Statistics: Class 10 Maths Chapter 13 Hinglish mein

Jab data classes mein ho, tab mean, mode aur median

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
31 practice questions
14 MCQs

Is chapter mein aap kya seekhenge

Grouped data ka mean direct, assumed-mean aur step-deviation methods se nikaalo, aur sabse jaldi wala method chuno
Samjhao ki class marks se nikala gaya mean sirf true mean ka ek estimate kyun hota hai
Modal class pehchano aur mode formula se grouped data ka mode calculate karo
Less-than aur more-than cumulative frequency tables banao aur unse median class dhoondho
Grouped data ka median calculate karo, less-than table se bhi aur missing frequencies ke saath bhi
Tay karo ki kisi sawaal ka sabse achha jawab mean, median ya mode mein se kaun deta hai, aur unke beech ka empirical relation use karo

NCERT Mathematics Class 10, Chapter 13, pp. 171–201 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

13.1 Introduction

Ek board result sheet mein hazaaron students ke marks hote hain. Ek weather office har din saikdon stations ki rainfall note karta hai. Ek cricket board ab tak khele gaye har match ka total store karta hai. Aise data ko koi ek-ek value karke nahi padhta. Hum pehle values ko classes mein sort karte hain, aur phir ek hi sawaal poochhte hain: kaun sa ek number poore data ko sabse achhe se represent karta hai?

Class 9 mein tumne teen jawab dekhe the, measures of central tendency (केंद्रीय प्रवृत्ति के माप): mean, median aur mode. Tab tumne inhe ungrouped data ke liye nikala tha, jahan har value pata hoti hai. Is chapter mein hum teeno ko grouped data ke liye nikalenge, jahan humein sirf classes pata hoti hain aur yeh ki har class mein kitni values hain. Raste mein tum cumulative frequency use karoge, ek running total jo batata hai ki data ka beech wala hissa kahan hai.

Warm-up: ungrouped data ke averages

3 mein se 1

6, 9, 9, 11 aur 15 ka mean nikaalo.

Answer dekho

Answer 10

Sum 50 hai aur 5 values hain, isliye mean hai

50 \div 5 = 10.

4 (yaani alag-alag values ki ginti) se divide karne par 12.5 aata hai, jo galat hai.

13.2 Grouped Data ka Mean

Maan lo paanch workers mein se har ek din ke ₹ 450 kamata hai. Saath mein woh $5 \times 450 = 2250$ rupees kamaate hain: jo value repeat hoti hai, use uski frequency se multiply karke ek hi baar mein jod dete hain. Yahi shortcut kisi bhi table ke liye kaam karta hai: har value $x_{i}$ ko uski frequency se multiply karo, jo hai $f_{i},$ phir products ko jod kar grand total nikaalo, aur total frequency se divide karo. “In sabko jodo” ke liye hum $Σ$ (Greek capital sigma) likhte hain, isse formula chhota rehta hai.

Frequencies wale data ka mean

\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}

(13.1)

jahan

$\overset{ˉ}{x}$: mean, ise “x bar” padhte hain
$f_{i}$: value $x_{i}$ ki frequency
$\sum$: $i = 1, 2, \dots, n$ par sum

Definition

Mean

Saare observations ka sum, observations ki ginti se divide kiya hua. Frequencies wale data ke liye yeh hai $Σ f_{i} x_{i} \div Σ f_{i} .$

ExampleTeen 4 aur ek 8 ka mean hai $(3 \times 4 + 8) \div 4 = 5.$

Example 13.1NCERT Example 1 · p. 172

Building site par daily wages

Lucknow mein ek contractor building site ke 40 workers ko unki skill ke hisaab se alag-alag daily wages deta hai. Table mein har wage aur use paane wale workers ki ginti di gayi hai. Mean daily wage nikaalo.

Table 13.140 workers ki daily wages

Daily wage (₹)	Workers
360	1
380	2
400	2
420	4
450	5
480	7
500	6
520	3
550	2
580	4
600	3
640	1

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Asli registers mein saikdon wages hoti hain, isliye unhe pehle classes mein group kiya jaata hai. Chalo inhi 40 wages ko ₹ 50 width ki classes mein group karte hain: 350–400, 400–450, aur aise hi 600–650 tak. Edges ke liye ek rule hai: jo wage kisi upper limit ke barabar ho, woh agli class mein jaati hai, isliye ₹ 450 wale paanch workers 450–500 mein jaate hain.

2/3Har ₹ 50 par lines wages ko chhe classes mein baant deti hain.

Fig. 13.1Grouping counts ko rakhti hai, par exact values bhool jaati hai.NCERT Table 13.2 · p. 173

Ek number line 350 se 650 rupees tak jaati hai. Step 1: chaalees dots, har worker ka ek, us wage ke upar stack kiye gaye hain jo woh worker kamata hai, 360 se 640 tak. Step 2: 350, 400, 450 waghairah par dashed lines number line ko 50 width ki chhe classes mein baant deti hain, aur har class par uske workers ka count likha hai: 3, 6, 12, 9, 6 aur 4. Step 3: har class ke dots side mein khisak kar class ke beech, yaani 375, 425, 475, 525, 575 aur 625 ke upar ek column mein aa jaate hain; isse dikhta hai ki grouped data har value ko aise treat karta hai jaise woh class mark par baithi ho.

Table 13.2₹ 50 width ki classes mein 40 wages

Daily wage (₹)	Workers
350–400	3
400–450	6
450–500	12
500–550	9
550–600	6
600–650	4
Total	40

Wages group hone ke baad individual values chali jaati hain: table sirf itna batata hai ki 12 workers ₹ 450 se ₹ 500 ke beech kuch kamaate hain. Formula (13.1) ko phir bhi har class ke liye ek value $x_{i}$ chahiye. Sabse sahi ek guess class ka beech hai, yaani uska class mark (वर्ग चिह्न). Ise use karne ka matlab hai yeh maan lena ki har class ke workers beech ke aas-paas barabar faile hue hain: yeh hai mid-point assumption.

Definition

Class mark

Kisi class ka mid-point: $\frac{1}{2} (lower limit + upper limit) .$

Example450–500 ka class mark hai $\frac{1}{2} (450 + 500) = 475.$

Ab class marks ban jaate hain $x_{i},$ aur mean bilkul pehle ki tarah nikalta hai. Is tarah seedha $Σ f_{i} x_{i}$ aur $Σ f_{i}$ se mean nikaalne ko direct method kehte hain.

Table 13.3Grouped wages ke liye direct method

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
350–400	3	375	1125
400–450	6	425	2550
450–500	12	475	5700
500–550	9	525	4725
550–600	6	575	3450
600–650	4	625	2500
Total	40	—	20,050

To $\overset{ˉ}{x} = 20,050 \div 40 = 501.25.$ Direct method se mean wage ₹ 501.25 hai.

Assumed mean method

Direct method mein jaldi hi bade multiplications aa jaate hain: $12 \times 475$ hi 5700 hai. Yahan ek trick hai. Beech ke paas ka ek convenient class mark chuno, use assumed mean $a$ (कल्पित माध्य) kaho, aur har class mark ko usse naapo. Yeh distances $d_{i} = x_{i} - a,$ jinhe deviations kehte hain, chhote numbers hote hain, kuch negative bhi, jabki frequencies bilkul waisi hi rehti hain.

DerivationDeviations wapas mean tak kaise le jaate hain

Deviations ka average lo, har ek ko uski frequency se weight karke:

\overset{ˉ}{d} = \frac{\sum f_{i} (x_{i} - a)}{\sum f_{i}}

Upar wale hisse ko do sums mein todo. Constant $a$ doosre sum se bahar aa jaata hai, aur phir $Σ f_{i}$ cancel ho jaata hai:

\begin{aligned} \overset{ˉ}{d} & = \frac{Σ f_{i} x_{i} - a Σ f_{i}}{Σ f_{i}} \\ = \frac{\sum f_{i} x_{i}}{\sum f_{i}} - a \\ = \overset{ˉ}{x} - a \end{aligned}

To average deviation mean se kam hai, aur yeh fark hai $a .$ $a$ wapas jodne par assumed mean method milta hai:

Assumed mean method

\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}

(13.2)

Wages ke liye lo $a = 525.$ Deviations −150, −100, −50, 0, 50 aur 100 hain. Inhe frequencies 3, 6, 12, 9, 6 aur 4 se multiply karke jodne par milta hai $Σ f_{i} d_{i} = - 950.$ To $\overset{ˉ}{x} = 525 - 950 \div 40 = 525 - 23.75 = 501.25,$ yaani pehle jaisa hi ₹ 501.25, aur products bahut chhote.

Activity 13.1

Kya a ki choice se fark padta hai?

Tumhe chahiye

grouped wages wala table
ek calculator (optional)

Pehle guess karo

Grouped wages ke liye Ravi

a = 375

leta hai aur Sana

a = 625

leti hai. Unke means kya aayenge?

Ravi ka chhota hogaSana ka chhota hogadono ₹ 501.25 honge

$a = 375$ lo aur har deviation nikaalo, $d_{i} = x_{i} - 375.$
Har deviation ko uski frequency se multiply karke jodo; isse milta hai $Σ f_{i} d_{i} .$
Ab nikaalo $a + Σ f_{i} d_{i} \div 40.$
Baaki class marks ke saath dohrao, har ek ko bari-bari se banao $a .$

Table 13.4

$a$	$Σ f_{i} d_{i}$	Mean
375
425
475
525
575
625

Tumhari entries apne aap save hoti hain.

Step-deviation method

Deviations ko phir se dekho: −150, −100, −50, 0, 50, 100. Har ek 50 ka multiple hai, jo class size $h$ (वर्ग माप) hai. $h$ se divide karne par yeh chhote whole numbers ban jaate hain:

u_{i} = \frac{x_{i} - a}{h}

Derivationu's ke average se mean tak

$u_{i}$ ka frequencies ke saath average lo, aur factor $\frac{1}{h}$ ko bahar nikaalo:

\begin{aligned} \overset{ˉ}{u} & = \frac{\sum f_{i} u_{i}}{\sum f_{i}} \\ = \frac{1}{h} \cdot \frac{\sum f_{i} d_{i}}{\sum f_{i}} \\ = \frac{1}{h} (\overset{ˉ}{x} - a) \end{aligned}

Doosri line assumed mean derivation ka result use karti hai. $h$ se multiply karo aur $a$ jodo, to milta hai $\overset{ˉ}{x} = a + h \overset{ˉ}{u} :$

Step-deviation method

\overset{ˉ}{x} = a + h \frac{\sum f_{i} u_{i}}{\sum f_{i}}

(13.3)

Table 13.5Wages ke liye step-deviation method,

a = 525

aur

h = 50

ke saath

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
375	3	−3	−9
425	6	−2	−12
475	12	−1	−12
525	9	0	0
575	6	1	6
625	4	2	8
Total	40	—	−19

To $\overset{ˉ}{u} = - 19 \div 40 = - 0.475$ aur $\overset{ˉ}{x} = 525 + 50 \times (- 0.475) = 501.25.$ Ab products single digits hain, aur answer bilkul nahi hila. Fig. 13.2 dono shortcuts ko number line par scale ke change ki tarah dikhata hai.

3/4 $h = 50$ se divide karne par milte hain chhote whole numbers $u_{i} .$

Fig. 13.2Pehle subtract karo

a,

phir divide karo by

h,

average lo, aur aakhir mein dono steps ulte karo.NCERT Table 13.5 · p. 176

Dots ke chhe columns wage classes ki frequencies 3, 6, 12, 9, 6 aur 4 dikhate hain. Step 1 columns par unke class marks 375 se 625 tak likhta hai. Step 2 assumed mean 525 subtract karta hai, to labels deviations −150, −100, −50, 0, 50 aur 100 ban jaate hain, aur 0 chauthe column ke neeche hai. Step 3 50 se divide karta hai, jisse −3, −2, −1, 0, 1 aur 2 milte hain. Step 4 balance point mark karta hai: u-values ka average −0.475 hai, jo wapas 501.25 rupees ka mean banta hai, aur yeh 525 ke thoda left mein ek marker se dikhaya gaya hai.

Example 13.2NCERT Example 2 · p. 178

T20 totals, teen tarah se

Table mein ek state T20 league ke 40 matches ke first-innings totals hain. Teeno methods se mean total nikaalo.

Table 13.6

Total (runs)	Matches
100–120	3
120–140	5
140–160	9
160–180	12
180–200	7
200–220	4

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.3NCERT Example 3 · p. 180

Unequal classes wali rainfall

Maharashtra ke 50 stations par rain gauges ne July ke pehle hafte mein yeh rainfall record ki. Saari classes ki width same nahi hai. Mean rainfall nikaalo.

Table 13.7

Rainfall (mm)	Stations
0–40	4
40–80	7
80–120	12
120–200	15
200–320	8
320–400	4

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Activity 13.2

Collect karo, group karo aur average nikaalo

Tumhe chahiye

ek notebook
ek measuring tape
ek bathroom ya luggage scale

Teen teams banao.
Team A har classmate se poochhti hai ki school aane mein kitne minute lagte hain.
Team B har classmate ka school bag tolti hai, kg mein, ek decimal place tak.
Team C har classmate ka arm span cm mein naapti hai, haath faila kar ek fingertip se doosri fingertip tak.
Har team barabar width ki 5 se 7 classes chunti hai, apna data tally karti hai aur ek grouped frequency table banati hai.
Har team apne numbers ke hisaab se suitable method se mean nikaalti hai, aur batati hai ki usne woh method kyun chuna.

Table 13.8

Team	Method	Mean
A
B
C

Tumhari entries apne aap save hoti hain.

Grouped data ka mean

4 mein se 1

Class 24.5–29.5 ka class mark kya hai?

Answer dekho

Answer 27

(24.5 + 29.5) \div 2 = 27.

Class size, yaani 5, ek alag quantity hai.

Practice · 9 questions

Practice 13.1: answers ke saath

1
NumericalEasyNCERT Exercise 13.1, Q1
Ek panchayat ne 25 gaon mein ek hafte ke power cut ke ghante record kiye.
Power cut (ghante) Villages
0–2 3
2–4 6
4–6 8
6–8 4
8–10 3
10–12 1
1. Har gaon mein power cut ke ghanton ka mean nikaalo.
  h
  Solution dikhao
  Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0–2 3 1 3
  2–4 6 3 18
  4–6 8 5 40
  6–8 4 7 28
  8–10 3 9 27
  10–12 1 11 11
  Total 25 — 127
  $\overset{ˉ}{x} = 127 \div 25 = 5.08$ ghante.
2. Tumne kaun sa method use kiya, aur kyun?
  Solution dikhao
  Direct method. Class marks aur frequencies chhote hain, isliye products $f_{i} x_{i}$ aasaan hain aur shortcut se kuch nahi bachta.
NumericalMediumNCERT Exercise 13.1, Q2
Ek college ke paas paying-guest rooms ke liye 40 students jo monthly rent dete hain, woh neeche group kiya gaya hai. Suitable method se mean monthly rent nikaalo.
Rent (₹) Students
4000–5000 5
5000–6000 9
6000–7000 12
7000–8000 8
8000–9000 6
₹
Solution dikhao
Class marks 4500, 5500, …, 8500 bade hain aur barabar doori par hain, isliye step deviation use karo, saath mein $a = 6500,$ $h = 1000.$
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
4500 5 −2 −10
5500 9 −1 −9
6500 12 0 0
7500 8 1 8
8500 6 2 12
Total 40 — 1
$\overset{ˉ}{x} = 6500 + 1000 \times \frac{1}{40} = 6525.$ Mean rent ₹ 6525 hai.
NumericalMediumNCERT Exercise 13.1, Q3
Kuch senior citizens ne note kiya ki woh har subah park mein kitne minute walk karte hain. Mean time 37 minutes hai. Missing frequency nikaalo, $k .$
Time (min) People
10–20 5
20–30 8
30–40 $k$
40–50 12
50–60 7
Solution dikhao
Class marks: 15, 25, 35, 45, 55. To $Σ f_{i} = 32 + k$ aur $Σ f_{i} x_{i} = 75 + 200 + 35 k + 540 + 385 = 1200 + 35 k .$
$1200 + 35 k = 37 (32 + k)$
$1200 + 35 k = 1184 + 37 k$
To $2 k = 16$ aur $k = 8.$
NumericalMediumNCERT Exercise 13.1, Q4
Sports day par 36 students ne 100 m race daudi. Unke times neeche group kiye gaye hain. Suitable method chun kar mean time nikaalo.
Time (s) Students
12–13 2
13–14 4
14–15 7
15–16 9
16–17 8
17–18 4
18–19 2
s
Solution dikhao
Class marks 12.5, 13.5, …, 18.5 decimals hain jinmein 1 ka fark hai, isliye step deviation, saath mein $a = 15.5,$ $h = 1$ inhe −3 se 3 tak bana deta hai.
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
12.5 2 −3 −6
13.5 4 −2 −8
14.5 7 −1 −7
15.5 9 0 0
16.5 8 1 8
17.5 4 2 8
18.5 2 3 6
Total 36 — 1
$\overset{ˉ}{x} = 15.5 + 1 \times \frac{1}{36} \approx 15.53$ s.
5
NumericalMediumNCERT Exercise 13.1, Q5
Ek quality inspector ne 160 notebooks ke pages gine.
Pages Notebooks
96–98 12
99–101 38
102–104 60
105–107 34
108–110 16
1. Har notebook mein pages ka mean nikaalo.
  Solution dikhao
  Class marks: 97, 100, 103, 106, 109, jinmein 3 ka fark hai. Lo $a = 103,$ $h = 3.$
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  97 12 −2 −24
  100 38 −1 −38
  103 60 0 0
  106 34 1 34
  109 16 2 32
  Total 160 — 4
  $\overset{ˉ}{x} = 103 + 3 \times \frac{4}{160} = 103.075 \approx 103.08$ pages.
2. Tumne kaun sa method chuna, aur kyun?
  Solution dikhao
  Step deviation: frequencies badi hain aur class marks barabar doori par hain, isliye deviations ko 3 se divide karne par har product chhota rehta hai.
NumericalMediumNCERT Exercise 13.1, Q6
Ek town ke 30 dairy booths par ek din mein bika doodh (litres mein) neeche group kiya gaya hai. Har booth par bikne wali quantity ka mean nikaalo.
Milk (L) Booths
40–60 3
60–80 5
80–100 9
100–120 7
120–140 4
140–160 2
L
Solution dikhao
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
50 3 −2 −6
70 5 −1 −5
90 9 0 0
110 7 1 7
130 4 2 8
150 2 3 6
Total 30 — 10
$\overset{ˉ}{x} = 90 + 20 \times \frac{10}{30} \approx 96.67$ litres.
7
NumericalMediumNCERT Exercise 13.1, Q7
40 gaon ke kuon ke paani mein fluoride ki matra neeche group ki gayi hai, mg per litre mein.
Fluoride (mg/L) Wells
0.0–0.3 5
0.3–0.6 9
0.6–0.9 12
0.9–1.2 7
1.2–1.5 4
1.5–1.8 3
1. Mean fluoride content nikaalo.
  mg/L
  Solution dikhao
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0.15 5 −2 −10
  0.45 9 −1 −9
  0.75 12 0 0
  1.05 7 1 7
  1.35 4 2 8
  1.65 3 3 9
  Total 40 — 5
  $\overset{ˉ}{x} = 0.75 + 0.3 \times \frac{5}{40} = 0.7875 \approx 0.79$ mg/L.
2. Peene ke paani ke liye safe limit 1.5 mg/L hai. Mean usse kam hai. Kya iska matlab hai ki har kuan safe hai?
  Solution dikhao
  Nahi. Mean sirf saare kuon ko ek group ki tarah describe karta hai. Teen kuen class 1.5–1.8 mein hain, yaani limit par ya usse upar, isliye unki checking zaroori hai, bhale hi mean kam ho.
NumericalHardNCERT Exercise 13.1, Q8
Ek bank branch ke 50 employees ne ek saal mein jitne din chhutti li, woh neeche group kiya gaya hai. Classes barabar nahi hain. Chhutti ke dinon ka mean nikaalo.
Chhutti ke din Employees
0–4 6
4–8 12
8–12 14
12–20 10
20–30 5
30–36 3
din
Solution dikhao
Class marks 2, 6, 10, 16, 25, 33 chhote hain aur unke deviations mein koi kaam ka common factor nahi hai, isliye direct method sabse simple hai.
Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
0–4 6 2 12
4–8 12 6 72
8–12 14 10 140
12–20 10 16 160
20–30 5 25 125
30–36 3 33 99
Total 50 — 608
$\overset{ˉ}{x} = 608 \div 50 = 12.16$ din.
NumericalEasyNCERT Exercise 13.1, Q9
Ek district ke 40 schools ke Class 10 pass percentages neeche group kiye gaye hain. Mean pass percentage nikaalo.
Pass percentage Schools
50–60 3
60–70 8
70–80 12
80–90 11
90–100 6
%
Solution dikhao
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
55 3 −2 −6
65 8 −1 −8
75 12 0 0
85 11 1 11
95 6 2 12
Total 40 — 9
$\overset{ˉ}{x} = 75 + 10 \times \frac{9}{40} = 77.25,$ isliye mean pass percentage 77.25% hai.

Power cut (ghante)	Villages
0–2	3
2–4	6
4–6	8
6–8	4
8–10	3
10–12	1

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–2	3	1	3
2–4	6	3	18
4–6	8	5	40
6–8	4	7	28
8–10	3	9	27
10–12	1	11	11
Total	25	—	127

Rent (₹)	Students
4000–5000	5
5000–6000	9
6000–7000	12
7000–8000	8
8000–9000	6

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
4500	5	−2	−10
5500	9	−1	−9
6500	12	0	0
7500	8	1	8
8500	6	2	12
Total	40	—	1

Time (min)	People
10–20	5
20–30	8
30–40	$k$
40–50	12
50–60	7

Time (s)	Students
12–13	2
13–14	4
14–15	7
15–16	9
16–17	8
17–18	4
18–19	2

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
12.5	2	−3	−6
13.5	4	−2	−8
14.5	7	−1	−7
15.5	9	0	0
16.5	8	1	8
17.5	4	2	8
18.5	2	3	6
Total	36	—	1

Pages	Notebooks
96–98	12
99–101	38
102–104	60
105–107	34
108–110	16

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
97	12	−2	−24
100	38	−1	−38
103	60	0	0
106	34	1	34
109	16	2	32
Total	160	—	4

Milk (L)	Booths
40–60	3
60–80	5
80–100	9
100–120	7
120–140	4
140–160	2

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
50	3	−2	−6
70	5	−1	−5
90	9	0	0
110	7	1	7
130	4	2	8
150	2	3	6
Total	30	—	10

Fluoride (mg/L)	Wells
0.0–0.3	5
0.3–0.6	9
0.6–0.9	12
0.9–1.2	7
1.2–1.5	4
1.5–1.8	3

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0.15	5	−2	−10
0.45	9	−1	−9
0.75	12	0	0
1.05	7	1	7
1.35	4	2	8
1.65	3	3	9
Total	40	—	5

Chhutti ke din	Employees
0–4	6
4–8	12
8–12	14
12–20	10
20–30	5
30–36	3

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–4	6	2	12
4–8	12	6	72
8–12	14	10	140
12–20	10	16	160
20–30	5	25	125
30–36	3	33	99
Total	50	—	608

Pass percentage	Schools
50–60	3
60–70	8
70–80	12
80–90	11
90–100	6

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
55	3	−2	−6
65	8	−1	−8
75	12	0	0
85	11	1	11
95	6	2	12
Total	40	—	9

13.3 Grouped Data ka Mode

Ek shoe shop ke liye yeh jaanna bekaar hai ki uske customers ka average size 7.4 hai; aisa koi joota hota hi nahi. Use woh size chahiye jo sabse zyada bikta hai. Yahi mode (बहुलक) hai, yaani sabse zyada frequency wali value. Kabhi-kabhi do ya zyada values ki top frequency barabar hoti hai, aur aise data ko bimodal ya multimodal kehte hain. Is chapter ke har data set ka ek hi mode hai.

Example 13.4NCERT Example 4 · p. 183

Shoe sizes

Ek dukaan ne ek Saturday ko school shoes ke 15 pairs beche, in sizes mein: 8, 6, 9, 7, 8, 10, 7, 8, 5, 9, 8, 6, 7, 8, 9. Modal size nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Data group hone ke baad koi ek value repeat hoti hui nahi dikhti. Table jo dikhata hai, woh hai sabse bhari class, yaani sabse zyada frequency wali class: modal class (बहुलक वर्ग). Mode isi ke andar kahin hai, aur padosi classes tay karti hain ki kahan. Agar modal class se theek pehle wali class lagbhag utni hi bhari ho aur uske baad wali patli ho, to values shayad lower end ki taraf jama hain, isliye mode lower limit ke zyada paas hota hai.

Grouped data ka mode

\begin{aligned} Mode = l + \\ \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h \end{aligned}

(13.4)

jahan

$l$: modal class ki lower limit
$h$: class size (saari classes barabar)
$f_{1}$: modal class ki frequency
$f_{0}$: usse pehle wali class ki frequency
$f_{2}$: usse baad wali class ki frequency

Fig. 13.3 dikhata hai ki formula kahan se aata hai. Modal bar ke har top corner ko doosri taraf wale padosi bar ke top se jodo, taaki dono lines ek doosre ko cross karein. Crossing point mode ke upar hota hai. $f_{0}$ se $f_{1}$ tak ki chadhaai ko us girawat se compare karo jo jaati hai $f_{2},$ chadhaai jitni zyada steep hogi, mode utna hi right mein hoga.

3/4Har top corner ko door wale padosi ke top se jodo.

Fig. 13.3Dono crossing lines lambe padosi ki taraf jhukti hain.

Ek histogram 20 se 50 minutes tak ke finishing times 5 minute ki classes mein dikhata hai, heights 4, 9, 16, 11, 6 aur 4 hain. Step 1 sirf bars dikhata hai. Step 2 sabse lambe bar, 30 se 35, ko modal class ki tarah highlight karta hai aur heights par label lagata hai: usse pehle f0 = 9, f1 = 16 aur uske baad f2 = 11. Step 3 ek line 25 se 30 wale bar ke top se, 30 par, modal bar ke top right corner tak kheenchta hai, aur doosri line modal bar ke top left corner se neeche 35 se 40 wale bar ke top tak, 35 par; dono modal bar ke andar cross karti hain. Step 4 crossing se axis tak ek dashed line girata hai, lagbhag 32.9 minutes par, jo mode hai.

Example 13.5NCERT Example 5 · p. 184

Fun-run ke finishing times

Bengaluru mein pachaas runners ne 5 km fun run poori ki. Unke times neeche group kiye gaye hain. Modal finishing time nikaalo.

Table 13.9

Time (min)	Runners
20–25	4
25–30	9
30–35	16
35–40	11
40–45	6
45–50	4

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.6NCERT Example 6 · p. 185

Wages: mode banaam mean

Table 13.2 ki grouped wages ka mode nikaalo. Use mean, ₹ 501.25, se compare karo aur batao ki dono contractor ko kya batate hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh mat socho ki mode hamesha mean se neeche hoga: jo data left ki taraf failta jaata hai, usmein mode mean se upar hota hai, aur symmetric data mein dono barabar hote hain. Dono mein se kaun chunna hai, yeh sawaal par depend karta hai. Total wage bill ka budget banane wale contractor ko mean chahiye. Site ke ek typical worker ko kitna milta hai, yeh poochhne wala labour officer mode batayega. Ek aur limit: formula maan kar chalta hai ki saari classes ki width same hai. Unequal classes ka mode bhi nikal sakta hai, lekin woh is chapter ke bahar hai.

Grouped data ka mode

3 mein se 1

Ek class ke marks: 0–10: 5 students, 10–20: 12, 20–30: 9, 30–40: 3. Modal class kaun si hai?

10–2030–4020–300–10

Answer dekho

Answer (A) 10–20

10–20 ki frequency sabse zyada hai, 12. 30–40 mein marks sabse zyada hain lekin students sabse kam.

Practice · 6 questions

Practice 13.2: answers ke saath

1
NumericalMediumNCERT Exercise 13.2, Q1
Ek free eye camp mein check kiye gaye 80 logon ki umar neeche group ki gayi hai.
Age (years) People
10–20 5
20–30 8
30–40 10
40–50 14
50–60 19
60–70 16
70–80 8
1. Mode nikaalo.
  saal
  Solution dikhao
  Modal class 50–60: $l = 50,$ $f_{1} = 19,$ $f_{0} = 14,$ $f_{2} = 16,$ $h = 10.$
  $\begin{aligned} Mode \\ = 50 + \frac{5}{38 - 30} \times 10 \\ = 56.25 \end{aligned}$
2. Mean nikaalo.
  saal
  Solution dikhao
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  15 5 −3 −15
  25 8 −2 −16
  35 10 −1 −10
  45 14 0 0
  55 19 1 19
  65 16 2 32
  75 8 3 24
  Total 80 — 34
  $\overset{ˉ}{x} = 45 + 10 \times \frac{34}{80} = 49.25$ saal.
3. Dono values ko compare karo aur unka matlab samjhao.
  Solution dikhao
  Kisi bhi aur umar ke mukable lagbhag 56 saal ki umar ke log sabse zyada aaye, jabki average umar lagbhag 49 thi. Kam umar ke visitors mean ko mode se neeche kheench lete hain. Reading glasses ka stock plan karne ke liye mode zyada kaam ka figure hai.
NumericalEasyNCERT Exercise 13.2, Q2
Ek dukaan ne 180 inverter batteries ka backup time test kiya. Modal backup time nikaalo.
Backup (min) Batteries
60–90 12
90–120 28
120–150 46
150–180 52
180–210 30
210–240 12
min
Solution dikhao
$l = 150,$ $h = 30,$ $f_{1} = 52,$ $f_{0} = 46,$ $f_{2} = 30.$
$\begin{aligned} Mode \\ = 150 + \frac{6}{104 - 76} \times 30 \end{aligned}$
$= 150 + \frac{180}{28} \approx 156.43$ minutes.
3
NumericalMediumNCERT Exercise 13.2, Q3
Ek colony ke 150 gharon ke monthly electricity bills neeche group kiye gaye hain.
Bill (₹) Homes
500–1000 14
1000–1500 30
1500–2000 38
2000–2500 27
2500–3000 20
3000–3500 13
3500–4000 8
1. Modal monthly bill nikaalo.
  ₹
  Solution dikhao
  Modal class 1500–2000: $l = 1500,$ $f_{1} = 38,$ $f_{0} = 30,$ $f_{2} = 27,$ $h = 500.$
  $\begin{aligned} Mode \\ = 1500 + \frac{8}{76 - 57} \times 500 \end{aligned}$
  $= 1500 + \frac{4000}{19} \approx 1710.53,$ yaani lagbhag ₹ 1710.53.
2. Mean monthly bill nikaalo.
  ₹
  Solution dikhao
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  750 14 −3 −42
  1250 30 −2 −60
  1750 38 −1 −38
  2250 27 0 0
  2750 20 1 20
  3250 13 2 26
  3750 8 3 24
  Total 150 — −70
  $\overset{ˉ}{x} = 2250 + 500 \times \frac{- 70}{150} \approx 2016.67,$ yaani lagbhag ₹ 2016.67.
4
NumericalMediumNCERT Exercise 13.2, Q4
40 schools ke ek survey mein record kiya gaya ki har computer ko kitne students share karte hain. Do classes khaali hain.
Per computer Schools
5–10 4
10–15 9
15–20 13
20–25 8
25–30 3
30–35 0
35–40 0
40–45 3
1. Mode nikaalo.
  Solution dikhao
  Modal class 15–20: $l = 15,$ $f_{1} = 13,$ $f_{0} = 9,$ $f_{2} = 8,$ $h = 5.$
  $\begin{aligned} Mode \\ = 15 + \frac{4}{26 - 17} \times 5 \\ \approx 17.22 \end{aligned}$
2. Mean nikaalo.
  Solution dikhao
  Class marks 7.5, 12.5, …, 42.5. Products $f_{i} x_{i}$ hain 30, 112.5, 227.5, 180, 82.5, 0, 0 aur 127.5, jinka sum 760 hai.
  $\overset{ˉ}{x} = 760 \div 40 = 19.$
3. Dono measures ka matlab samjhao.
  Solution dikhao
  Sabse common situation mein lagbhag 17 students ek computer share karte hain, lekin average 19 hai. Teen bahut bhide schools (40–45) mean ko upar kheench lete hain; unhi ko naye computers ki sabse zyada zaroorat hai.
NumericalEasyNCERT Exercise 13.2, Q5
Pune mein 60 autorickshaws ne ek din mein jitni doori tay ki, woh neeche group ki gayi hai. Mode nikaalo.
Distance (km) Autos
40–60 5
60–80 11
80–100 18
100–120 14
120–140 8
140–160 4
km
Solution dikhao
$l = 80,$ $h = 20,$ $f_{1} = 18,$ $f_{0} = 11,$ $f_{2} = 14.$
$\begin{aligned} Mode \\ = 80 + \frac{7}{36 - 25} \times 20 \end{aligned}$
$= 80 + \frac{140}{11} \approx 92.73$ km.
NumericalEasyNCERT Exercise 13.2, Q6
Ek busy metro station par ek student ne 120 trains mein se har train ke ek coach mein chadhne wale passengers gine. Mode nikaalo.
Passengers Trains
0–10 6
10–20 11
20–30 18
30–40 27
40–50 24
50–60 17
60–70 10
70–80 7
Solution dikhao
$l = 30,$ $h = 10,$ $f_{1} = 27,$ $f_{0} = 18,$ $f_{2} = 24.$
$\begin{aligned} Mode \\ = 30 + \frac{9}{54 - 42} \times 10 \\ = 37.5 \end{aligned}$
Ek coach mein chadhne wale passengers ki sabse common ginti lagbhag 37 ya 38 hai.

Age (years)	People
10–20	5
20–30	8
30–40	10
40–50	14
50–60	19
60–70	16
70–80	8

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
15	5	−3	−15
25	8	−2	−16
35	10	−1	−10
45	14	0	0
55	19	1	19
65	16	2	32
75	8	3	24
Total	80	—	34

Backup (min)	Batteries
60–90	12
90–120	28
120–150	46
150–180	52
180–210	30
210–240	12

Bill (₹)	Homes
500–1000	14
1000–1500	30
1500–2000	38
2000–2500	27
2500–3000	20
3000–3500	13
3500–4000	8

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
750	14	−3	−42
1250	30	−2	−60
1750	38	−1	−38
2250	27	0	0
2750	20	1	20
3250	13	2	26
3750	8	3	24
Total	150	—	−70

Per computer	Schools
5–10	4
10–15	9
15–20	13
20–25	8
25–30	3
30–35	0
35–40	0
40–45	3

Distance (km)	Autos
40–60	5
60–80	11
80–100	18
100–120	14
120–140	8
140–160	4

Passengers	Trains
0–10	6
10–20	11
20–30	18
30–40	27
40–50	24
50–60	17
60–70	10
70–80	7

13.4 Grouped Data ka Median

Data ko order mein lagane ke baad beech wali value median (माध्यक) kehlaati hai: aadhe observations usse neeche hote hain aur aadhe upar. Ascending order mein $n$ ungrouped values ke liye, jab $n$ odd ho, to median $(\frac{n + 1}{2})$ th value hoti hai. Jab $n$ even ho, to yeh $\frac{n}{2}$ th aur $(\frac{n}{2} + 1)$ th values ka average hota hai.

Frequency table ho to har value likhne ki zaroorat nahi. Frequencies ka running total rakho, yaani cumulative frequency (संचयी बारंबारता), aur dekho ki woh beech ko kahan paar karta hai. Yeh rahe Class 10 ke 60 students ke bataye neend ke ghante, nearest hour tak round kiye hue.

Table 13.1060 students ke neend ke ghante, cumulative frequency (cf) ke saath

Hours	Students	cf
5	4	4
6	9	13
7	17	30
8	16	46
9	10	56
10	4	60

$n = 60$ even hai, isliye humein 30th aur 31st values chahiye. Pehle 13 students 6 ghante ya usse kam sote hain. Agle 17, yaani 14th se 30th tak, 7 ghante sote hain, isliye 30th value 7 hai. 31st agle group mein hai, 8 ghante. Median $\frac{7 + 8}{2} = 7.5$ ghante hai: lagbhag aadhe students 7.5 ghante se kam sote hain aur aadhe zyada. Sirf pehla aur aakhri column milkar ek cumulative frequency table banate hain.

Grouped data ki cumulative frequency

Ab grouped data lo: ek pre-board Science paper mein 70 students ke 80 mein se marks.

Table 13.1170 students ke marks, cumulative frequency (cf) ke saath

Marks	Students	cf
0–10	3	3
10–20	5	8
20–30	8	16
30–40	11	27
40–50	15	42
50–60	13	55
60–70	9	64
70–80	6	70

cf column ek running total rakhta hai. 20–30 ke liye iski entry hai $3 + 5 + 8 = 16 :$ solah students ne 30 se kam score kiya, isliye us class ki cumulative frequency 16 hai. Har entry upar wali entry mein class ki apni frequency jodti hai, isliye aakhri entry hamesha hoti hai $n .$ Har total ek upper limit se neeche ke students ginta hai (10 se kam, 20 se kam, …), isliye is table ko less than cumulative frequency distribution kehte hain.

3/335 wali line pehli baar 40–50 bar tak nahi pahunchti.

Fig. 13.4

\frac{n}{2}

se upar jaane wala pehla running total median class batata hai.NCERT Table 13.13 · p. 191

Classes 0–10 se 70–80 ke upar aath bars khade hain. Step 1: bars class frequencies 3, 5, 8, 11, 15, 13, 9 aur 6 dikhate hain. Step 2: har bar us class tak ki saari frequencies ke running total tak badh jaata hai, 3, 8, 16, 27, 42, 55, 64 aur 70, isliye aakhri bar n = 70 tak pahunchta hai. Step 3: 70 ke aadhe, yaani 35, par ek horizontal dashed line chauthe bar (27) ke upar se lekin paanchve bar (42) ke neeche se guzarti hai, isliye paanchvi class, 40–50, median class ki tarah shade ki gayi hai.

Ulti taraf se ginna bhi kaam karta hai: saare 70 se shuru karo aur ek-ek class hatate jao. 0–10 ke 3 students hatao, to 67 bachte hain jinhone kam se kam 10 score kiya; agle 5 hatao, to 62 ne kam se kam 20 score kiya, aur aise hi aage. Yeh totals lower limits se upar ki taraf ginte hain, aur isse more than or equal to cumulative frequency distribution milta hai.

Table 13.12Wahi marks “more than or equal to” distribution ke roop mein

Marks	Students
0 ya zyada	70
10 ya zyada	67
20 ya zyada	62
30 ya zyada	54
40 ya zyada	43
50 ya zyada	28
60 ya zyada	15
70 ya zyada	6

Median formula

70 ka aadha 35 hai. Less-than totals hain 3, 8, 16, 27, 42, …: 35th student pehle 27 mein nahi hai lekin pehle 42 mein hai, isliye 35th mark 40–50 mein hai. Yahi median class (माध्यक वर्ग) hai, yaani woh pehli class jiski cumulative frequency isse zyada ho: $\frac{n}{2} .$

Median ko class ke andar rakhne ke liye maan lo ki iske 15 students 40 se 50 tak barabar faile hain (Fig. 13.5). Humein inmein se $35 - 27 = 8$ chahiye, jo humein class ke across $\frac{8}{15}$ hissa aage le jaata hai: $40 + \frac{8}{15} \times 10 \approx 45.33.$ Yaani lagbhag aadhe students ne 45.33 se kam score kiya aur aadhe ne zyada.

3/315 mein se 8 yaani width ka $\frac{8}{15}$ : median ≈ 45.33.

Fig. 13.5Median, median class ke across ek fraction

\frac{n / 2 - c f}{f}

jitna aage hota hai.NCERT Median formula · p. 193

Ek strip 40 se 50 marks tak jaati hai. Step 1: pandrah barabar doori wale dots median class ke 15 students hain, aur left mein ek note kehta hai ki 27 students ne 40 se kam score kiya. Step 2: pehle 8 dots highlight hote hain, kyunki 35 minus 27 yaani 8 aur students chahiye 35th tak pahunchne ke liye. Step 3: ek arrow 40 se aathve dot tak jaata hai, uspar likha hai 10 ki class width ka 8/15, aur us point ko median mark kiya gaya hai, lagbhag 45.33.

Grouped data ka median

\begin{aligned} Median = l + \\ \frac{\frac{n}{2} - c f}{f} \times h \end{aligned}

(13.5)

jahan

$l$: median class ki lower limit
$n$: observations ki ginti
$c f$: median class se pehle wali class ki cumulative frequency
$f$: median class ki frequency
$h$: class size (saari classes barabar)

Mode formula ki tarah yeh formula bhi barabar class sizes maan kar chalta hai. Unequal classes wale data ka median bhi nikal sakta hai, lekin woh is chapter ke bahar hai.

Example 13.7NCERT Example 7 · p. 194

School-bag ka weight

Bag-check day par students ne 60 school bags tole. Koi bhi bag 2 kg se halka nahi tha. Median weight nikaalo.

Table 13.13

Weight (kg)	Bags
3 se kam	5
4 se kam	14
5 se kam	28
6 se kam	44
7 se kam	54
8 se kam	60

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.8NCERT Example 8 · p. 195

Do mitee hui frequencies

Ek hospital OPD mein 60 patients ke waiting times neeche group kiye gaye hain. Do frequencies mit gayi thi. Median waiting time 35 minutes hai. Nikaalo $x$ aur $y .$

Table 13.14

Intezaar ka time (min)	Patients
0–10	3
10–20	$x$
20–30	12
30–40	16
40–50	$y$
50–60	7
60–70	4

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/2Neeche 0.5 ghatao aur upar 0.5 jodo.

Fig. 13.6Har limit ko aadha gap bahar khiskao taaki classes mil jaayein.NCERT Exercise 13.3 Q4 ka hint · p. 199

Step 1 teen histogram bars dikhata hai jin par 41–45, 46–50 aur 51–55 likha hai, aur 45 aur 46 ke beech tatha 50 aur 51 ke beech patle khaali gaps hain. Step 2 har bar ko dono taraf aadha unit chauda karta hai, to ab bars ek doosre ko chhoote hain, aur nayi boundaries 40.5, 45.5, 50.5 aur 55.5 neeche likhi hain. Ek note kehta hai ki ab class width 5 hai.

Measure chunna

Ab tum kisi bhi grouped data ke teen averages nikaal sakte ho. Report kaun sa karein? Mean ki kamzori se shuru karte hain. Fig. 13.7 mein nau daily wages ₹ 500 par balance hoti hain. Inmein se ek ko manager ke ₹ 1300 se badlo, to mean ₹ 580 par uchhal jaata hai, ek aisi wage jo nau mein se aath log nahi kamaate. Phir bhi mean sabse zyada kaam aane wala measure hai: har observation usmein jaata hai, isliye poore groups ko compare karne ka yahi fair tareeka hai, jaise do sections, do saal ya do districts.

2/3Ek wage ₹ 1300 ho jaati hai: mean ₹ 580 par uchhal jaata hai.

Fig. 13.7Ek extreme value mean ko kheenchti hai; median aur mode apni jagah rehte hain.NCERT Kaun sa measure · p. 197

Ek number line 400 se 1350 rupees tak jaati hai. Step 1 wages 420, 450, 480, 500, 500, 500, 520, 550 aur 580 ke nau dots dikhata hai; upar ek mean marker aur neeche ek median-aur-mode marker, dono 500 par point karte hain. Step 2 580 wale dot ko door right mein, 1300 par le jaata hai; mean marker 580 par khisak jaata hai jabki median aur mode markers 500 par hi rehte hain. Step 3 ek note jodta hai ki ab nau mein se aath wages mean se neeche hain, isliye median aur mode typical worker ko behtar describe karte hain.

Median sirf beech wali position par depend karta hai, isliye ₹ 1300 use hila nahi paata. Jab kuch values extreme hon, jaise incomes, house prices ya queue mein waiting time, to median typical case ko behtar describe karta hai. Mode is sawaal ka jawab deta hai ki “sabse common kya hai?”: sabse zyada bikne wala shoe size, ice-cream counter par favourite flavour, ration shop ka sabse busy ghanta. Teeno mein se sirf yahi un categories ke liye bhi kaam karta hai jo numbers nahi hain.

Compare

Kaun sa average report karein

Mean

har value use karta hai
groups compare karne ke liye best
extreme values se kheench jaata hai

Median

beech wali position
extremes se nahi hilta
incomes aur prices ke liye best

Mode

sabse common value
“sabse popular” wale sawaalon ke liye best
categories ke liye bhi kaam karta hai

Jo data bahut zyada ek taraf jhuka na ho, uske liye teeno measures ek empirical relation se jude hote hain, jo prove karke nahi balki experience se mila hai:

Empirical relation

\begin{aligned} 3 Median & = Mode \\ + 2 Mean \end{aligned}

(13.6)

Ise grouped wages par check karo. Unka median hai $450 + \frac{20 - 9}{12} \times 50 \approx 495.83,$ isliye $3 \times 495.83 \approx 1487.5.$ Mode aur mean se milta hai $483.33 + 2 \times 501.25 \approx 1485.8.$ Paas hai, lekin barabar nahi. Exam questions mein tum is relation ko equation ki tarah use karte ho, baaki do measures se teesra nikaalne ke liye.

InteractiveGrouped-data explorer

Ek grouped frequency table ka histogram, jismein chaar data sets mein se chun sakte ho: daily wages, T20 totals, fun-run times aur marks. Kisi class ki frequency badalne ke liye bar ko tap ya drag karo, ya minus aur plus buttons use karo. Vertical markers mean ko solid line, median ko dashed line aur mode ko dotted line se dikhate hain. Ek table har class ko uske class mark, frequency aur cumulative frequency ke saath dikhata hai, aur ek panel direct method se mean, apne formulas se median aur mode nikaalta hai, aur 3 times median ko mode plus do times mean se compare karta hai.

Fig. 13.8Frequencies badlo aur dekho mean, median aur mode kaise khiskte hain.

Median aur measure chunna

4 mein se 1

Classes 0–10, 10–20, 20–30, … ki cumulative frequencies 6, 15, 24, 37, 46, 50 hain. Median class hai

20–3030–4010–2040–50

Answer dekho

Answer (B) 30–40

\frac{n}{2} = 25.

25 se upar pehli cumulative frequency 37 hai, 30–40 ki. Total 24, 25 ke paas hai lekin usse kam hai, isliye 20–30 woh galat answer hai jo sahi lagta hai.

Practice · 7 questions

Practice 13.3: answers ke saath

1
Long answerMediumNCERT Exercise 13.3, Q1
Ek housing society ke 70 gharon ka daily paani ka use neeche group kiya gaya hai.
Water used (L) Households
100–150 4
150–200 9
200–250 15
250–300 19
300–350 12
350–400 7
400–450 4
1. Median nikaalo.
  L
  Solution dikhao
  Cumulative frequencies: 4, 13, 28, 47, 59, 66, 70. $\frac{n}{2} = 35,$ isliye median class 250–300 hai, saath mein $c f = 28,$ $f = 19.$
  $\begin{aligned} Median \\ = 250 + \frac{35 - 28}{19} \times 50 \\ \approx 268.42 \end{aligned}$
2. Mean nikaalo.
  L
  Solution dikhao
  $a = 275,$ $h = 50,$ ke saath $u_{i}$ −3 se 3 tak hain aur products $f_{i} u_{i}$ hain −12, −18, −15, 0, 12, 14, 12, isliye $Σ f_{i} u_{i} = - 7.$
  $\overset{ˉ}{x} = 275 + 50 \times \frac{- 7}{70} = 270$ litres.
3. Mode nikaalo.
  L
  Solution dikhao
  Modal class 250–300: $f_{1} = 19,$ $f_{0} = 15,$ $f_{2} = 12.$
  $\begin{aligned} Mode \\ = 250 + \frac{4}{38 - 27} \times 50 \\ \approx 268.18 \end{aligned}$
4. Teeno measures compare karo.
  Solution dikhao
  Teeno 270 litres ke paas hain, isliye data lagbhag symmetric hai aur koi extreme ghar nahi hai. Inmein se koi bhi ek typical ghar ka daily use describe karta hai.
2
NumericalHardNCERT Exercise 13.3, Q2
80 students ke 70 mein se marks neeche group kiye gaye hain. Median 36 hai. Nikaalo $x$ aur $y .$
Marks Students
0–10 6
10–20 $x$
20–30 14
30–40 20
40–50 $y$
50–60 10
60–70 5
1. Nikaalo $x .$
  Solution dikhao
  Total: $55 + x + y = 80,$ isliye $x + y = 25.$ Median 36, 30–40 mein hai, isliye $l = 30,$ $f = 20,$ $h = 10$ aur $c f = 6 + x + 14 = 20 + x .$
  $36 = 30 + \frac{40 - (20 + x)}{20} \times 10$
  To $6 = \frac{20 - x}{2},$ jisse milta hai $x = 8.$
2. Nikaalo $y .$
  Solution dikhao
  $y = 25 - 8 = 17.$
NumericalMediumNCERT Exercise 13.3, Q3
Ek RTO office ne learner's licence ke liye apply karne wale 150 logon ki umar record ki. 18 se kam umar ka koi apply nahi kar sakta. Median age nikaalo.
Age (years) Applicants
20 se kam 22
25 se kam 61
30 se kam 92
35 se kam 112
40 se kam 126
45 se kam 136
50 se kam 143
55 se kam 147
60 se kam 150
saal
Solution dikhao
Classes hain 18–20, 20–25, 25–30, …, 55–60. $\frac{n}{2} = 75.$ 75 se upar pehla running total 92 hai, isliye median class 25–30 hai, saath mein $c f = 61,$ $f = 92 - 61 = 31,$ $h = 5.$
$\begin{aligned} Median \\ = 25 + \frac{75 - 61}{31} \times 5 \\ \approx 27.26 \end{aligned}$
Median age lagbhag 27.26 saal hai.
NumericalMediumNCERT Exercise 13.3, Q4
Ek poultry farm ne 50 ande nearest gram tak tole. Median weight nikaalo.
Weight (g) Eggs
41–45 4
46–50 9
51–55 15
56–60 12
61–65 7
66–70 3
g
Solution dikhao
Class $f$ $c f$
40.5–45.5 4 4
45.5–50.5 9 13
50.5–55.5 15 28
55.5–60.5 12 40
60.5–65.5 7 47
65.5–70.5 3 50
$\frac{n}{2} = 25,$ isliye median class 50.5–55.5 hai: $l = 50.5,$ $c f = 13,$ $f = 15,$ $h = 5.$
$\begin{aligned} Median \\ = 50.5 + \frac{25 - 13}{15} \times 5 \\ = 54.5 \end{aligned}$
NumericalMediumNCERT Exercise 13.3, Q5
Ek dealer ne test kiya ki 250 electric scooters mein se har ek full charge par kitna chala. Median range nikaalo.
Range (km) Scooters
60–70 12
70–80 28
80–90 45
90–100 62
100–110 51
110–120 34
120–130 18
km
Solution dikhao
Cumulative frequencies: 12, 40, 85, 147, …. 125 se upar pehli 147 hai, isliye median class 90–100 hai, saath mein $c f = 85,$ $f = 62.$
$\begin{aligned} Median \\ = 90 + \frac{125 - 85}{62} \times 10 \\ \approx 96.45 \end{aligned}$
Aadhe scooters lagbhag 96.45 km se kam chale.
6
Long answerMediumNCERT Exercise 13.3, Q6
Ek school ne 100 students se survey kiya ki unhe school pahunchne mein kitna time lagta hai.
Time (min) Students
0–10 8
10–20 22
20–30 34
30–40 20
40–50 10
50–60 6
1. Median time nikaalo.
  min
  Solution dikhao
  Cumulative frequencies: 8, 30, 64, 84, 94, 100. Median class 20–30 hai, saath mein $c f = 30,$ $f = 34.$
  $\begin{aligned} Median \\ = 20 + \frac{50 - 30}{34} \times 10 \\ \approx 25.88 \end{aligned}$
2. Mean time nikaalo.
  min
  Solution dikhao
  Class marks 5, 15, …, 55. Products $f_{i} x_{i}$ hain 40, 330, 850, 700, 450, 330, jinka sum 2700 hai.
  $\overset{ˉ}{x} = 2700 \div 100 = 27$ minutes.
3. Modal time nikaalo.
  min
  Solution dikhao
  Modal class 20–30: $f_{1} = 34,$ $f_{0} = 22,$ $f_{2} = 20.$
  $\begin{aligned} Mode \\ = 20 + \frac{12}{68 - 42} \times 10 \\ \approx 24.62 \end{aligned}$
NumericalEasyNCERT Exercise 13.3, Q7
May ke 31 dinon mein Jaipur ke daily maximum temperatures neeche group kiye gaye hain. Median maximum temperature nikaalo.
Temperature (°C) Days
36–38 2
38–40 5
40–42 9
42–44 8
44–46 5
46–48 2
°C
Solution dikhao
Cumulative frequencies: 2, 7, 16, 24, 29, 31. 15.5 se upar pehli 16 hai, isliye median class 40–42 hai, saath mein $c f = 7,$ $f = 9,$ $h = 2.$
$\begin{aligned} Median \\ = 40 + \frac{15.5 - 7}{9} \times 2 \\ \approx 41.89 \end{aligned}$

Water used (L)	Households
100–150	4
150–200	9
200–250	15
250–300	19
300–350	12
350–400	7
400–450	4

Marks	Students
0–10	6
10–20	$x$
20–30	14
30–40	20
40–50	$y$
50–60	10
60–70	5

Age (years)	Applicants
20 se kam	22
25 se kam	61
30 se kam	92
35 se kam	112
40 se kam	126
45 se kam	136
50 se kam	143
55 se kam	147
60 se kam	150

Weight (g)	Eggs
41–45	4
46–50	9
51–55	15
56–60	12
61–65	7
66–70	3

Class	$f$	$c f$
40.5–45.5	4	4
45.5–50.5	9	13
50.5–55.5	15	28
55.5–60.5	12	40
60.5–65.5	7	47
65.5–70.5	3	50

Range (km)	Scooters
60–70	12
70–80	28
80–90	45
90–100	62
100–110	51
110–120	34
120–130	18

Time (min)	Students
0–10	8
10–20	22
20–30	34
30–40	20
40–50	10
50–60	6

Temperature (°C)	Days
36–38	2
38–40	5
40–42	9
42–44	8
44–46	5
46–48	2

13.5 Summary

Recap

Summary aur key points

Direct method: $\overset{ˉ}{x} = Σ f_{i} x_{i} \div Σ f_{i},$ jismein har class ko uske class mark par maana jaata hai (Section 13.2).
Assumed mean method: $\overset{ˉ}{x} = a + Σ f_{i} d_{i} \div Σ f_{i},$ jahan $d_{i} = x_{i} - a$ (assumed mean).
Step-deviation method: $\overset{ˉ}{x} = a + h \overset{ˉ}{u},$ jahan $u_{i} = (x_{i} - a) \div h$ (step deviation).
Teeno methods same mean dete hain; mid-point assumption ki wajah se yeh ek estimate hai.
Modal class ki frequency sabse zyada hoti hai, aur mode usi ke andar hota hai: formula (13.4).
Kisi class ki cumulative frequency uski apni frequency plus usse pehle ki saari classes ki frequencies hoti hai (cumulative frequency).
Less-than tables upper limits use karti hain; more-than-or-equal tables lower limits use karti hain.
Median class woh pehli class hai jiski cumulative frequency isse zyada ho: $\frac{n}{2};$ median formula (13.5) se aata hai.
Mode ya median formula use karne se pehle gaps wali classes ko continuous banao (gaps band karo).
Groups compare karne ke liye mean, extremes hon to median, sabse common value ke liye mode (measure chunna).
Empirical relation (13.6) kehta hai: $3 Median = Mode + 2 Mean .$

2/3Median area ko do barabar hisson mein baant deta hai.

Fig. 13.9Mean: balance point. Median: area ko aadha karta hai. Mode: crossing ke neeche.

40 grouped wages ka histogram, 350 se 650 rupees tak, bar heights 3, 6, 12, 9, 6 aur 4. Step 1 lagbhag 501 rupees ko ek pointer aur dashed line se mark karta hai: mean woh point hai jahan histogram balance hoga. Step 2 lagbhag 496 rupees par ek dashed line kheenchta hai aur uske left ka area shade karta hai; har taraf 20 workers hain, isliye median area ko aadha-aadha baant deta hai. Step 3 sabse lambe bar ke andar do crossing lines kheenchta hai aur crossing se lagbhag 483 rupees tak ek line girata hai, jo mode hai.

Flash cards · 7

Quick recall

1 / 7

Board-exam ki practice

Yeh questions CBSE board-paper ke formats follow karte hain: multiple choice, assertion–reason, short answer, long answer aur ek case study. Har calculation question mein working table dikhao; examiners uske marks dete hain.

Practice · 9 questions

Practice questions answers ke saath (NCERT “Board-exam ki practice”)

Quick answerEasy
Class 15.5–20.5 ka class mark hai
51817.518.5
Answer dekho
Answer (B) 18
$(15.5 + 20.5) \div 2 = 18.$ 5 class size hai.
Quick answerEasy
Step-deviation method mein $u_{i}$ kiske barabar hai?
$(x_{i} + a) \div h$ $h (x_{i} - a)$ $(x_{i} - a) \div h$ $(a - x_{i}) \div h$
Answer dekho
Answer (C) $(x_{i} - a) \div h$
$u_{i} = (x_{i} - a) \div h .$ Aakhri option mein sign ulta hai.
Quick answerEasy
Ek distribution ka mode 12 hai aur mean 15 hai. Empirical relation se uska median hai
13.5161413
Answer dekho
Answer (C) 14
$3 Median = 12 + 30 = 42,$ isliye median 14 hai.
Quick answerEasy
Less-than cumulative frequency table mein aakhri cumulative frequency kiske barabar hoti hai?
aakhri class ki frequencyaakhri class ki upper limit $\frac{n}{2}$ total frequency $n$
Answer dekho
Answer (D) total frequency $n$
Aakhir tak har observation gina jaata hai, isliye aakhri running total hai $n .$
Quick answerMedium
Assertion (A): Classes 1–5, 6–10, 11–15, … ke liye median class 6–10 ki jo lower limit median formula mein use hoti hai, woh 5.5 hai. Reason (R): Median formula ko aisi classes chahiye jinke beech koi gap na ho.
A aur R dono true hain, aur R, A ki sahi explanation hai.A aur R dono true hain, lekin R, A ki sahi explanation nahi hai.A true hai lekin R false hai.A false hai lekin R true hai.
Answer dekho
Answer (A) A aur R dono true hain, aur R, A ki sahi explanation hai.
Gaps band karne par 6–10 ban jaata hai 5.5–10.5, isliye $l = 5.5.$ Dono true hain aur R hi iski exact wajah hai.
Quick answerMedium
Assertion (A): Modal class woh class hai jiski cumulative frequency sabse badi ho. Reason (R): Mode woh value hai jo sabse zyada baar aati hai.
A aur R dono true hain, aur R, A ki sahi explanation hai.A aur R dono true hain, lekin R, A ki sahi explanation nahi hai.A true hai lekin R false hai.A false hai lekin R true hai.
Answer dekho
Answer (D) A false hai lekin R true hai.
Sabse badi cumulative frequency hamesha aakhri class ki hoti hai, yaani $n .$ Modal class ki frequency sabse badi hoti hai. A false hai aur R true hai.
Short answerMedium
Step-deviation method se Class 10 ke 40 students ka mean weight nikaalo.
Weight (kg) Students
35–40 3
40–45 7
45–50 12
50–55 9
55–60 6
60–65 3
kg
Solution dikhao
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
37.5 3 −2 −6
42.5 7 −1 −7
47.5 12 0 0
52.5 9 1 9
57.5 6 2 12
62.5 3 3 9
Total 40 — 17
$\overset{ˉ}{x} = 47.5 + 5 \times \frac{17}{40} = 49.625 \approx 49.63$ kg.
8
Long answerHard
Ek cycling club ke 120 members ne ek hafte mein jitni doori cycle ki, woh neeche group ki gayi hai. Mean distance 50 km hai. Nikaalo $f_{1}$ aur $f_{2} .$
Distance (km) Members
0–20 17
20–40 $f_{1}$
40–60 32
60–80 $f_{2}$
80–100 19
1. Nikaalo $f_{1} .$
  Solution dikhao
  Total: $68 + f_{1} + f_{2} = 120,$ isliye $f_{1} + f_{2} = 52.$
  Class marks 10, 30, 50, 70, 90, isliye $Σ f_{i} x_{i} = 170 + 30 f_{1} + 1600 + 70 f_{2} + 1710.$ Isko barabar hona chahiye isse: $50 \times 120 = 6000.$
  $30 f_{1} + 70 f_{2} = 2520$
  10 se divide karo: $3 f_{1} + 7 f_{2} = 252.$ Rakho $f_{2} = 52 - f_{1}$ : $3 f_{1} + 364 - 7 f_{1} = 252,$ isliye $4 f_{1} = 112$ aur $f_{1} = 28.$
2. Nikaalo $f_{2} .$
  Solution dikhao
  $f_{2} = 52 - 28 = 24.$ Check: $Σ f_{i} x_{i} = 170 + 840 + 1600 + 1680 + 1710 = 6000.$
9
Long answerMedium
Case study. Chennai ki ek cricket academy ne batters ka squad chunne ke liye ek season mein apne 50 players mein se har ek ke runs record kiye.
Runs Players
0–100 6
100–200 9
200–300 15
300–400 11
400–500 6
500–600 3
1. Median class kaun si hai?
  100–200200–300300–400400–500
  Answer dekho
  Answer (B) 200–300
  $\frac{n}{2} = 25;$ 25 se upar pehli cumulative frequency 30 hai, 200–300 ki.
2. Median runs nikaalo.
  runs
  Solution dikhao
  $\begin{aligned} Median \\ = 200 + \frac{25 - 15}{15} \times 100 \\ \approx 266.67 \end{aligned}$
3. Step-deviation method se mean runs nikaalo.
  runs
  Solution dikhao
  $a = 250,$ $h = 100,$ ke saath $u_{i}$ −2 se 3 tak hain aur products $f_{i} u_{i}$ hain −12, −9, 0, 11, 12, 9, isliye $Σ f_{i} u_{i} = 11.$
  $\overset{ˉ}{x} = 250 + 100 \times \frac{11}{50} = 272$ runs.
4. Modal runs nikaalo.
  runs
  Solution dikhao
  $\begin{aligned} Mode \\ = 200 + \frac{15 - 9}{30 - 9 - 11} \times 100 \\ = 260 \end{aligned}$
  Check: $3 \times 266.67 = 800$ aur $260 + 2 \times 272 = 804,$ paas-paas, jaisa empirical relation predict karta hai.

Weight (kg)	Students
35–40	3
40–45	7
45–50	12
50–55	9
55–60	6
60–65	3

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
37.5	3	−2	−6
42.5	7	−1	−7
47.5	12	0	0
52.5	9	1	9
57.5	6	2	12
62.5	3	3	9
Total	40	—	17

Distance (km)	Members
0–20	17
20–40	$f_{1}$
40–60	32
60–80	$f_{2}$
80–100	19

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise 13.1, Q4	Practice 13.1 · Q4
NCERT Exercise 13.1, Q5	Practice 13.1 · Q5
NCERT Exercise 13.1, Q6	Practice 13.1 · Q6
NCERT Exercise 13.1, Q7	Practice 13.1 · Q7
NCERT Exercise 13.1, Q8	Practice 13.1 · Q8
NCERT Exercise 13.1, Q9	Practice 13.1 · Q9
NCERT Exercise 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise 13.3, Q2	Practice 13.3 · Q2
NCERT Exercise 13.3, Q3	Practice 13.3 · Q3
NCERT Exercise 13.3, Q4	Practice 13.3 · Q4
NCERT Exercise 13.3, Q5	Practice 13.3 · Q5
NCERT Exercise 13.3, Q6	Practice 13.3 · Q6
NCERT Exercise 13.3, Q7	Practice 13.3 · Q7

Quick check: MCQs

Warm-up: ungrouped data ke averages3 questions
Grouped data ka mean4 questions
Grouped data ka mode3 questions
Median aur measure chunna4 questions

Glossary · 12

Key terms (English + Hindi)

assumed meanकल्पित माध्यkalpit maadhya: Ek convenient value $a,$ aam taur par beech ka class mark, jisse class marks naape jaate hain taaki calculation chhoti ho.Chapter mein pehli baar yahan aaya →

class markवर्ग चिह्नvarg chihn: Kisi class ka mid-point, (lower limit + upper limit) ÷ 2. Yeh class ke har observation ki jagah leta hai.Chapter mein pehli baar yahan aaya →
class sizeवर्ग मापvarg maap: Kisi class ki width, upper limit minus lower limit, jise likhte hain $h .$ Chapter mein pehli baar yahan aaya →
cumulative frequencyसंचयी बारंबारताsanchayi baarambaarta: Frequencies ka running total: kisi class ki frequency mein usse pehle ki saari classes ki frequencies jodi hui.Chapter mein pehli baar yahan aaya →

deviationविचलनvichalan: Class mark assumed mean se kitna door hai: $d_{i} = x_{i} - a .$ Class mark agar is value se chhota ho to yeh negative hota hai: $a .$ Chapter mein pehli baar yahan aaya →

frequencyबारंबारताbaarambaarta: Koi value kitni baar aati hai, ya kisi class mein kitne observations aate hain.Chapter mein pehli baar yahan aaya →

meanमाध्यmaadhya: Saare observations ka sum, unki ginti se divide kiya hua; frequencies wale data ke liye, $Σ f_{i} x_{i} \div Σ f_{i} .$
measure of central tendencyकेंद्रीय प्रवृत्ति का मापkendriya pravritti ka maap: Ek single number jo poore data set ko represent karta hai: mean, median ya mode.Chapter mein pehli baar yahan aaya →
medianमाध्यकmaadhyak: Order mein lagaye data ki beech wali value, taaki aadhe observations usse neeche hon aur aadhe upar.Chapter mein pehli baar yahan aaya →
median classमाध्यक वर्गmaadhyak varg: Woh pehli class jiski cumulative frequency isse badi ho: $\frac{n}{2};$ median isi ke andar hota hai.Chapter mein pehli baar yahan aaya →
modal classबहुलक वर्गbahulak varg: Grouped frequency distribution mein sabse zyada frequency wali class.Chapter mein pehli baar yahan aaya →
modeबहुलकbahulak: Woh value jo sabse zyada baar aati hai. Grouped data ke liye ise modal class ke andar mode formula se estimate kiya jaata hai.Chapter mein pehli baar yahan aaya →

Runs	Players
0–100	6
100–200	9
200–300	15
300–400	11
400–500	6
500–600	3

13.1 Introduction

13.2 Grouped Data ka Mean

Assumed mean method

Step-deviation method

13.3 Grouped Data ka Mode

13.4 Grouped Data ka Median

Grouped data ki cumulative frequency

Median formula

Measure chunna

13.5 Summary

Board-exam ki practice

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye