f_{i} x_{i} = 4 (5) + 6 (15) + 10 (25) + 6 (35) + 4 (45) = 20 + 90 + 250 + 210 + 180 = 750$ . Toh $\overset{x}{ˉ} = 750/30 = 25$ . Yeh step kyun? Har class $f_{i} x_{i}$ contribute karta hai (count × uska representative midpoint); total ÷ count = fair share.
Assumed-mean. $a = 25$ lo, $d_{i} = x_{i} - 25 = - 20, - 10, 0, 10, 20$ . $\sum f_{i} d_{i} = 4 (- 20) + 6 (- 10) + 0 + 6 (10) + 4 (20) = - 80 - 60 + 0 + 60 + 80 = 0$ . $\overset{x}{ˉ} = 25 + \frac{0}{30} = 25$ . Yeh step kyun? Ek guess $a$ se shift karne se arithmetic choti ho jaati hai; yahan total correction $0$ hai kyunki table $25$ ke aas-paas balance hai.
Step-deviation. $h = 10$ , $u_{i} = \frac{x _{i} - 25}{10} = - 2, - 1, 0, 1, 2$ . $\sum f_{i} u_{i} = 4 (- 2) + 6 (- 1) + 0 + 6 (1) + 4 (2) = - 8 - 6 + 0 + 6 + 8 = 0$ . $\overset{x}{ˉ} = 25 + 10 \cdot \frac{0}{30} = 25$ . Yeh step kyun? Deviations ko $h$ se divide karne par woh tiny integers ban jaati hain — sabse kam error-prone.

Verify: Teenon $25$ dete hain, jo hamari forecast ki symmetry ke centre ke barabar hai. Same mean, alag-alag size ki arithmetic. ✓

Ex 5 — Case E: inclusive classes with a hidden gap

Worked example Pehle limits ko boundaries mein convert karo

Inclusive classes mein recorded marks:

Class	$f_{i}$
10–19	3
20–29	7
30–39	12
40–49	8

Median class boundary $l$ aur median nikalo.

Forecast: 19 aur 20 ke beech ek gap hai. Agar tum naively $l = 30$ use karo to galat hoga. 30–39 class ki true lower boundary guess karo.

Gap spot karo. $19$ se $20$ mein width-1 gap hai. Use split karo: har lower limit se $0.5$ ghataao, har upper limit mein $0.5$ joddo → continuous boundaries $9.5 - 19.5, 19.5 - 29.5, 29.5 - 39.5, 39.5 - 49.5$ . Yeh step kyun? Interpolation assume karta hai ki classes touch karti hain. Gap width galat kar deta. Ab har class width $h = 10$ hai.
Cumulative frequencies. $3, 10, 22, 30$ . $n = 30$ , toh $\frac{n}{2} = 15$ . Yeh step kyun? Median class = pehli class jiska $c f \geq 15$ ho. Yahan $c f = 22$ (29.5–39.5 class) pehla hai jo 15 cross karta hai.
Pieces read karo. $l = 29.5$ (boundary, na ki 30), $c f = 10$ (pehle waali class), $f = 12$ , $h = 10$ . Yeh step kyun? Grouped-median formula ko exactly yeh chaar inputs chahiye — starting edge $l$ , kitne items pehle se use ho gaye ( $c f$ ), median class mein kitne hain ( $f$ ), aur woh class kitni wide hai ( $h$ ). Inhe abhi gather karne ka matlab Step 4 mein interpolation sirf substitution hoga bina table dobara padhhe.
Formula apply karo. $Median = 29.5 + \frac{15 - 10}{12} \times 10 = 29.5 + \frac{5}{12} \cdot 10 = 29.5 + 4.17 = 33.67.$ Yeh step kyun? Hume 12 ke class mein $15 - 10 = 5$ aur items chahiye jo width 10 par spread hain; har item $10/12$ width occupy karta hai.

Verify: Median $33.67$ median class $(29.5, 39.5)$ ke andar hai ✓. Agar hum $l = 30$ use karte to $34.17$ milta — exactly $0.5$ boundary correction ka difference, jo dikhata hai yeh fix kyun matter karta hai. ✓

Ex 6 — Case F: grouped median ek geometric "middle person" ke roop mein

Worked example Ogive se median

Parent ka table reuse karo: classes $0$ – $10, \dots, 40$ – $50$ with $f = 5, 8, 12, 7, 3$ ; $n = 35$ . Dikhao ki median $= 23.75$ geometrically kaise aata hai.

Forecast: $\frac{n}{2} = 17.5$ . Guess karo: cumulative frequency vs. class ke graph par, horizontal line " $c f = 17.5$ " curve ko kahan cross karegi?

Ise kaise padhein: horizontal axis marks hai (upper class boundary), vertical axis cumulative frequency

c f

hai. Black staircase-curve ogive hai; red horizontal line height

17.5 = \frac{n}{2}

par hai, aur red dashed drop dikhata hai woh height curve se kahan milti hai aur marks axis par

23.75

par kahan girti hai.

Cumulative frequencies build karo. $5, 13, 25, 32, 35$ . Har ek ko apni class ki upper boundary ke against plot karo: points $(10, 5), (20, 13), (30, 25), (40, 32), (50, 35)$ . Unhe join karo — yeh rising line ogive hai (dekho Cumulative Frequency & Ogives). Hum upper boundaries kyun use karte hain: $c f$ sab logon ko count karta hai including ek class, toh woh running total class ki right edge par hi "achieve" hoti hai. Midpoint ya lower edge par plot karne se claim hoga ki class puri ho gayi pehle hi, poori curve left shift ho jaati.
Half-way line draw karo. Height $\frac{n}{2} = 17.5$ par red horizontal line (figure dekho). Yeh $x = 20$ aur $x = 30$ ke beech ogive ko kaatti hai. Yeh step kyun? Cumulative rank $17.5$ par woh person hi median hai; ogive humein unki data value batata hai.
Axis par drop karo. Crossing point se seedha neeche drop karo. Yeh $x = 23.75$ par land karta hai. Yeh step kyun? $x$ -value padhne se "rank 17.5" ek actual marks value mein convert hoti hai.
Algebraically confirm karo. $l = 20, c f = 13, f = 12, h = 10$ : $20 + \frac{17.5 - 13}{12} \cdot 10 = 20 + 3.75 = 23.75$ . Yeh step kyun? Formula us red-line crossing ki algebra hai: points $(20, 13)$ aur $(30, 25)$ ke beech ogive ek straight segment hai, aur formula exactly compute karta hai kahan horizontal line $c f = 17.5$ us segment se milti hai. Dono tarafon se same number milna confirm karta hai ki picture aur algebra ek hi idea hain.

Verify: Graph reading ( $23.75$ ) aur formula ( $23.75$ ) exactly agree karte hain — formula sirf us red-line crossing ki algebra hai. ✓

Ex 7 — Case G: mode left ya right lean karti hai

Worked example Peak kis taraf lean karti hai?

Same table $f = 5, 8, 12, 7, 3$ . Mode ko modal-class centre ke against compare karo.

Forecast: modal class $20$ – $30$ hai (highest $f = 12$ ), centre $25$ . Class pehle ka $f_{0} = 8$ hai, class baad ka $f_{2} = 7$ hai. Kyunki $f_{0} > f_{2}$ , kya true peak $25$ ke left mein hogi ya right mein?

Ise kaise padhein: horizontal axis marks hai (class), vertical axis frequency

f

hai. Har bar ek class hai; red bar modal class

20

–

30

hai (sabse tall,

f = 12

). Dotted vertical line class centre

25

mark karti hai; black dashed line computed mode

24.44

mark karti hai, jo thodi si left mein hai.

Pieces identify karo. $l = 20, f_{1} = 12, f_{0} = 8, f_{2} = 7, h = 10$ . Yeh step kyun? Mode individual frequency use karta hai, kabhi cumulative nahi — aur dono neighbours chahiye.
Rise fractions compute karo. Left neighbour ke upar rise: $f_{1} - f_{0} = 4$ . Total excess: $2 f_{1} - f_{0} - f_{2} = 24 - 8 - 7 = 9$ . Yeh step kyun? Fraction $\frac{4}{9}$ kehta hai peak class mein $\frac{4}{9}$ raste par hai — left ke kareeb kyunki left neighbour heavy hai aur us side ko "prop up" kar raha hai.
Apply karo. $Mode = 20 + \frac{4}{9} \cdot 10 = 20 + 4.44 = 24.44$ .
Lean interpret karo. $24.44 < 25$ (class centre) → peak left lean karti hai, exactly kyunki $f_{0} = 8 > f_{2} = 7$ hai. Figure mein mode marker dotted centre line ke left mein hai. Yeh step kyun? Geometry confirm karta hai: heavier left neighbour → mode left boundary ke kareeb.

Verify: $\frac{4}{9} \in (0, \frac{1}{2})$ "halfway se kam" confirm karta hai → centre se left ✓. Agar hum $f_{0} = 7, f_{2} = 8$ swap karein, fraction $\frac{5}{9} > \frac{1}{2}$ hoga → mode right lean karega. ✓

Ex 8 — Case H: degenerate inputs

Worked example Saari equal, aur peak ke liye tie

(a) Data $7, 7, 7, 7, 7$ . (b) Data $2, 2, 5, 5, 9$ .

Forecast: (a) mein har summary $7$ par collapse honi chahiye. (b) mein do values "most frequent" ke liye tie karti hain — mode ka kya hoga?

(a) Mean. $\frac{7 \cdot 5}{5} = 7$ . Median: sorted middle $= 7$ . Mode: $7$ (5 baar aata hai). Yeh step kyun? Zero spread ke saath, "typical" unambiguous hai — teenon tools ko agree karna chahiye. Yeh ek aise distribution ka limiting case hai jisme koi variability nahi (dekho Measures of Dispersion — Variance & Standard Deviation, jahan yahan variance $= 0$ hai).
(b) Mode. Dono $2$ aur $5$ do baar aate hain → data bimodal hai; koi single mode nahi hai. Yeh step kyun? Mode highest count se define hota hai; jab do counts tie karein, dono modes hain. Sirf ek report karna structure hide kar deta.
(b) Contrast ke liye Mean & median. Mean $= \frac{2 + 2 + 5 + 5 + 9}{5} = \frac{23}{5} = 4.6$ . Median ( $2, 2, 5, 5, 9$ mein se 3rd) $= 5$ . Yeh step kyun? Mean aur median single-valued rehte hain chahe mode split ho jaye.

Verify (a): Teenon $7$ ke equal hain aur total deviation $\sum (x_{i} - 7) = 0$ ✓ (perfect balance point). Verify (b): counts ${2 : 2, 5 : 2, 9 : 1}$ — do maxima bimodality confirm karte hain; mean $4.6$ ≠ median $5$ mild left skew dikhata hai. ✓

Ex 9 — Case I: word problem + empirical relation

Worked example Mean aur median se mode estimate karo

Daily screen-time (hours) ka ek survey ek moderately skewed distribution ke liye mean $3.2$ aur median $3.5$ deta hai. Raw data kho gaya. Mode estimate karo.

Forecast: yahan median > mean hai. Guess karo: kya mode median se upar aayega ya mean se neeche?

Tool choose karo. Empirical relation $Mode \approx 3 Median - 2 Mean$ use karo. Yeh step kyun? Yeh akela formula hai jo ek measure doosre dono se recover karta hai — perfect jab raw data unavailable ho, moderate skew ke liye valid hai.
Substitute karo. $Mode \approx 3 (3.5) - 2 (3.2) = 10.5 - 6.4 = 4.1$ hours. Yeh step kyun? Do knowns plug karo; "3 aur 2" weighting empirical hai, typical skewed shapes ke liye tuned hai.
Interpret karo. Ordering: Mean $3.2 <$ Median $3.5 <$ Mode $4.1$ . Mode se neeche mean ek left-skewed long tail signal karta hai very-low-usage logon ka jo mean ko neeche kheench rahe hain (compare Skewness).

Verify: Rearrange karne par, $3 (3.5) - 2 (3.2) = 4.1$ ✓. Sanity: moderate skew ke liye teenon measures roughly evenly spaced hone chahiye; yahan gaps $0.3$ (mean→median) aur $0.6$ (median→mode) hain — same direction, one-sided skew ke consistent. ✓

(Ek data value ke total ko weight karne ka idea Weighted Mean aur expected value mein bhi aata hai, jahan har outcome ko frequency ki jagah probability se weight kiya jaata hai.)

Recall check

Recall Ex 3 mein median mean ko kyun beat karta hai?

Kyunki median sirf position par depend karta hai, isliye ek single extreme value (₹210k wala owner) use nahi kheench sakta; mean totals balance karta hai aur har ordinary worker se bahut upar kheench jata hai.

Recall Ex 5 mein

$l = 30$ kyun nahi hai? Inclusive classes (20–29, 30–39) mein ek hidden gap hai. Continuous boundaries mein convert karne par $l = 29.5$ milta hai; $30$ use karne par har answer $0.5$ se shift ho jaata.

Recall Ex 7 mein mode left kyun lean karta hai?

Class pehle ( $f_{0} = 8$ ) class baad ( $f_{2} = 7$ ) se heavier hai, isliye fraction $\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} = \frac{4}{9} < \frac{1}{2}$ hai — halfway se kam, matlab class centre ke left mein.

Kaunsa measure ek extreme outlier se unaffected rehta hai?

Median (aur aksar mode); mean outlier ki taraf kheench jaata hai.

Inclusive classes 20–29, 30–39 ke liye, doosri class ki lower boundary kya hai?

29.5

(gap close karne ke liye 0.5 ghataao).

Agar do values highest frequency ke liye tie karein, toh data kaisa hai?

Bimodal — koi single mode nahi hota.

Mode estimate karne ka empirical relation?

Mode \approx 3 Median - 2 Mean