Statistics: Class 11 Maths Chapter 13 Hinglish mein

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

Published 22 Oct 2026 · Updated 22 Oct 2026

78 min
1 simulation
33 practice questions
25 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki sirf ek measure of central tendency kisi data set ko poori tarah describe kyun nahi kar sakta, aur data sets ko unke spread se compare karo.
Data set ki range nikalo, uska matlab samjho aur uski kamzoriyan pehchano.
Raw data aur discrete va continuous frequency distributions ke liye mean aur median ke about mean deviation calculate karo.
Grouped distribution ka median cumulative frequencies se aur median-class formula se nikalo.
Step deviations (ek assumed mean aur scale ka change) use karke means aur standard deviations ka hisaab aasaan banao.
Variance aur standard deviation define karo, unke computing formulas derive karo aur ungrouped va grouped data ke liye unhe calculate karo.
Predict karo ki har observation ko shift ya scale karne par mean, variance aur standard deviation kaise badalte hain, aur recording error ke baad unhe correct karo.

NCERT Mathematics Class 11, Chapter 13, pp. 257–288 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

13.1 Introduction

Pune mein ek hi stop se do buses, Route P aur Route Q, Meera ke college tak jaati hain. Aath subah tak woh note karti hai ki use har bus ke liye kitne minute wait karna pada. Route P: 15, 3, 21, 11, 1, 19, 7, 11. Route Q: 12, 10, 11, 13, 9, 11, 12, 10. Exam wali subah use kis bus par bharosa karna chahiye?

Pichhli classes ke averages se shuru karte hain, yaani measures of central tendency. $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ ka mean unka total hai jo sabmein barabar baanta gaya ho:

n observations ka mean

\overset{ˉ}{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}

(13.1)

jahan

$\sum_{i = 1}^{n} x_{i}$: sum $x_{1} + x_{2} + \dots + x_{n}$

Median ke liye pehle observations ko increasing (ya decreasing) order mein lagao. Phir:

agar $n$ odd hai, to median $(\frac{n + 1}{2})$ th observation hai;
agar $n$ even hai, to yeh $\frac{n}{2}$ th aur $(\frac{n}{2} + 1)$ th observations ka mean hai.

Har route ke aath waits ka total 88 minute hai, isliye dono means 11 minute hain. Order mein Route P 1, 3, 7, 11, 11, 15, 19, 21 hai aur Route Q 9, 10, 10, 11, 11, 12, 12, 13 hai. Dono lists mein 4th aur 5th values 11 aur 11 hain, isliye dono medians bhi 11 hain. Averages ke hisaab se dono routes bilkul ek jaise hain.

3/3Dono routes ka mean 11 aur median 11 hai, phir bhi Q kahin zyada predictable hai.

Fig. 13.1Centre same, spread bilkul alag.NCERT Fig 13.1 · p. 258

0 se 24 minute tak do number lines. Pehle step mein Route P ke aath waits gol dots ke roop mein 1, 3, 7, 11, 11, 15, 19 aur 21 par dikhte hain, dono 11 ek ke upar ek; yeh line ke zyadatar hisse par faile hain. Doosre step mein Route Q ke waits neeche wali line par square dots ke roop mein 9, 10, 10, 11, 11, 12, 12 aur 13 par dikhte hain, ek doosre ke bilkul paas, aur repeat hone wali values ek ke upar ek. Teesre step mein 11 par ek dashed vertical line dono plots ko kaatti hai aur uspar mean = median = 11 likha hai, jo dikhata hai ki same centre ke peeche bahut alag spreads chhupe ho sakte hain.

Phir bhi jo bhi dono buses mein baitha ho, woh kahega ki dono bahut alag hain. Route Q ka wait kabhi bhi 11 se 2 minute se zyada door nahi gaya. Route P ne ek baar use 21 minute wait karaya aur ek baar sirf 1 minute. Q ki values centre ke aas-paas simti hui hain jabki P ki values poori line par bikhri hain (Fig. 13.1). Average hume batata hai ki data kahan hai. Hume ek aur number chahiye jo bataye ki data kitna faila hua hai.

Definition

Measure of dispersion

Ek single number jo batata hai ki observations ka set kitna faila (bikhra) hua hai, aam taur par kisi central value ke about.

ExampleRange, mean deviation aur standard deviation.

Yeh chapter aise measures of dispersion (प्रकीर्णन की माप) banata hai aur dikhata hai ki raw lists aur frequency tables ke liye inhe kaise compute karte hain.

13.2 Measures of Dispersion

Spread ko number dene ka koi ek tareeka nahi hai. Kuch measures sirf do extremes ko dekhte hain; baaki poochhte hain ki har observation mean ya median jaise centre se kitni door hai. Aam taur par use hone wale measures hain:

range;
quartile deviation;
mean deviation;
standard deviation, aur uske saath uska square, variance.

Inmein se quartile deviation aage ke courses ke liye chhoda gaya hai; yeh chapter baaki teen ko ek-ek karke develop karta hai.

13.3 Range

Spread ko jaldi samajhne ka sabse aasaan tareeka hai do extremes ko dekhna. Route P ke waits 1 se 21 minute tak the; Route Q ke 9 se 13 tak.

Definition

Range

Kisi data set ke sabse bade aur sabse chhote observation ka difference.

ExampleRoute P ke liye, $21 - 1 = 20$ minute.

range

Range = x_{\max} - x_{\min}

(13.2)

To Route P ki range (परिसर) 20 minute hai aur Route Q ki range 4 minute: is measure par P paanch guna zyada bikhra hai. Range data ki units mein hi hoti hai.

3/3Agar 13 wala ek wait 40 ho jaaye, to Q ki range uchhal kar 31 ho jaati hai.

Fig. 13.2Range sirf do extreme values use karti hai.

0 se 40 minute tak do number lines. Pehle step mein Route P ke dots ke upar ek bracket 1 se 21 tak jaata hai aur uspar range 20 likha hai. Doosre step mein Route Q ke paas-paas wale dots ke upar ek bracket 9 se 13 tak jaata hai aur uspar range 4 likha hai. Teesre step mein Route Q ka ek dot, 13, khisak kar door right mein 40 par chala jaata hai, jaise us din bus kharab ho gayi ho; bracket 9 se 40 tak khinch jaata hai aur uspar range 31 likha hai, jabki baaki saat dots hile bhi nahi.

Range nikalna aasaan hai, lekin yeh sirf do observations ki sunti hai. Maan lo day 4 ko Route Q kharab ho gayi aur Meera ne 13 ki jagah 40 minute wait kiya. Aath mein se saat subah wahi hain, phir bhi range 4 se uchhal kar 31 ho jaati hai (Fig. 13.2). Range yeh bhi nahi batati ki values mean ya median ke aas-paas kaise baithi hain. Uske liye hume aise measures chahiye jo har observation ki centre se doori se bane hon.

Quick check

3 mein se 1

Ek hafte mein Jaisalmer ke sabse zyada temperatures 41, 44, 39, 46, 43, 40 aur 45 °C the. Range nikalo.

°C

Answer dekho

Answer 7 °C

46 - 39 = 7

°C. Dino ka order matter nahi karta.

13.4 Mean Deviation

Ek central value $a$ chuno, aam taur par mean ya median. Kisi observation $x$ ka $a$ se deviation (विचलन) $x - a$ hai: $a$ se upar wali values ke liye positive aur neeche wali ke liye negative. Agla seedha kadam hai deviations ka average lena. Mean ke about yeh poori tarah fail ho jaata hai. Neeche ke sums mein $\sum$ saare observations par chalta hai.

mean ke about deviations

\begin{aligned} \sum (x_{i} - \overset{ˉ}{x}) \\ = \sum x_{i} - n \overset{ˉ}{x} \\ = n \overset{ˉ}{x} - n \overset{ˉ}{x} = 0 \end{aligned}

(13.3)

Doosri line sum ko todti hai; teesri $\sum x_{i} = n \overset{ˉ}{x}$ use karti hai, jo bas mean ki definition hai. Positive deviations negative deviations ko bilkul cancel kar dete hain, kyunki mean data ka balance point hai. Zero average deviation ka matlab hoga ki kisi bhi data set mein spread nahi hai, jo bekaar hai.

Iska ilaaj hai signs ko ignore karna. Absolute value $∣ x - a ∣$ number line par $x$ aur $a$ ke beech ki doori hai, aur dooriyan kabhi cancel nahi hotin.

Definition

a ke about mean deviation

Observations ke ek fixed value $a$ se absolute deviations $∣ x_{i} - a ∣$ ka mean; ise M.D. $(a)$ likhte hain.

a ke about mean deviation

\begin{aligned} M.D. (a) \\ = \frac{1}{n} \sum ∣ x_{i} - a ∣ \end{aligned}

(13.4)

Koi bhi central value $a$ ka role nibha sakti hai, lekin practice mein hum mean $\overset{ˉ}{x}$ ya median use karte hain. Ab se $M$ ka matlab median hai. Mean deviation (माध्य विचलन) data ki units mein naapa jaata hai.

13.4.1 Ungrouped data ke liye mean deviation

Raw observations $x_{1}, x_{2}, \dots, x_{n}$ ke liye kaam ke chaar steps hain:

centre $a$ nikalo, mean $\overset{ˉ}{x}$ ya median $M$ ;
har observation mein se $a$ ghatao;
absolute values lo, har deviation ko ek doori $∣ x_{i} - a ∣$ mein badal do;
dooriyan jodo aur $n$ se divide karo.

3/4Dooriyan 5, 2, 1, 2, 6 ka total 16 hai.

Fig. 13.3Signed deviations cancel ho jaate hain; dooriyan nahi.

0 se 14 tak ki number line par 2, 5, 6, 9 aur 13 par paanch dots hain aur mean 7 par ek marker hai. Doosre step mein paanch horizontal arrows mean se shuru hokar har dot par khatam hote hain: teen left ki taraf hain aur un par minus 5, minus 2 aur minus 1 likha hai, do right ki taraf hain aur un par plus 2 aur plus 6 likha hai; left arrows ka total 8 hai aur right walon ka bhi. Teesre step mein teeno left arrows palat kar right ki taraf ho jaate hain, to ab paanchon dooriyan 5, 2, 1, 2 aur 6 hain jinka total 16 hai. Chauthe step mein mean se 3.2 lambai ki ek moti bar par M.D. = 16 ÷ 5 = 3.2 likha hai.

Do aam centres ke saath likhne par definition yeh ban jaati hai (Fig. 13.3):

mean ke about aur median ke about

\begin{aligned} M.D. (\overset{ˉ}{x}) \\ = \frac{1}{n} \sum ∣ x_{i} - \overset{ˉ}{x} ∣ \\ M.D. (M) \\ = \frac{1}{n} \sum ∣ x_{i} - M ∣ \end{aligned}

(13.5)

Example 13.1NCERT Example 1 · p. 261

Quiz scores ka mean deviation

Aath dost ek 20-mark quiz mein 14, 9, 17, 6, 11, 20, 8 aur 11 score karte hain. Mean ke about mean deviation nikalo.

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

Thodi practice ke baad tum signed deviations likhna chhod kar seedhe dooriyan list kar sakte ho.

Example 13.2NCERT Example 2 · p. 261

Ek lambi list

Solah bachchon ne ek din mein bahar khelne ke yeh minute note kiye: 25, 40, 10, 35, 45, 15, 30, 45, 20, 50, 30, 5, 40, 55, 25, 10. Mean ke about mean deviation nikalo.

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

Example 13.3NCERT Example 3 · p. 262

Median ke about

Nau students ne ek hafte mein ek novel ke 42, 37, 55, 48, 29, 61, 50, 38 aur 45 pages padhe. Median ke about mean deviation nikalo.

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

Quick check

3 mein se 1

3, 7, 8, 10, 12 ka mean ke about mean deviation nikalo.

Answer dekho

Answer 2.4

8 se dooriyan hain 5, 1, 0, 2, 4, jinka total 12 hai, isliye M.D.

= \frac{12}{5} = 2.4.

Signed deviations ka average lene par 0 aata.

13.4.2 Grouped data ke liye mean deviation

Bade data sets aam taur par frequency tables ke roop mein aate hain, do mein se kisi ek form mein:

discrete frequency distribution har alag value $x_{i}$ ko uski frequency $f_{i}$ ke saath list karta hai, yaani woh kitni baar aati hai;
continuous frequency distribution values ko bina gap wale class intervals mein baantta hai, jaise 0–20, 20–40, 40–60, aur har class ki frequency deta hai.

Discrete frequency distribution

Maan lo alag values $x_{1}, x_{2}, \dots, x_{n}$ kramshah $f_{1}, f_{2}, \dots, f_{n}$ baar aati hain, aur $N = \sum f_{i}$ total frequency hai. Jo value $f_{i}$ baar aati hai woh total mein $f_{i} x_{i}$ jodti hai aur dooriyon ke sum mein $f_{i} ∣ x_{i} - a ∣$ .

(i) Mean ke about. Mean aur mean deviation hain

frequency distribution ka mean

\begin{aligned} \overset{ˉ}{x} & = \frac{1}{N} \sum f_{i} x_{i} \\ N & = \sum f_{i} \end{aligned}

(13.6)

mean ke about

\begin{aligned} M.D. (\overset{ˉ}{x}) \\ = \frac{1}{N} \sum f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣ \end{aligned}

(13.7)

(ii) Median ke about. Values ko increasing order mein likho aur ek cumulative frequency (c.f.) column jodo, yaani frequencies ka chalta hua total. Kisi value ki c.f. woh aakhri position hai jo woh bharti hai jab saare $N$ observations order mein list kiye jaate hain. Median position $\frac{N + 1}{2}$ par hai; even $N$ ke liye yeh positions $\frac{N}{2}$ aur $\frac{N}{2} + 1$ ke theek beech mein hai. To woh pehli value dhoondho jiski c.f. us position tak pahunchti hai (Fig. 13.4). Zyadatar tables mein dono middle positions ek hi value par padti hain.

median ke about

\begin{aligned} M.D. (M) \\ = \frac{1}{N} \sum f_{i} ∣ x_{i} - M ∣ \end{aligned}

(13.8)

4/4Size 7 positions 17 se 30 tak bharta hai, isliye $M = 7$ .

Fig. 13.4Cumulative frequency beech wali position dhoondhti hai.NCERT Table 13.2 · p. 264

Shoe sizes 4 se 10 ke upar bars khade hain. Pehle step mein unki heights frequencies 2, 5, 9, 14, 10, 6 aur 4 hain. Doosre step mein har bar badh kar apni cumulative frequency tak pahunchti hai, jisse 2, 7, 16, 30, 40, 46 aur 50 ki chadhti staircase banti hai. Teesre step mein 25 par, yaani 50 observations ke aadhe par, ek dashed horizontal line staircase ko kaatti hai. Chauthe step mein size 7 ki bar highlight hoti hai: uski cumulative frequency 16 se 30 par jump karti hai, isliye positions 17 se 30, jinmein 25th aur 26th bhi hain, sab size 7 hain, aur median 7 hai.

Example 13.4NCERT Example 4 · p. 263

Ek term mein li gayi books

Ek school library ne record kiya ki 40 students mein se har ek ne ek term mein kitni books li. Mean ke about mean deviation nikalo. Pehle working table bharne ki koshish karo.

Table 13.1Li gayi books (working table)

Books $x_{i}$	Students $f_{i}$	$f_{i} x_{i}$	$∣ x_{i} - \overset{ˉ}{x} ∣$	$f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣$
4	3
6	8
8	8
10	12
12	5
14	4

Tumhari entries apne aap save hoti hain.

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

Example 13.5NCERT Example 5 · p. 264

Ek din mein bike shoe sizes

Agra ki ek shoe shop ne ek din mein 50 pairs beche. Median size ke about mean deviation nikalo.

Table 13.2

Size	Bike pairs
4	2
5	5
6	9
7	14
8	10
9	6
10	4

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

Continuous frequency distribution

Continuous frequency distribution mein individual values kho jaati hain: hume sirf yeh pata hota hai ki har class mein kitni values hain. Upper limit ke barabar value, jaise class 20–40 ke liye 40, agli class 40–60 mein gini jaati hai.

Kuch bhi compute karne ke liye hum maan lete hain ki class ka har observation class ke mid-point par baitha hai, yaani uske class mark par. 20–40 ke liye class mark hai $\frac{20 + 40}{2} = 30.$ Is tareeke se hone wali errors kuch had tak cancel ho jaati hain, kyunki class ki kuch values uske mid-point se upar hoti hain aur kuch neeche (Fig. 13.5).

2/3Har class apne mid-point par values ka ek dher ban jaati hai.

Fig. 13.5Class ki har value ko uska mid-point maana jaata hai.NCERT p. 265

Pehle step mein ek histogram mein 0–20, 20–40, 40–60, 60–80, 80–100 aur 100–120 kilometre ki classes par chhah judi hui bars hain jinki frequencies 2, 10, 15, 11, 8 aur 4 hain. Doosre step mein bars dhundhli hokar dashed outlines ban jaati hain aur har class ki jagah uske mid-point 10, 30, 50, 70, 90 aur 110 par usi height ka ek single spike aa jaata hai, jo dikhata hai ki class ki har value ko uska class mark maana gaya hai. Teesre step mein axis ke neeche ek triangle mean 60 ko mark karta hai, jo spikes ka balance point hai.

(i) Mean ke about. Har class ki jagah uska class mark $x_{i}$ rakho aur bilkul discrete distribution ki tarah kaam karo, (13.6) aur (13.7) ke saath.

Example 13.6NCERT Example 6 · p. 266

Delivery riders ki chalayi gayi dooriyan

Bengaluru ki ek food-delivery company ne record kiya ki uske 50 riders mein se har ek ne ek din mein kitni doori (km mein) tay ki. Mean ke about mean deviation nikalo.

Table 13.3Riders ki dooriyan (working table)

Doori (km)	Riders $f_{i}$	Mid-point $x_{i}$	$f_{i} x_{i}$	$∣ x_{i} - \overset{ˉ}{x} ∣$	$f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣$
0–20	2
20–40	10
40–60	15
60–80	11
80–100	8
100–120	4

Tumhari entries apne aap save hoti hain.

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

Mean ka ek shortcut: step deviations. Products $f_{i} x_{i}$ jaldi bade ho jaate hain. Do kadam numbers ko chhota kar dete hain. Pehla, origin shift karo: har mid-point ko beech ke paas ki ek suvidhajanak value $a$ se naapo, jise assumed mean kehte hain. Doosra, scale badlo: agar saari shifted values mein ek common factor $h$ hai (aksar class width), to usse divide karo. Jo results $d_{i}$ milte hain woh step deviations hain (Fig. 13.6).

3/3 $h = 20$ se divide karne par chhote numbers $- 2$ se $3$ bachte hain.

Fig. 13.6Origin shift karo, phir scale chhota karo.NCERT Figs 13.3, 13.4 · p. 267

Teen parallel rulers par chhah tick positions same hain. Upar wale ruler par mid-points x likha hai aur woh 10, 30, 50, 70, 90 aur 110 padhta hai. Doosre step mein neeche ek ruler aata hai, jispar 50 ghatao likha hai, jo −40, −20, 0, 20, 40 aur 60 padhta hai, aur purane 50 ke neeche wala zero naye origin ke roop mein highlight hai. Teesre step mein ek teesra ruler aata hai, jispar phir 20 se divide likha hai, jo −2, −1, 0, 1, 2 aur 3 padhta hai: wahi points ek chhote scale par.

step deviations se mean

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

(13.9)

jahan

$d_{i}$: step deviation $\frac{x_{i} - a}{h}$
$a$: assumed mean
$h$: common factor

Yeh kaam kyun karta hai? $d_{i} = \frac{x_{i} - a}{h}$ se hume $x_{i} = a + h d_{i}$ milta hai. $f_{i}$ se guna karke, jod kar aur $N$ se divide karke milta hai $\overset{ˉ}{x} = a + h \overset{ˉ}{d} :$ origin shift karne se mean shift hota hai, aur scale badalne se mean bhi usi scale se badalta hai.

Example

Step deviations se Example 13.6

Example 13.6 ke riders ki mean doori step-deviation method se nikalo.

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

Shortcut sirf yeh badalta hai ki tum $\overset{ˉ}{x}$ kaise nikalte ho. Dooriyan $∣ x_{i} - \overset{ˉ}{x} ∣$ ab bhi original scale par naapi jaati hain, isliye mean deviation phir se 22 km hai.

(ii) Median ke about. Continuous distribution mein median ka andaaza ek class ke andar lagaya jaata hai. Median class woh class hai jismein c.f. pehli baar $\frac{N}{2}$ tak pahunchti hai. Maan lo usse pehle $C$ observations aate hain, woh $l$ se shuru hoti hai, uski width $h$ hai aur usmein $f$ observations hain, jo usmein barabar faile hain. Hume un $f$ observations mein se $\frac{N}{2} - C$ chahiye, isliye hum class mein utne hi fraction tak aage jaate hain (Fig. 13.7):

continuous distribution ka median

M = l + \frac{\frac{N}{2} - C}{f} \times h

(13.10)

jahan

$l$: median class ki lower limit
$C$: usse pehle wali class ki c.f.
$f$: median class ki frequency
$h$: median class ki width

4/4 $M = 50 + \frac{6}{15} \times 10 = 54$

Fig. 13.7Median apni class mein sahi fraction jitna aage baitha hota hai.NCERT p. 269

Upar ek strip 60 observations ko order mein teen blocks ke roop mein dikhati hai: 50 se neeche 24, class 50–60 mein 15 aur 60 se upar 21. Doosre step mein position 30 par, yaani 60 ke aadhe par, ek marker beech wale block ke andar padta hai. Teesre step mein beech wala block neeche bada karke 50 se 60 tak ki ek bar ke roop mein dikhaya jaata hai jismein barabar doori par 15 dots hain, aur pehle 6 dots highlight hote hain, kyunki 30 minus 24 = 6. Chauthe step mein bar ke saath 50 se 54 tak ek arrow chalta hai, width 10 ka chhah-pandrahwaan hissa, aur uspar M = 54 likha hai.

Phir class marks ki $M$ se dooriyan naapo aur (13.8) use karo.

Example 13.7NCERT Example 7 · p. 269

Yoga camp mein umrein

Rishikesh ke ek yoga camp mein 60 logon ki umrein neeche grouped hain. Median umar ke about mean deviation nikalo.

Table 13.4Umrein (working table)

Umar (saal)	Log $f_{i}$	c.f.	Mid-point $x_{i}$	$∣ x_{i} - M ∣$	$f_{i} ∣ x_{i} - M ∣$
20–30	3
30–40	5
40–50	16
50–60	15
60–70	11
70–80	10

Tumhari entries apne aap save hoti hain.

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

Quick check

4 mein se 1

Values 1, 2 aur 3 kramshah 5, 3 aur 2 baar aati hain. Mean ke about mean deviation nikalo.

Answer dekho

Answer 0.7

\overset{ˉ}{x} = \frac{5 + 6 + 6}{10} = 1.7.

Dooriyan 0.7, 0.3, 1.3 ko 5, 3, 2 se weight karne par

3.5 + 0.9 + 2.6 = 7

milta hai, isliye M.D.

= \frac{7}{10} = 0.7.

Practice · 12 questions

Practice 13.1: answers ke saath

NumericalEasyNCERT Exercise 13.1, Q1
Ek student ke aath weekly spelling tests mein scores 5, 9, 12, 18, 7, 10, 14, 5 the. Average mein yeh apne mean se kitni door hain? Mean ke about mean deviation batao.
Solution dikhao
Total 80 hai, isliye $\overset{ˉ}{x} = 10$ .
10 se dooriyan hain 5, 1, 2, 8, 3, 0, 4, 5, jinka total 28 hai.
M.D. $(\overset{ˉ}{x}) = \frac{28}{8} = 3.5.$
NumericalEasyNCERT Exercise 13.1, Q2
Das students ek test mein 100 mein se 62, 48, 75, 55, 81, 59, 66, 70, 52 aur 62 score karte hain. Mean ke about mean deviation nikalo.
Solution dikhao
$\overset{ˉ}{x} = \frac{630}{10} = 63.$
63 se dooriyan hain 1, 15, 12, 8, 18, 4, 3, 7, 11, 1, jinka total 80 hai.
M.D. $(\overset{ˉ}{x}) = \frac{80}{10} = 8$ marks.
NumericalMediumNCERT Exercise 13.1, Q3
Ek shop ne baarah dinon tak ek magazine ki daily sales record ki: 21, 13, 26, 16, 31, 22, 19, 25, 15, 28, 20, 24. Median ke about mean deviation nikalo.
Solution dikhao
Order mein: 13, 15, 16, 19, 20, 21, 22, 24, 25, 26, 28, 31. Median hai $\frac{21 + 22}{2} = 21.5.$
21.5 se dooriyan hain 8.5, 6.5, 5.5, 2.5, 1.5, 0.5, 0.5, 2.5, 3.5, 4.5, 6.5, 9.5, jinka total 52 hai.
M.D. $(M) = \frac{52}{12} \approx 4.33.$
NumericalMediumNCERT Exercise 13.1, Q4
Ek kabaddi club ke das members ke weights (kg) hain 64, 38, 55, 47, 73, 52, 49, 58, 44 aur 52. Median ke about mean deviation nikalo.
kg
Solution dikhao
Order mein: 38, 44, 47, 49, 52, 52, 55, 58, 64, 73. 5th aur 6th dono values 52 hain, isliye $M = 52$ .
Dooriyan hain 14, 8, 5, 3, 0, 0, 3, 6, 12, 21, jinka total 72 hai.
M.D. $(M) = \frac{72}{10} = 7.2$ kg.
NumericalMediumNCERT Exercise 13.1, Q5
Neeche ki frequency table ke liye mean ke about mean deviation nikalo.
Table 13.5
$x_{i}$ $f_{i}$
3 2
6 8
9 12
12 4
15 4
Solution dikhao
$N = 30$ aur products $f_{i} x_{i}$ hain 6, 48, 108, 48, 60, jinka total 270 hai, isliye $\overset{ˉ}{x} = 9$ .
Dooriyan $∣ x_{i} - 9 ∣$ hain 6, 3, 0, 3, 6; $f_{i}$ se guna karke: 12, 24, 0, 12, 24, total 72.
M.D. $(\overset{ˉ}{x}) = \frac{72}{30} = 2.4.$
NumericalMediumNCERT Exercise 13.1, Q6
Is table ki values average mein apne mean se kitni door hain? Mean ke about mean deviation nikalo.
Table 13.6
$x_{i}$ $f_{i}$
20 4
40 12
60 14
80 15
100 5
Solution dikhao
$N = 50$ aur $\sum f_{i} x_{i} = 80 + 480 + 840 + 1200 + 500 = 3100,$ isliye $\overset{ˉ}{x} = 62$ .
62 se dooriyan hain 42, 22, 2, 18, 38; $f_{i}$ se guna karke: 168, 264, 28, 270, 190, total 920.
M.D. $(\overset{ˉ}{x}) = \frac{920}{50} = 18.4.$
NumericalMediumNCERT Exercise 13.1, Q7
Neeche ki frequency table ke liye median ke about mean deviation nikalo.
Table 13.7
$x_{i}$ $f_{i}$
4 9
6 3
8 12
11 4
13 8
16 4
Solution dikhao
c.f. column hai 9, 12, 24, 28, 36, 40. 20th aur 21st values positions 13 se 24 mein hain, jo sab 8 hain, isliye $M = 8$ .
8 se dooriyan hain 4, 2, 0, 3, 5, 8; $f_{i}$ se guna karke: 36, 6, 0, 12, 40, 32, total 126.
M.D. $(M) = \frac{126}{40} = 3.15.$
NumericalMediumNCERT Exercise 13.1, Q8
Median ko centre maan kar is table ki values ka mean deviation nikalo.
Table 13.8
$x_{i}$ $f_{i}$
12 2
18 12
24 10
28 3
33 3
Solution dikhao
c.f. column hai 2, 14, 24, 27, 30. Positions 15 se 24 mein 24 hai, isliye $M = 24$ .
Dooriyan hain 12, 6, 0, 4, 9; $f_{i}$ se guna karke: 24, 72, 0, 12, 27, total 135.
M.D. $(M) = \frac{135}{30} = 4.5.$
NumericalHardNCERT Exercise 13.1, Q9
Ek market ke 50 street vendors ki daily kamai neeche grouped hai. Mean ke about mean deviation nikalo.
Table 13.9
Kamai (₹) Vendors
100–200 5
200–300 8
300–400 12
400–500 9
500–600 6
600–700 6
700–800 4
₹
Solution dikhao
Mid-points: 150, 250, …, 750. $a = 450$ aur $h = 100$ ke saath $d_{i}$ $- 3$ se $3$ tak jaata hai aur $\sum f_{i} d_{i} = - 15 - 16 - 12 + 0 + 6 + 12 + 12 = - 13.$
$\overset{ˉ}{x} = 450 + 100 \times \frac{- 13}{50} = 424.$
424 se dooriyan hain 274, 174, 74, 26, 126, 226, 326; $f_{i}$ se guna karke: 1370, 1392, 888, 234, 756, 1356, 1304, total 7300.
M.D. $(\overset{ˉ}{x}) = \frac{7300}{50} = 146,$ yaani ₹146.
NumericalMediumNCERT Exercise 13.1, Q10
Ek sports academy ki 60 ladkiyon ke weights neeche grouped hain. Mean ke about mean deviation nikalo.
Table 13.10
Weight (kg) Ladkiyan
40–44 3
44–48 9
48–52 13
52–56 16
56–60 13
60–64 6
kg
Solution dikhao
$\sum f_{i} x_{i} = 126 + 414 + 650 + 864 + 754 + 372 = 3180,$ isliye $\overset{ˉ}{x} = \frac{3180}{60} = 53$ kg.
53 se dooriyan hain 11, 7, 3, 1, 5, 9; $f_{i}$ se guna karke: 33, 63, 39, 16, 65, 54, total 270.
M.D. $(\overset{ˉ}{x}) = \frac{270}{60} = 4.5$ kg.
NumericalMediumNCERT Exercise 13.1, Q11
Median ko centre maan kar neeche ke grouped data ka mean deviation nikalo.
Table 13.11
Class Frequency
10–20 4
20–30 8
30–40 13
40–50 10
50–60 9
60–70 6
Solution dikhao
c.f. column hai 4, 12, 25, 35, 44, 50. $\frac{N}{2} = 25$ class 30–40 mein pahunchta hai, isliye $l = 30$ , $C = 12$ , $f = 13$ , $h = 10$ .
$M = 30 + \frac{25 - 12}{13} \times 10 = 40.$
Mid-points 15, 25, …, 65 ki 40 se dooriyan 25, 15, 5, 5, 15, 25 hain; $f_{i}$ se guna karke: 100, 120, 65, 50, 135, 150, total 620.
M.D. $(M) = \frac{620}{50} = 12.4.$
NumericalHardNCERT Exercise 13.1, Q12
Ek science museum aane wale 80 visitors ki umrein neeche grouped hain. Classes ke beech gaps hain (15 aur 16 alag classes mein hain). Median umar ke about mean deviation nikalo.
Table 13.12
Umar (saal) Visitors
11–15 5
16–20 12
21–25 17
26–30 20
31–35 18
36–40 6
41–45 2
saal
Solution dikhao
Umrein whole numbers hain, isliye 15 aur 16 padosi hain aur pehli do classes ke beech asli boundary 15.5 hai. Isi tarah classes 10.5–15.5, 15.5–20.5, …, 40.5–45.5 ban jaati hain. Mid-points 13, 18, …, 43 nahi badalte.
c.f. column hai 5, 17, 34, 54, 72, 78, 80. $\frac{N}{2} = 40$ class 25.5–30.5 mein padta hai, jahan $C = 34$ , $f = 20$ , $h = 5$ .
$M = 25.5 + \frac{40 - 34}{20} \times 5 = 27.$
27 se dooriyan hain 14, 9, 4, 1, 6, 11, 16; $f_{i}$ se guna karke: 70, 108, 68, 20, 108, 66, 32, total 472.
M.D. $(M) = \frac{472}{80} = 5.9$ saal.

$x_{i}$	$f_{i}$
3	2
6	8
9	12
12	4
15	4

$x_{i}$	$f_{i}$
20	4
40	12
60	14
80	15
100	5

$x_{i}$	$f_{i}$
4	9
6	3
8	12
11	4
13	8
16	4

$x_{i}$	$f_{i}$
12	2
18	12
24	10
28	3
33	3

Kamai (₹)	Vendors
100–200	5
200–300	8
300–400	12
400–500	9
500–600	6
600–700	6
700–800	4

Weight (kg)	Ladkiyan
40–44	3
44–48	9
48–52	13
52–56	16
56–60	13
60–64	6

Class	Frequency
10–20	4
20–30	8
30–40	13
40–50	10
50–60	9
60–70	6

Umar (saal)	Visitors
11–15	5
16–20	12
21–25	17
26–30	20
31–35	18
36–40	6
41–45	2

13.4.3 Mean deviation ki limitations

Mean deviation samajhna aasaan hai, lekin iske teen kamzor points hain.

Answer is baat par depend karta hai ki hum kaun sa centre chunte hain. Total doori median ke about sabse kam hoti hai, $\sum ∣ x_{i} - M ∣ \leq \sum ∣ x_{i} - \overset{ˉ}{x} ∣,$ isliye mean ke about mean deviation dono mein kabhi chhota nahi hota.
Bahut zyada bikhre data ke liye koi bhi centre values ko achchhi tarah describe nahi karta. 1, 2, 3, 40, 41 mein median 3 hai, jo paanch mein se do values se bahut door hai, isliye usse naapa gaya spread hume zyada kuch nahi batata.
Algebra mein absolute values mushkil hoti hain. $(x - a)^{2}$ ke ulat, expression $∣ x - a ∣$ ko expand nahi kar sakte, isliye combined ya transformed data ke mean deviation ka koi seedha rule nahi hai.

SimulationDooriyon ka sum kahan sabse kam hai?

0 se 14 tak ek coordinate grid, jismein x-axis par 1, 2, 3, 7 aur 12 par paanch fixed points A, B, C, D aur E hain, yaani observations. Ek chhatha point M, yaani centre, mean 5 se shuru hota hai aur aadhe-aadhe unit ke steps mein drag kiya ja sakta hai; panel dooriyan MA, MB, MC, MD aur ME list karta hai. 5 par inka total 18 hai, yaani mean deviation 3.6. M ko median 3 par khiskane se total ghat kar 16 ho jaata hai, yaani mean deviation 3.2, aur M ko 3 se kisi bhi taraf hilane par total phir badh jaata hai.

Fig. 13.8Observations 1, 2, 3, 7, 12

x

-axis par hain. Centre M ko axis par khiskao aur paanchon dooriyan jodo.

Try karo

M mean, 5, se shuru hota hai. Paanchon dooriyan jodo, phir M ko median, 3, par le jaao aur phir se jodo. Kaun sa total chhota hai?
M ko 2 par rakho, phir 4 par. Kya 3 ke dono taraf total badhta hai?
M ko $x$ -axis par hi rakho: axis se hatne par dooriyan $∣ x_{i} - a ∣$ nahi rehtin.

To hume aisa measure chahiye jo ab bhi har observation ki centre se doori use kare lekin algebra mein achchha behave kare. Deviations ka square lena bilkul yahi karta hai.

Quick check

2 mein se 1

Data 1, 2, 3, 7, 12 ke liye sum

\sum ∣ x_{i} - a ∣

sabse kam tab hai jab

a

3, median5, mean6.5, range ka beech1, sabse chhoti value

Answer dekho

Answer (A) 3, median

Sums hain: 5 par 18, 3 par 16, 6.5 par 19.5 aur 1 par 20. Dooriyon ka sum hamesha median ke about sabse kam hota hai.

13.5 Variance aur Standard Deviation

Deviations ko cancel hone se rokne ka doosra tareeka square lena hai, kyunki koi bhi square negative nahi hota. Pehla candidate hai total $\sum (x_{i} - \overset{ˉ}{x})^{2} .$ Yeh ek baat mein theek behave karta hai: yeh theek tab 0 hota hai jab har value mean ke barabar ho.

Lekin total mein ek kami hai jo aasaani se nazar nahi aati: yeh sirf isliye badhta hai kyunki zyada observations jud gaye. Same mean, 30, wale do sets ise saamne laate hain. Set A mein paanch observations hain: 10, 20, 30, 40, 50. Set B mein pachchees: 18 se 42 tak har whole number. A ke liye squared deviations 400, 100, 0, 100, 400 hain, jinka total 1000 hai. B ke liye deviations −12 se 12 tak jaate hain, isliye

\begin{aligned} \sum (y_{i} - 30)^{2} \\ = 2 (1^{2} + \dots + 12^{2}) \\ = 2 \times \frac{12 \cdot 13 \cdot 25}{6} \\ = 1300 \end{aligned}

jahan humne yeh use kiya: $1^{2} + 2^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} .$ Sums ke hisaab se B, A se zyada bikhra lagta hai. Phir bhi B ki har value mean se 12 ke andar hai, jabki A ki values 20 door tak jaati hain (Fig. 13.9). B ka total isliye bada hai kyunki woh paanch guna zyada squares jodta hai.

3/3 $n$ se divide karne par: A se 200, B se 52.

Fig. 13.9Sum observations ki sankhya ke saath badhta hai; mean nahi.NCERT Figs 13.5, 13.6 · p. 273

0 se 60 ke scale par do dot plots. Set A mein 10, 20, 30, 40 aur 50 par paanch dots hain; set B mein pachchees dots hain, 18 se 42 tak har whole number par ek. Doosre step mein do bars squared deviations ke sums ko compare karti hain: A ki bar 1000 hai aur B ki lambi bar 1300, jo galat tarah se dikhati hai ki B zyada faila hua hai. Teesre step mein bars squares ke mean mein badal jaati hain: A ki 200 ho jaati hai aur B ki ghat kar 52, to ab A sahi tarah se zyada bikhra set dikhta hai.

Ilaaj wahi hai jo humne dooriyon ke liye use kiya tha: mean lo. Squared deviations ka mean A ke liye $\frac{1000}{5} = 200$ hai aur B ke liye $\frac{1300}{25} = 52$ , jo ab sets ko sahi order mein rakhta hai.

Definition

Variance

Observations ke unke mean se squared deviations ka average; ise $σ^{2}$ likhte hain (sigma squared padhte hain).

variance

σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{ˉ}{x})^{2}

(13.11)

Set A ka variance (प्रसरण) 200 hai aur set B ka 52.

13.5.1 Standard Deviation

Variance mein ek kami hai: uski units data ki units ka square hoti hain. Minutes mein waiting times se variance square minutes mein aata hai, jise koi imagine nahi kar sakta. Positive square root lene se hum original units mein wapas aa jaate hain.

Definition

Standard deviation

Variance ka positive square root, $σ = \sqrt{σ^{2}} .$ Yeh mean ke about spread ko data ki units mein naapta hai.

standard deviation

σ = \sqrt{\frac{1}{n} \sum (x_{i} - \overset{ˉ}{x})^{2}}

(13.12)

Standard deviation (मानक विचलन) ke peeche ek saaf picture hai (Fig. 13.10). Har deviation par ek square banao. Variance in squares ka average area hai, aur $σ$ us square ki side hai jiska area yeh average ho.

3/3Mean area $σ^{2} = \frac{80}{5} = 16,$ isliye $σ = 4$ .

Fig. 13.10Variance average square hai; σ uski side hai.

2 se 18 tak ki number line par 4, 8, 10, 12 aur 16 par dots hain aur mean 10 par ek marker hai. Doosre step mein har deviation par, yaani har dot aur mean ke beech, ek square banta hai: 4 aur 16 ke liye side 6 aur area 36 ke do bade squares, 8 aur 12 ke liye side 2 aur area 4 ke do chhote squares, aur 10 ke liye kuch nahi. Teesre step mein yeh squares dhundhle ho jaate hain aur area 16 ka ek bold square, yaani 36, 4, 0, 4 aur 36 ka average, mean ke upar khada hota hai; uski side, 4, standard deviation hai.

Example 13.8NCERT Example 8 · p. 274

Puzzle times ka variance

Aath students ek sudoku khatam karne mein 12, 18, 21, 15, 27, 24, 9 aur 18 minute lete hain. Variance aur standard deviation nikalo.

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

Wapas do buses par chalte hain. Route P ke 11 se deviations hain $4, - 8, 10, 0, - 10, 8, - 4, 0,$ isliye $σ_{P}^{2} = \frac{360}{8} = 45$ aur $σ_{P} \approx 6.71$ minute. Route Q se $σ_{Q}^{2} = \frac{12}{8} = 1.5$ aur $σ_{Q} \approx 1.22$ minute milta hai. Ab ek number wahi batata hai jo dot plots ne dikhaya tha: Route Q kahin zyada bharosemand hai.

InteractiveSpread lab

Do modes. Dot plot mein 0 se 24 tak ki number line par aath dots ko drag kiya ja sakta hai ya arrow keys se hilaya ja sakta hai; buttons Route P, Route Q, ya 24 minute wali ek late bus ke saath Route Q ke waiting times load karte hain. Ek panel mean, median, range, mean ke about aur median ke about mean deviation, variance aur standard deviation deta hai, aur ek overlay mean se har doori dikha sakta hai, ya squared deviations ka bar chart jismein unke average, yaani variance, par ek dashed line ho. Shift and stretch mein sliders data 2, 4, 5, 6, 8 ke liye k ko −3 se 3 aur c ko −6 se 6 tak set karte hain; do dot plots x ko kx + c se compare karte hain, har ek mein uske mean ke dono taraf ek standard deviation ki band hoti hai, aur panel check karta hai ki naya mean purane mean ka k guna plus c hai aur naya standard deviation purane ka |k| guna.

Fig. 13.11Data badalne ke liye dots ko drag karo, ya data set ko shift aur stretch karo, aur dekho ki measures of spread kaise react karte hain.

13.5.2 Discrete frequency distribution ka standard deviation

Jab value $x_{i}$ kul $f_{i}$ baar aati hai, to uska squared deviation $f_{i}$ baar gina jaata hai. $N = \sum f_{i}$ ke saath:

discrete distribution

\begin{aligned} σ = \sqrt{V}, where \\ V = \frac{1}{N} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2} \end{aligned}

(13.13)

Example 13.9NCERT Example 9 · p. 275

Bus mein chadhne wale passengers

Ek trip mein ek city bus 40 stops par ruki. Table dikhati hai ki kitne stops par $x$ passengers chadhe. Variance aur standard deviation nikalo.

Table 13.13

Passengers $x_{i}$	Stops $f_{i}$
2	1
5	5
8	10
11	10
14	7
17	5
20	2

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

13.5.3 Continuous frequency distribution ka standard deviation

Classes ke liye class marks ko $x_{i}$ maano aur discrete distribution ki tarah aage badho. Lekin jab $\overset{ˉ}{x}$ whole number nahi hota, to har deviation ek uljha hua decimal hota hai. Ek equivalent formula deviations ki zaroorat hi khatam kar deta hai.

DerivationBina deviations ka formula

Square ko expand karo aur sum ko teen hisson mein todo:

\begin{aligned} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2} \\ = \sum f_{i} x_{i}^{2} \\ - 2 \overset{ˉ}{x} \sum f_{i} x_{i} \\ + {\overset{ˉ}{x}}^{2} \sum f_{i} \end{aligned}

Ab $\sum f_{i} x_{i} = N \overset{ˉ}{x}$ aur $\sum f_{i} = N$ , isliye aakhri do hisse hain $- 2 N {\overset{ˉ}{x}}^{2} + N {\overset{ˉ}{x}}^{2} = - N {\overset{ˉ}{x}}^{2} .$ $N$ se divide karne par:

squares ka mean minus mean ka square

\begin{aligned} σ^{2} & = \frac{1}{N} \sum f_{i} x_{i}^{2} \\ - {\overset{ˉ}{x}}^{2} \end{aligned}

(13.14)

Ab $\overset{ˉ}{x} = \frac{1}{N} \sum f_{i} x_{i}$ rakho aur poore ko $N^{2}$ se guna karo:

computing formula

\begin{aligned} N^{2} σ^{2} \\ = N \sum f_{i} x_{i}^{2} \\ - (\sum f_{i} x_{i})^{2} \end{aligned}

(13.15)

To $σ$ right-hand side ke square root ka $\frac{1}{N}$ guna hai. Raw data ke liye har $f_{i} = 1$ aur $N = n$ , aur (13.14) ban jaata hai $σ^{2} = \frac{1}{n} \sum x_{i}^{2} - {\overset{ˉ}{x}}^{2} :$ squares ka mean minus mean ka square.

Example 13.10NCERT Example 10 · p. 277

Highway par speeds

Highway par ek speed camera ne 40 vehicles ki speeds record ki. Mean, variance aur standard deviation nikalo.

Table 13.14

Speed (km/h)	Vehicles
20–30	2
30–40	5
40–50	7
50–60	7
60–70	9
70–80	8
80–90	2

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

Example 13.11NCERT Example 11 · p. 278

Fitness test mein push-ups

Ek fitness test mein 50 students ne neeche dikhaye gaye push-ups kiye. (13.15) use karke standard deviation nikalo.

Table 13.15

Push-ups $x_{i}$	Students $f_{i}$
5	3
10	7
15	12
20	14
25	9
30	5

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

13.5.4 Variance aur standard deviation nikalne ka shortcut method

Jab values badi hon, to $f_{i} x_{i}^{2}$ bhi bhaari ho jaata hai. Data ko step-deviation method ki tarah code karo, ek assumed mean $A$ aur ek scale $h$ ke saath.

DerivationData ko code karna

\begin{aligned} y_{i} = \frac{x_{i} - A}{h}, so \\ x_{i} = A + h y_{i} \end{aligned}

$f_{i}$ se guna karo, saare $i$ par jodo aur $N$ se divide karo:

coded mean

\overset{ˉ}{x} = A + h \overset{ˉ}{y}

(13.16)

Ghatane par milta hai $x_{i} - \overset{ˉ}{x} = h (y_{i} - \overset{ˉ}{y}) .$ To $x$ ka har squared deviation $y$ ke matching squared deviation ka $h^{2}$ guna hai; $f_{i}$ se weight karke average lene par milta hai $σ_{x}^{2} = h^{2} σ_{y}^{2},$ yaani,

σ, h ke saath scale hota hai

σ_{x} = h σ_{y}

(13.17)

Aakhir mein coded values $y_{i}$ par (13.15) lagao:

shortcut formula

\begin{aligned} N^{2} σ_{x}^{2} \\ = h^{2} [N \sum f_{i} y_{i}^{2} \\ - (\sum f_{i} y_{i})^{2}] \end{aligned}

(13.18)

Seedhe shabdon mein: origin ko $A$ se shift karne par spread nahi badalta, aur scale ko $h$ se badalne par $σ$ , $h$ se guna ho jaata hai. Example 13.13 aur Example 13.15 kisi bhi data ke liye yahi effects dikhate hain.

Example 13.12NCERT Example 12 · p. 280

Shortcut se Example 13.10

Example 13.10 ki speeds ka mean aur standard deviation shortcut method se nikalo.

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

Quick check

3 mein se 1

Ek distribution ke liye

N = 10

\sum f_{i} x_{i} = 60

aur

\sum f_{i} x_{i}^{2} = 400

hai. Variance nikalo.

Answer dekho

Answer 4

\overset{ˉ}{x} = 6

, isliye

σ^{2} = 40 - 36 = 4.

Mean ka square lena bhoolne par 34 aata hai.

Practice · 10 questions

Practice 13.2: answers ke saath

1
NumericalEasyNCERT Exercise 13.2, Q1
14, 9, 17, 6, 11, 15, 10, 6 ka mean aur variance nikalo.
1. mean
  Solution dikhao
  $\overset{ˉ}{x} = \frac{88}{8} = 11.$
2. variance
  Solution dikhao
  11 se deviations hain $3, - 2, 6, - 5, 0, 4, - 1, - 5;$ unke squares 9, 4, 36, 25, 0, 16, 1, 25 ka total 116 hai.
  $σ^{2} = \frac{116}{8} = 14.5.$
ProofHardNCERT Exercise 13.2, Q2
Pehle $n$ odd natural numbers ka mean aur variance nikalo: $1, 3, 5, \dots, 2 n - 1.$
Solution dikhao
Answermean $n$ ; variance $\frac{n^{2} - 1}{3}$
Pehle $n$ odd numbers ka sum $n^{2}$ hai, isliye $\overset{ˉ}{x} = \frac{n^{2}}{n} = n .$
$\begin{aligned} \sum (2 k - 1)^{2} \\ = 4 \sum k^{2} - 4 \sum k + n \\ = \frac{n (2 n - 1) (2 n + 1)}{3} \end{aligned}$
simplify karne ke baad ( $n = 2$ se check karo: $1 + 9 = 10$ ).
$σ^{2} = \frac{4 n^{2} - 1}{3} - n^{2} = \frac{n^{2} - 1}{3}$
Jaise, 1, 3, 5, 7, 9 ka mean 5 aur variance $\frac{24}{3} = 8$ hai. (Yahi method $1, 2, \dots, n$ ke liye variance $\frac{n^{2} - 1}{12}$ deta hai.)
3
NumericalEasyNCERT Exercise 13.2, Q3
6 ke pehle 9 multiples ka mean aur variance nikalo.
1. mean
  Solution dikhao
  Nau values barabar doori par hain, isliye mean beech wali value, 30, hai.
2. variance
  Solution dikhao
  Deviations hain $\pm 24, \pm 18, \pm 12, \pm 6$ aur 0. Squares ka sum hai $2 (576 + 324 + 144 + 36) = 2160.$
  $σ^{2} = \frac{2160}{9} = 240.$
NumericalMediumNCERT Exercise 13.2, Q4
Is distribution ka mean aur variance nikalo.
Table 13.16
$x_{i}$ $f_{i}$
4 2
8 3
11 8
15 10
20 11
23 3
27 3
Solution dikhao
Answer $\overset{ˉ}{x} = 16$ , $σ^{2} = 34.4$
Products $f_{i} x_{i}$ hain 8, 24, 88, 150, 220, 69, 81, jinka total 640 hai, isliye $\overset{ˉ}{x} = 16$ .
Squared deviations hain 144, 64, 25, 1, 16, 49, 121; $f_{i}$ se guna karke: 288, 192, 200, 10, 176, 147, 363, total 1376.
$σ^{2} = \frac{1376}{40} = 34.4.$
NumericalMediumNCERT Exercise 13.2, Q5
30 students ki heights (cm) neeche di gayi hain. Mean aur variance nikalo.
Table 13.17
Height (cm) Students
143 1
146 4
148 4
150 8
153 7
155 4
160 2
Solution dikhao
Answer $\overset{ˉ}{x} = 151$ cm, $σ^{2} = 15.4$
$y_{i} = x_{i} - 150$ ke saath: $- 7, - 4, - 2, 0, 3, 5, 10.$ Tab $\sum f_{i} y_{i} = 30$ aur $\sum f_{i} y_{i}^{2} = 492$ .
$\overset{ˉ}{x} = 150 + \frac{30}{30} = 151$ cm.
Shift karne se variance nahi badalta: $σ^{2} = \frac{492}{30} - 1^{2} = 15.4.$
NumericalMediumNCERT Exercise 13.2, Q6
60 andon ke weights (g) neeche diye gaye hain. Shortcut method se mean aur standard deviation nikalo.
Table 13.18
Weight (g) Ande
42 5
44 6
46 8
48 11
50 15
52 13
54 2
Solution dikhao
Answer $\overset{ˉ}{x} = 48.4$ g, $σ \approx 3.26$ g
$y_{i} = \frac{x_{i} - 48}{2}$ ki values $- 3$ se $3$ tak hain. Tab $\sum f_{i} y_{i} = 12$ aur $\sum f_{i} y_{i}^{2} = 162$ .
$\overset{ˉ}{x} = 48 + 2 \times \frac{12}{60} = 48.4$ g.
$\begin{aligned} 60^{2} σ^{2} & = 4 [60 (162) - 12^{2}] \\ = 4 \times 9576 \end{aligned}$
$σ^{2} = 10.64$ aur $σ \approx 3.26$ g.
NumericalMediumNCERT Exercise 13.2, Q7
Is distribution ka mean aur variance nikalo.
Table 13.19
Class Frequency
0–40 1
40–80 10
80–120 13
120–160 11
160–200 3
200–240 2
Solution dikhao
Answer $\overset{ˉ}{x} = 111$ , $σ^{2} = 2079$
Mid-points 20, 60, …, 220; $y_{i} = \frac{x_{i} - 100}{40}$ ki values $- 2, - 1, 0, 1, 2, 3$ hain. Tab $\sum f_{i} y_{i} = 11$ aur $\sum f_{i} y_{i}^{2} = 55$ .
$\overset{ˉ}{x} = 100 + 40 \times \frac{11}{40} = 111.$
$40^{2} σ^{2} = 40^{2} [40 (55) - 11^{2}],$ isliye $σ^{2} = 2200 - 121 = 2079.$
8
NumericalMediumNCERT Exercise 13.2, Q8
Is distribution ka mean aur variance nikalo.
Table 13.20
Class Frequency
10–20 6
20–30 7
30–40 17
40–50 11
50–60 9
1. mean
  Solution dikhao
  $\sum f_{i} x_{i} = 90 + 175 + 595 + 495 + 495 = 1850,$ isliye $\overset{ˉ}{x} = 37$ .
2. variance
  Solution dikhao
  37 se squared deviations hain 484, 144, 4, 64, 324; $f_{i}$ se guna karke: 2904, 1008, 68, 704, 2916, total 7600.
  $σ^{2} = \frac{7600}{50} = 152.$
NumericalHardNCERT Exercise 13.2, Q9
Ek nursery ke 60 paudhon ki heights (cm) neeche grouped hain. Shortcut method se mean, variance aur standard deviation nikalo.
Table 13.21
Height (cm) Paudhe
120–124 2
124–128 5
128–132 8
132–136 18
136–140 13
140–144 7
144–148 7
Solution dikhao
Answer $135.6$ cm; $37.44$ ; $σ \approx 6.12$ cm
$y_{i} = \frac{x_{i} - 134}{4}$ ki values $- 3$ se $3$ tak hain. $\sum f_{i} y_{i} = - 6 - 10 - 8 + 0 + 13 + 14 + 21 = 24$ aur $\sum f_{i} y_{i}^{2} = 150$ .
$\overset{ˉ}{x} = 134 + 4 \times \frac{24}{60} = 135.6$ cm.
$60^{2} σ^{2} = 16 [60 (150) - 24^{2}] = 16 \times 8424,$ isliye $σ^{2} = 37.44$ aur $σ \approx 6.12$ cm.
NumericalHardNCERT Exercise 13.2, Q10
Ek box ki 80 keelon ki lambai (mm) neeche ki classes mein baanti gayi, jinke beech gaps hain. Mean lambai aur standard deviation nikalo.
Table 13.22
Lambai (mm) Keelein
20–24 8
25–29 15
30–34 29
35–39 21
40–44 7
Solution dikhao
Answer $32.25$ mm; $σ \approx 5.47$ mm
Lambaiyan poore millimetres mein hain, isliye asli boundaries classes ke theek beech mein, 24.5, 29.5 aur aage, hain. Classes 19.5–24.5, …, 39.5–44.5 ban jaati hain, aur mid-points wahi 22, 27, 32, 37, 42 rehte hain. $A = 32$ , $h = 5$ lo: $y_{i}$ ki values $- 2, - 1, 0, 1, 2$ hain.
$\sum f_{i} y_{i} = - 16 - 15 + 0 + 21 + 14 = 4$ aur $\sum f_{i} y_{i}^{2} = 96$ .
$\overset{ˉ}{x} = 32 + 5 \times \frac{4}{80} = 32.25$ mm.
$80^{2} σ^{2} = 25 [80 (96) - 16] = 25 \times 7664,$ isliye $σ^{2} \approx 29.94$ aur $σ \approx 5.47$ mm.

Height (cm)	Students
143	1
146	4
148	4
150	8
153	7
155	4
160	2

Weight (g)	Ande
42	5
44	6
46	8
48	11
50	15
52	13
54	2

Class	Frequency
0–40	1
40–80	10
80–120	13
120–160	11
160–200	3
200–240	2

Class	Frequency
10–20	6
20–30	7
30–40	17
40–50	11
50–60	9

Height (cm)	Paudhe
120–124	2
124–128	5
128–132	8
132–136	18
136–140	13
140–144	7
144–148	7

Lambai (mm)	Keelein
20–24	8
25–29	15
30–34	29
35–39	21
40–44	7

Miscellaneous Examples

Yeh examples formulas ko ulti disha mein chalate hain aur dikhate hain ki jab har observation ko ek hi tarah badla jaata hai to spread kaise react karta hai (Fig. 13.12).

3/3 $2 x$ : mean 10 ho jaata hai, $σ$ 4 ho jaata hai.

Fig. 13.12Jodne se data khisakta hai; guna karne se failta hai.

0 se 16 tak teen number lines par vertical grid lines same hain. Upar wali line data 2, 4, 5, 6 aur 8 dikhati hai jiska mean 5 hai aur 3 se 7 tak ek shaded band hai, yaani har taraf 2 ka ek standard deviation. Doosre step mein beech wali line har value plus 6 dikhati hai: 8, 10, 11, 12 aur 14; band khisak kar 9 se 13 par aa jaati hai lekin uski width wahi rehti hai, isliye sigma ab bhi 2 hai. Teesre step mein neeche wali line har value ka doguna dikhati hai: 4, 8, 10, 12 aur 16; mean doguna hokar 10 ho jaata hai aur band chaudi hokar 6 se 14 ho jaati hai, isliye sigma 4 ho jaata hai aur variance 4 se badh kar 16 ho jaata hai.

Example 13.13NCERT Example 13 · p. 282

Celsius se Fahrenheit

Ek weather station ke 15 dinon ke dopahar ke temperatures ka variance °C mein 6 hai. Ek visitor har reading ko $F = 1.8 C + 32$ se Fahrenheit mein badalta hai. Fahrenheit readings ka variance kya hai?

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

Example 13.14NCERT Example 14 · p. 283

Do khoye hue observations

Paanch dost ek quiz khelte hain. Unke scores ka mean 6 aur variance 8.8 hai. Meera ko teen scores, 2, 7 aur 9, yaad hain, lekin baaki do nahi. Unhe nikalo.

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

Example 13.15NCERT Example 15 · p. 284

Har observation ko shift karna

Ek teacher class test ke har score $x_{1}, x_{2}, \dots, x_{n}$ mein same number $c$ grace marks jodti hai ( $c$ negative bhi ho sakta hai, yaani penalty). Prove karo ki scores ka variance nahi badalta.

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

Example 13.16NCERT Example 16 · p. 284

Copying error theek karna

Ek clerk ne 40 daily rainfall readings (mm) spreadsheet mein type karte samay ek reading jo asal mein 37 thi, uski jagah 21 daal diya. Spreadsheet ne mean 22 aur standard deviation 6.6 bataya. Sahi mean aur standard deviation nikalo.

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

Quick check

3 mein se 1

x_{1}, \dots, x_{n}

ka variance 5 hai.

- 2 x_{1} + 7, \dots, - 2 x_{n} + 7

ka variance kya hai?

- 3

10

20

27

Answer dekho

Answer (C) $20$

(- 2)^{2} \times 5 = 20

. Variance kabhi negative nahi hota, aur 10 factor ka square lena bhool jaata hai.

Recap

Summary aur key points

Measure of central tendency centre batata hai; measure of dispersion spread batata hai. Barabar means aur medians wala data bhi bahut alag tarah se faila ho sakta hai (Section 13.1).
Range = sabse bada − sabse chhota. Yeh jaldi nikalti hai lekin sirf do values use karti hai (Section 13.3).
Mean ke about deviations ka total zero hota hai, isliye hum dooriyan $∣ x_{i} - a ∣$ ya squares $(x_{i} - \overset{ˉ}{x})^{2}$ use karte hain ((13.3)).
Raw data: M.D. $(a) = \frac{1}{n} \sum ∣ x_{i} - a ∣,$ $\overset{ˉ}{x}$ ya $M$ ke about (Section 13.4.1).
Grouped data: M.D. $(a) = \frac{1}{N} \sum f_{i} ∣ x_{i} - a ∣$ jahan $N = \sum f_{i}$ ; classes ki jagah unke class marks rakhe jaate hain (Section 13.4.2).
Grouped data ka median cumulative frequencies se aata hai; classes ke liye $M = l + \frac{N / 2 - C}{f} \times h$ ((13.10)).
Step deviations $d_{i} = \frac{x_{i} - a}{h}$ se $\overset{ˉ}{x} = a + h \overset{ˉ}{d}$ milta hai ((13.9)).
Dooriyon ka sum median ke about sabse kam hota hai, aur algebra mein absolute values ko sambhalna mushkil hai (Section 13.4.3).
Variance $σ^{2} = \frac{1}{N} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2};$ standard deviation $σ = \sqrt{σ^{2}},$ data ki units mein (Section 13.5).
Computing form: $σ^{2} = \frac{1}{N} \sum f_{i} x_{i}^{2} - {\overset{ˉ}{x}}^{2},$ ya $N^{2} σ^{2} = N \sum f_{i} x_{i}^{2} - (\sum f_{i} x_{i})^{2}$ ((13.15)).
Shortcut: agar $y_{i} = \frac{x_{i} - A}{h},$ to $\overset{ˉ}{x} = A + h \overset{ˉ}{y}$ aur $σ_{x} = h σ_{y}$ (Section 13.5.4).
$y = k x + c$ ke liye: $\overset{ˉ}{y} = k \overset{ˉ}{x} + c,$ $σ_{y}^{2} = k^{2} σ_{x}^{2};$ $c$ jodne se spread nahi badalta (Miscellaneous Examples).

Timeline

Statistics kaise viksit hua

c. 300 BCE
Arthashastra
Mauryan state chalane par Kautilya ka granth aise village officials ke baare mein batata hai jo gharon, janm aur maut ka record rakhte the.
c. 1595
Ain-i-Akbari
Akbar ke empire ka Abul Fazl ka record har province ke land surveys, fasal ki paidavar aur revenue ko darj karta hai.
1662
John Graunt
London ke janm aur maut ke weekly records ko study karte hain aur unmein steady patterns paate hain, jisse population data ki study ki neenv padi.
1713
Jacob Bernoulli
Unki maut ke baad chhapi unki Ars Conjectandi law of large numbers ka ek shuruaati roop prove karti hai.
1880s
Francis Galton
Hazaaron logon ke measurements se inheritance ko study karte hain, aur regression aur correlation ki shuruaat karte hain.
1894
Karl Pearson
Pehli baar print mein standard deviation term use karte hain; baad mein unhone London mein pehla university statistics department shuru kiya.
1918
R. A. Fisher
Variance shabd ki shuruaat karte hain; unke methods genetics aur agriculture mein statistics ko naya roop dete hain.
1931
P. C. Mahalanobis
Kolkata mein Indian Statistical Institute ki sthapana karte hain; India unke birthday, 29 June, ko Statistics Day ke roop mein manata hai.

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 13.1, Q10	Practice 13.1 · Q10
NCERT Exercise 13.1, Q11	Practice 13.1 · Q11
NCERT Exercise 13.1, Q12	Practice 13.1 · Q12
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 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise 13.2, Q8	Practice 13.2 · Q8
NCERT Exercise 13.2, Q9	Practice 13.2 · Q9
NCERT Exercise 13.2, Q10	Practice 13.2 · Q10
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6

Quick check: MCQs

13.1 Introduction3 questions
13.3 Range3 questions
13.4.1 Ungrouped data ke liye mean deviation3 questions
Continuous frequency distribution4 questions
13.4.3 Mean deviation ki limitations2 questions
13.5.1 Standard Deviation4 questions
13.5.4 Variance aur standard deviation nikalne ka shortcut method3 questions
Miscellaneous Examples3 questions

Glossary · 14

Key terms (English + Hindi)

assumed meanकल्पित माध्यkalpit madhya: Data ke beech ke paas ki ek suvidhajanak value jisse deviations naape jaate hain taaki hisaab aasaan ho.Chapter mein pehli baar yahan aaya →

class markवर्ग चिह्नvarg chihn: Class interval ka mid-point, jo us class ke har observation ki jagah use hota hai.Chapter mein pehli baar yahan aaya →
continuous frequency distributionसतत बारंबारता बंटनsatat barambarta bantan: Ek table jo bina gap wale class intervals ke set mein se har ek mein observations ki sankhya batati hai.Chapter mein pehli baar yahan aaya →
cumulative frequencyसंचयी बारंबारताsanchayi barambarta: Kisi value ya class tak (use milakar) frequencies ka chalta hua total.Chapter mein pehli baar yahan aaya →

deviationविचलनvichalan: Kisi observation aur ek fixed value $a$ ke beech ka difference $x - a$ ; yeh positive ya negative ho sakta hai.Chapter mein pehli baar yahan aaya →
discrete frequency distributionअसतत बारंबारता बंटनasatat barambarta bantan: Ek table jo har alag value ko uske aane ki ginti ke saath list karti hai.Chapter mein pehli baar yahan aaya →

mean deviationमाध्य विचलनmadhya vichalan: Observations ki ek central value $a$ se dooriyon $∣ x_{i} - a ∣$ ka mean, aam taur par mean ya median ke about.Chapter mein pehli baar yahan aaya →
measure of central tendencyकेंद्रीय प्रवृत्ति की मापkendriya pravritti ki maap: Mean ya median jaisi ek single value, jiske aas-paas data kendrit hota hai.Chapter mein pehli baar yahan aaya →
measure of dispersionप्रकीर्णन की मापprakeernan ki maap: Ek number jo batata hai ki observations kitni door tak bikhre hain, jaise range ya standard deviation.Chapter mein pehli baar yahan aaya →
median classमाध्यिका वर्गmadhyika varg: Woh class jismein cumulative frequency pehli baar total frequency ke aadhe tak pahunchti hai.Chapter mein pehli baar yahan aaya →

rangeपरिसरparisar: Sabse bada observation minus sabse chhota observation.Chapter mein pehli baar yahan aaya →

standard deviationमानक विचलनmanak vichalan: Variance ka positive square root, jise $σ$ likhte hain; iski units data wali hi hoti hain.Chapter mein pehli baar yahan aaya →
step deviationपद विचलनpad vichalan: Assumed mean se deviation ko ek common factor $h$ se divide karke milne wali value: $d = \frac{x - a}{h}$ .Chapter mein pehli baar yahan aaya →

varianceप्रसरणprasaran: Mean se squared deviations ka mean, jise $σ^{2}$ likhte hain.Chapter mein pehli baar yahan aaya →

Statistics: Class 11 Maths Chapter 13 Hinglish mein

Is chapter mein aap kya seekhenge

13.1 Introduction

13.2 Measures of Dispersion

13.3 Range

13.4 Mean Deviation

13.4.1 Ungrouped data ke liye mean deviation

13.4.2 Grouped data ke liye mean deviation

Discrete frequency distribution

Continuous frequency distribution

Practice 13.1: answers ke saath

13.4.3 Mean deviation ki limitations

13.5 Variance aur Standard Deviation

13.5.1 Standard Deviation

13.5.2 Discrete frequency distribution ka standard deviation

13.5.3 Continuous frequency distribution ka standard deviation

13.5.4 Variance aur standard deviation nikalne ka shortcut method

Practice 13.2: answers ke saath

Miscellaneous Examples

Summary aur key points

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye