NCERT Solutions for Class 11 Mathematics Chapter 15 Statistics

If you’re a Class 11 student studying Mathematics from the NCERT Textbook, you would have  probably come across Chapter 15 Statistics. Statistics is concerned with the gathering of information for specific purposes. Decisions can be made through analysing and interpreting facts. This chapter discusses measures of dispersion, range, mean deviation, variance, and standard deviation, as well as frequency distribution analysis.

You’ve already learned how to express data graphically and tabularly in previous classes. In Chapter 15 Class 11 Mathematics, you will learn about data representation, as well as specific prominent elements and qualities of the data. This is a chapter with a medium weightage for JEE Main.


Class 11 Mathematics Chapter 15 – Topics 

In the financial and economic industries, statistics is one of the most extensively used disciplines for planning and forecasting. They assist in the development of mathematical models and the analysis of time-series data. Therefore, students should extensively practise with the NCERT Solutions Class 12 Mathematics Chapter 15 and revise these concepts. The  solutions include answers, illustrations, and explanations for the entire chapter 15 titled Statistics, which is taught in Class 11.

Let’s go over the sections in this chapter before diving into the NCERT Solutions for Class 11 Mathematics Chapter 15:

Exercise Topics
15.1 Introduction
15.2 Measures of Dispersion
15.3 Range
15.4 Mean Deviation
15.4.1 Mean Deviation for Ungrouped Data
15.4.2 Mean Deviation for Grouped Data
15.4.3 Limitations of Mean Deviation
15.5 Variance and Standard Deviation
15.5.1 Standard Deviation
15.5.2 Standard Deviation of a Discrete Frequency Distribution
15.5.3 Standard Deviation of a Continuous Frequency Distribution
15.5.4 Shortcut method to find Variance and Standard Deviation
15.6 Analysis of Frequency Distributions
15.6.1 Comparison of Two Frequency Distributions with same Mean


15.1 Introduction

With examples, this section discusses the concepts of central tendency, mean, and median [during even and odd number of observations]. It introduces the notion of dispersion measurement. Measures of central tendency are values that cluster around the middle or centre of the distribution. It includes the mean, the median, and the mode. In a class, mean can be used to calculate the average of the students’ grades.

15.2 Measures of Dispersion

This section discusses measurements of dispersion, including range, quartile deviation, mean deviation, and standard deviation.

Measures of dispersion explain the link with measures of central tendency. The spread of data, for example, indicates how well the data is represented by the mean. If the spread is high, the mean is not indicative of the data.

15.3 Range

This section explains the range, its calculation, and provides an example. The range represents the variety of scores by utilising the set’s highest and minimum values.


15.4 Mean Deviation

This section defines mean deviation and its formula. For example, biologists can utilise the notion of mean deviation to compare different animal weights and determine the weight that  is considered healthy.

15.4.1 Mean Deviation – Ungrouped  Data 

This section describes how to calculate the mean deviation for ungrouped data. Students will have to calculate the mean, deviations from the mean, and absolute deviations, then enter the numbers into the mean deviation formula to get the result.

15.4.2 Mean Deviation – Grouped Data 

This section explains how to calculate mean deviation for continuous and discrete frequency distributions using solved examples.

15.4.3 Mean deviation limitations

This section will cover the limitations of using mean deviation. Students will learn that: 

  • If there is a lot of variation in a series, the median will not be a good representation of the data. As a result, the mean deviation derived around such a median cannot be trusted.
  • The mean deviation from the mean is not very specific if the sum of deviations from the mean is larger than the sum of deviations from the median.
  • Further algebraic handling of the obtained absolute mean deviation is not possible. It cannot be used as an accurate measure of dispersion.


15.5 Standard Deviation and Variance

15.5.1 Standard Deviation

This section defines variance and standard deviation, as well as provides formulae  and solved examples for discrete and continuous frequency distributions. 

15.5.2 Standard Deviation of a Discrete Frequency Distribution

Students will learn that standard deviation can be determined by one of two methods based on the frequency distribution of the dataset. One of such methods is when the pieces of the data presented are discontinuous (discrete) in which there are no observation groups, and each element is assigned a specific frequency value.

15.5.3. Standard Deviation of a Continuous Frequency Distribution

The second method is when the data is provided, i.e., the elements or observations are displayed in a continuous or grouped format. It means that instead of providing a specific value to each frequency, a ‘class’ or ‘group’ of a specific element range is provided.

15.5.4 Quick Technique for Calculating Variance and Standard Deviation

With a few illustrations, this section discusses the simpler method of determining the standard deviation. 

15.6 Frequency Distribution Analysis

This section explains how to compare the variability of two series with the same mean, coefficient of variation, and a few solved issues.

15.6.1 Comparison of Two Frequency Distributions with Same Mean

To compare the variability or dispersion of two series, the coefficient of variance for each series should be estimated. It is calculated by dividing the standard deviation by the mean and finding its percentage. 


NCERT Solutions for Class 11 Mathematics Chapter 15 – Statistics 

You must be seeking answers to the questions after you have completed Chapter 15 Mathematics Class 11. The solutions provide alternative solutions and clarifications, making students feel more confident when appearing in the exam. In case, students aren’t able to derive the answer or are unsure if their answer/approach is correct, solutions by Extramarks will prove to be of great help. The subject matter experts from Extramarks have designed the Class 11 Mathematics NCERT Solutions Chapter 15 based on the most recent CBSE Syllabus 2022-23, keeping in mind the question paper setting and the distribution of marks throughout the chapters. You can access and download these solutions and practise them offline as well. 

Access NCERT Solutions for Class 11 Mathematics Chapter 15 – Statistics


Class 11 Mathematics NCERT Solutions Chapter 15

As students move through their educational journey, the concepts provided in this chapter will serve as the foundation for the introduction of more complex topics. As a result, students should make it a point to be well-versed in the formulas and theories of Statistics, as these will undoubtedly make their way into future board exams. These NCERT Solutions offer a variety of practical examples to deliver an engaging lesson, guaranteeing that students gain a solid statistical foundation and also download the exercise-by-exercise solutions given in the links below.


Mean Deviation About Mean


In the following chapter, we go over all of the exercises:

How is Statistics Class 11 Mathematics NCERT Solutions Advantageous to Students?

If you ask any top student what their secret to success is, they will undoubtedly answer that they extensively read the NCERT textbooks. All of the fundamentals are covered in these textbooks. After you have finished the theories, it is critical that you solve the NCERT questions thoroughly. This is essential for both – board exams and competitive tests such as JEE Main and JEE Advanced. 

The NCERT Solutions for Class 11 Mathematics Chapter 15 have exact explanations in easy language to assist students in performing well in their first term exams. Before moving on to more difficult topics, basic concepts are thoroughly explained so that students’ doubts or confusion are cleared.

The step-by-step manner of problem-solving offers students a thorough understanding of the mark weightage as per the new CBSE Syllabus 2022-23. Students will be able to identify their areas of weakness and try to improve them in order to improve their academic performance.

You now have access to all of the Class 11 Mathematics NCERT Solutions Chapter 15 on Statistics. Use these solutions to your advantage and master the chapter.

Q.1 Find the mean deviation about the mean for the data in Exercises 1 and 2.
Q.1: 4, 7, 8, 9, 10, 12, 13, 17
Q.2: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44



Mean=4+7+8+9+10+12+13+178        =808        =10Mean deviation={|410|+|710|+|810|+|910|+|1010|+|1210|+|1310|+|1710|8}       =6+3+2+1+0+2+3+78       =248       =3

Thus, the mean deviation is 3.


Mean=38+70+48+40+42+55+63+46+54+4410       =50010       =50Mean deviation={|3850|+|7050|+|4850|+|4050|+|4250|+|5550|+|6350|+|4650|+|5450|+|4450|10}        =12+20+2+10+8+5+13+4+4+610        =8410=8.4Thus, the mean deviation is 8.4.

Q.2 Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49


3. Ascending order of data:10,11,11,12,13,13,14,16,16,17,17,18Here, n=12(even)Median=(n2)thterm+(n2+1)thterm2        =(122)thterm+(122+1)thterm2       =6thterm+7thterm2       =13+142       =272Median=13.5Mean deviation about median      ={|1013.5|+|1113.5|+|1113.5|+|1213.5|+|1313.5|+|1313.5|+|1413.5|+|1613.5|+|1613.5|+|1713.5|+|1713.5|+|1813.5|12}      =(3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5)12      =2812      =2.33(Approx.) 4. Ascending order of data:36,42,45,46,46,49,51,53,60,72 Here, n=10(even)Median=(n2)thterm+(n2+1)thterm2      =(102)thterm+(102+1)thterm2      =5thterm+6thterm2      =46+492      =952Median=47.5Mean deviation about median      ={|3647.5|+|4247.5|+|4547.5|+|4647.5|+|4647.5|+|4947.5|+|5147.5|+|5347.5|+|6047.5|+|7247.5|10}      =(11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5)10      =7010=7

Q.3 5. Find the mean deviation about the mean for the data in Exercises 5 and 6.

Xi 5 10 15 20 25
fi 7 4 6 3 5


Xi 10 30 50 70 90
fi 4 24 28 16 8


xififixixix¯fixix¯5735514=963104401014=416156901514=16203602014=6182551252514=1155Total25350158 Meanx¯=fixifi           =35025           =14Mean deviation about mean           =fixix¯fi           =15825=6.32Thus, the mean deviation about mean is 6.32. xififixixix¯fixix¯104401050=4016030247203050=20480502814005050=00701611207050=203209087209050=40320Total8040001280 Meanx¯=fixifi          =400080          =50Mean deviation about mean          =fixix¯fi          =128080=16Thus, the mean deviation about mean is 16.

Q.4 Find the mean deviation about the median for the data in Exercises 7 and 8.

Xi 5 7 9 10 12 15
fi 8 6 2 2 2 6
Xi 15 21 27 30 35
fi 3 5 6 7 8



n = 26evenMedian = n2th term+n2+1th term2         = 262th term+262+1th term2         = 13th term+14th term2         = 7+72         = 7Mean deviation about median         = fixi-Mefi         = 8426         = 3.23Thus, Mean deviation about median is 3.23.


xifiC.f.xiMedianfixiMedian15331530=154521582130=945276142730=318307213030=00358293530=540Total29148 n = 29oddMedian= n+12th term        = 29+12th term        = 15th term        = 30Mean deviation about Median        = fixi-Mefi        = 14829        = 5.1Thus, the Mean deviation about Median is 5.1.

Q.5 Find the mean deviation about the mean for the data in Exercises 9 and 10.


Income per day 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Number of persons 4 8 9 10 7 5 4 3
Height in cms 95-105 105-115 115-125 125-135 135-145 145-155
Number of boys 9 13 26 30 12 10



Incomefixiui=xAhfiuixix¯fixix¯010045041650358=30812321002008150324150358=20816642003009250218250358=10897230040010350110350358=8804005007450=A00450358=92644500600555015550358=192960600700465028650358=2921168700800375039750358=3921176Total50467896 Mean=A+h×fiuifi, where ui=xiAh and h=100      =450+100×4650      =4502×46      =45092      =358Mean deviation about mean        =fixix¯fi        =789650        =157.92


Classintervalfixiui=xiAhfiuixix¯fixix¯951059100327100125.3=25.3227.710511513110226110125.3=15.3196.911512526120126120125.3=5.3137.812513530130=A00130125.3=4.714113514512140112140125.3=14.7176.414515510150220150125.3=24.7247Total100471128.8 Mean=A+h×fiuifi, where ui=xiAh and h=10      =130+10×47100      =1304.7      =125.3Mean deviation about Mean      =fixix¯fi      =1128.8100      =11.288      =11.29ApproxThus, Mean deviation about Mean is 11.29.

Q.6 Find the mean deviation about median for the following data :

Marks 0-10 Oct-20 20-30 30-40 40-50 50-60
Number of Girls 6 8 14 16 4 2


Marksfixic.f.|xiMe|fi|xix¯|010656|527.85|=22.85137.1102081514|1527.85|=12.85102.82030142528|2527.85|=2.8539.93040163544|3527.85|=7.15114.4405044548|4527.85|=17.1568.6506025550|5527.85|=27.1554.3Total50517.1N=50(even)N2=502=25So, the median class is 2030.Median=l+(N2)Cf×h       =20+251414×10[C=14,f=14​  and h=10]       =20+7.85=27.85Mean deviation about Meadian    =517.150   =10.34Thus, Mean deviation about Meadian is 10.34.

Q.7 Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age 16-20 21-25 36-30 31-35 36-40 41-45 46-50 51-55
Number 5 6 12 14 26 12 16 9


The given table can be arranged as given below:

Marksfixic.f.|xiMe|fi|xix¯|15.520.55185|1838|=2010020.525.562311|2338|=159025.530.5122823|2838|=1012030.535.5143337|3338|=57035.540.5263863|3838|=0040.545.5124375|4338|=56045.550.5164891|4838|=1016050.555.5953100|5338|=15135Total100735 N=100(even)N2=1002=50So, the median class is 35.540.5. Median=l+(N2)Cf×h       =35.5+503726×5[C=37,f=26​  and h=5]       =35.5+2.5=38Mean deviation about Meadian   =fi|xiMe|fi   =735100   =7.35Thus, Mean deviation about Meadian is 7.35.

Q.8 Find the mean and variance for each of the data in Exercies 1 to 5.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
Q2. First n natural numbers.
Q3. First 10 multiples of 3.

Xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
Xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3



Mean(x¯)=6+7+10+12+13+4+8+128 =728 =9 xi(xix¯)(xix¯)2669=39779=2410109=1112129=3913139=416449=525889=1112129=39Total74Variance(σ2)=(xix¯)2N =748 =9.25


Mean of n natural numbers=i=1nxin=(n(n+1)2)n=n+12Variance(σ2)=i=1nxi2(x¯)2n =(12+22+32+...+n2)n(n+12)2=1n×n(n+1)(2n+1)6(n2+2n+14)Variance(σ2)=2n2+3n+16(n2+2n+14)=4n2+6n+23n26n312=n2112


Mean of 10 multiples of 3          =(3+6+9+12+15+18+21+24+27+30)10          =16510=16.5Variance(σ2)=i=1nxi2(x¯)2n          =(32+62+92+122+152+182+212+242+272+302)10(16.5)2           =346510272.25          =346.5272.25         Variance(σ2)=74.25


xififixi(xix¯)(xix¯)2fi(xix¯)26212619=13169338104401019=981324147981419=52517518122161819=11122481922419=5252002841122819=981324303903019=11121363Total407601736Mean (x¯)=i=1nfixin =76040 =19 Variance (σ2)=i=1nfi(xix¯)2n Variance(σ2)=173640 =43.4


xififixi(xix¯)(xix¯)2fi(xix¯)292327692100=86419293218693100=7499897329197100=392798219698100=2481026612102100=24241043312104100=416481093327109100=981243Total222200640Mean(x¯)=i=1nfixin = 220022 =100Variance(σ2)=i=1nfi(xix¯)2n Variance(σ2)=64022 =29.09

Q.9 Find the mean and standard deviation using short-cut method.

Xi 60 61 62 63 64 65 66 67 68
fi 2 1 12 29 25 12 10 4 5


xifidi=xiA1fidixix¯xix¯2fixix¯2602484163261133399621222424486329129112964=A25000006512112111266102202440674312393668542041680Total1000286 Meanx¯=A+i=1nfidin         =64+0100=64Varianceσ2=i=1nfixix¯2n              =286100              =2.86Standard deviation              =2.86              =1.69 Approx.

Q.10 Find the mean and variance for the following frequency distributions in Exercises 7 and 8.


Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2
Classes 0-10 10-20 20-30 30-40 40-50
Frequencies 5 8 15 16 6



Classesfixiui=xiA30ui2fiuifiui203021539618306034524612609057511559012010105=A0000120150313511331501805165241020180210219539618Total30276 Mean=A+ i=1 n f i u i n ×h =105+ 2 30 ×30 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaxMaacaWLjaGaamytaiaadwgacaWGHbGaamOBaiabg2da9iaadgeacqGHRaWkdaWcaaqaamaaqahabaGaamOzamaaBaaaleaacaWGPbaabeaakiaadwhadaWgaaWcbaGaamyAaaqabaaabaGaamyAaiabg2da9iaaigdaaeaacaWGUbaaniabggHiLdaakeaacaWGUbaaaiabgEna0kaadIgaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7cqGH9aqpcaaIXaGaaGimaiaaiwdacqGHRaWkdaWcaaqaaiaaikdaaeaacaaIZaGaaGimaaaacqGHxdaTcaaIZaGaaGimaaaaaa@5F1B@ =107   Variance(σ2)=h2n2[nin(fiui2)(infiui)2]    =(30)2(30)2[30×76(2)2]   =22804=2276


Classesfixiui=xiA10ui2fiuifiui2010552410201020815118820301525=A0000304016351116164050645241224Total501068Mean=A+i=1nfiuin×h    =25+1050×10    =27   Variance(σ2)=h2n2[nin(fiui2)(infiui)2]     =(10)2(50)2[50×68(10)2] =125[3400100]   =132

Q.11 Find the mean, variance and standard deviation using short-cut method

Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of children 3 4 7 7 15 9 6 6 3


Classes f i x i u i = x i A 5 u i 2 f i u i f i u i 2 7075 3 72.5 4 16 12 48 7580 4 77.5 3 9 12 36 8085 7 82.5 2 4 14 28 8590 7 87.5 1 1 7 7 9095 15 92.5=A 0 0 0 0 95100 9 97.5 1 1 9 9 100105 6 102.5 2 4 12 24 105110 6 107.5 3 9 18 54 110115 3 112.5 4 16 12 78 Total 60 6 254 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD5F@ Mean=A+i=1nfiuin×h =92.5+660×5    =92.5+0.5    =93Variance(σ2)=h2n2[nin(fiui2)(infiui)2]    =(5)2(60)2[60×254(6)2]    =253600[1524036]    =253600×15204    =105.52Standard deviation   =105.52   =10.27

Q.12 The diameters of circles (in mm) drawn in a design are given below:

Diameters 33-36 37-40 41-44 45-48 49-52
No. of circles 15 17 21 22 25

Calculate the standard deviation and mean diameter of the circles.


Classe-intervafixiui=xi-A4ui2fiufiui232.5-36.51534.5-24-306036.5-40.51738.5-11-17-1740.5-44.52142.5=A000044.5-48.52246.511222248.5-52.52550.52450100Total10025199  Varianceσ2 = h2n2ninfiui2infiui2                  = 421002100×199-252                  = 162519900-625                  = 1625×19275                  = 30.84Standard deviation                  = 30.84                  = 5.55           Mean = A+i=1nfiuin×h                  = 42.5+25100×4                  = 42.5+1                  = 43.5

Q.13 From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7


For group A:

Marks(A)fixiui=xiA10ui2fiuifiui21020915392781203017252434683040323511323240503345=A0000506040551140406070106524204070809753927811506342  Mean(x¯)=A+i=1nfiuin×h=45+6150×10=450.4=44.6Variance(σ2)=h2n2[nin(fiui2)(infiui)2] Variance(σ2)=(10)2(150)2[150×342(6)2]          =1225(5130036)      =1225×51264          =227.84Standard deviation(σ)          =227.84          =15.09 Marks(A)fixiui=xiA10ui2fiuifiui210201015393090203020252440803040303511303040502545=A0000506043551143436070156524306070807753921631506366 Mean(x¯)=A+i=1nfiuin×h=45+6150×10=450.4=44.6 Variance(σ2)=h2n2 [nin(fiui2)(infiui)2]Variance(σ2)=(10)2(150)2 [150×366(6)2]         =1225(5490036)         =1225×54864         =243.84Standard deviation(σ)         =243.84         =15.61

Since, both groups have equal mean. Group B has greater standard deviation. So, group B is more variable than group A.

Q.14 From the prices of shares X and Y below, find out which is more stable in value:

X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101


For prices of share X:

Mean(x¯)=35+54+52+53+56+58+52+50+51+4910         =51010         =51 xi(xix¯)(xix¯)2353551=16256545451=39525251=11535351=24565651=525585851=749525251=11505051=11515151=00494951=24350Variance(σ2)=i=1n(xix¯)2N=35010=35Standard deviation(σ)=35=5.91C.V.(share X)=σx¯×100=5.9151×100=11.59(Appox.)

For prices of share Y:

Mean(y¯)=108+107+105+105+106+107+104+103+104+10110         =105010         =105yi(yiy¯)(yiy¯)2108108105=39107107105=24105105105=00105105105=00106106105=11107107105=24104104105=11103103105=24104104105=11101101105=41640Variance(σ2)=i=1n(xix¯)2N=4010=4Standard deviation(σ)=4 =2C.V.(share Y)=σy¯×100=2105×100=1.9<11.59C.V. of prices of share Y is lesser than that of share X.Thus, the prices of share Y are more stable than that of share X.

Q.15 An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A Firm B
No. of wage earners 586 648
Mean of monthly wages Rs. 5253 Rs. 5253
Variance of distribution of wages 100 121

1. Which firm A or B pays larger amount as monthly wages?
2. Which firm, A or B, shows greater variability in individual wages?


i.         Monthly wage of firm A = Rs. 5253     Number of workers in firm A = 586     Total paid amount in wages = Rs. 5253 × 586                                       = Rs. 3078258           Monthly wage of firm B = Rs. 5253     Number of workers in firm A = 648      Total paid amount in wages = Rs. 5253 x 648                                        = Rs.3403944Thus, firm B pays the larger amount as monthly wages.ii. Variance of distribution of wages in firm A = 100               Standard deviation of wages in firm A = 100                                                      = 10       Variance of distribution of wages in firm B = 121                 Standard deviation of wages in firm B = 121                                                        = 11Since the average monthly wages in both the firms is same,i.e., Rs. 5253, therefore, the firm with greater standard deviationwill have more variability.Thus, the firm B has greater variability in the individual wages.

Q.16 The following is the record of goals scored by team A in a football session:

No. of goals scored 0 1 2 3 4
No. of matched 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?


No. of goals x i No. of matches f i x i x ¯ x i x ¯ 2 f i x i x ¯ 2 0 1 02=2 4 4 1 9 12=1 1 9 2 7 22=0 0 0 3 5 32=1 1 5 4 3 42=2 4 12 25 30 Mean = i=1nfixin       = 5025       = 2Variationσ2= i=1nfixix¯2n= 3025= 1.2Standard deviation= 1.2= 1.09Standard deviation of team B= 1.25 goalsMean of team B= 2The team with lower standard deviation is more consistent.Therefore, team A is more consistent.

Q.17 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

i=150xi=212,  i=150xi2=902.8,   i=150yi=261,  i=150yi2=1457.6Which is more varying, the length or weight?


i=150xi=212,  i=150xi2=902.8,Mean=i=150xi50            =21250            =4.24Variance(σ2)=i=150xi250(i=150xi50)2                    =902.850(21250)2                    =18.05617.978                    =0.078Standandard deviation(σ)                   =0.078                   =0.28C.V. of length=σx¯×100                     =6.6i=150yi=261,  i=150yi2=1457.6Mean(y¯)=i=150yi50                  =26150  =5.22Variance(σ2)=i=150yi250(i=150yi50)2                    =1457.650(26150)2                    =29.1527.25                    =1.89Standandard deviation (σ)                    =1.89                    =1.37C.V. of weight=σy¯×100                    =1.375.22×100                    =26.24 Since, C.V. of weights is greater than the C.V. of lengths.Therefore, weights are more varing than lengths.

Q.18 The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.


Let remaining two observations be x and y.

Therefore, the series is 6, 7, 10, 12,12,13,x and y.Then, Mean=6+7+10+12+12+13+x+y8 9=60+x+y8        72=60+x+y  x+y=7260  x+y=12 ...(i)Variance=i=18(xix¯)28    9.25={(69)2+(79)2+(109)2+(129)2+(129)2+(139)2+(x9)2+(y9)2}8    74=9+4+1+9+9+16+x218x+81+y218y+81    74=210+x218(x+y)+y2    74=210+x218(12)+y2 [From eqation (i)]    74=210+x2216+y2    80=x2+y2     x2+y2=80 ...(ii) Squarring equation (i), we get MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWGtbGaamyCaiaadwhacaWGHbGaamOCaiaadkhacaWGPbGaamOBaiaadEgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gadaqadaqaaiaadMgaaiaawIcacaGLPaaacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaaa@52AF@     (x+y)2=(12)2x2+2xy+y2=144         2xy=144(x2+y2)         2xy=14480         2xy=64 ...(iii)Now,(xy)2=x2+y22xy       =8064(xy)2=16      xy=±4 ...(iv)From equation (i) and (iv), we get      x=8 and y=4(By taking, xy=4)orx=4 and y=8(By taking, xy=4)Thus, the remaining two observations are 4 and 8.

Q.19 The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.


Let remaining two observations be x and y.

Therefore, the series is 2, 4, 10, 12,14,x and y.Then, Mean=2+4+10+12+14+x+y7 8=42+x+y7      56=42+x+y    x+y=5642             x+y=14 ...(i)         Variance=i=18(xix¯)27        16={(28)2+(48)2+(108)2+(128)2+(148)2+(x8)2+(y8)2}7  112=36+16+4+16+36+x216x+64+y216y+64  112=236+x216(x+y)+y2  112=236+x216(14)+y2[From eqation (i)]  112=236+x2224+y2  100=x2+y2     x2+y2=100 ...(ii)Squarring equation (i), we get   (x+y)2=(14)2x2+2xy+y2=196       100+2xy=196 [From equation (i)]          2xy=96 ...(iii)Now,(xy)2=x2+y22xy       =10096(xy)2=4      xy=±2 ...(iv) From equation (i) and (iv), we get      x=8 and y=6(By taking, xy=2)or  x=6 and y=8(By taking, xy=2)Thus, the remaining two observations are 6 and 8.

Q.20 The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.


       Mean of 6 observations=8          Sum of 6 observations=6×8          =48When each observation is multiplied by 3,New mean=(3x6)=3×486          =24Standard deviation=4Variance of 6 observations =16       x26(x6)2=16              x26(8)2=16                    x26=16+64       x2=6×80           =480When each observation is multiplied by 3,  (3x)2=9x2          =9×480New variance of 6 observations         =(3x)26(3x6)2         =9×4806(24)2         =720576         =144New standard deviation=144         =12Thus, new mean and new standard deviatiion of 6 observations are 24 and 12 respectively.


Given that x¯ is the mean and σ2 is the variance of n observations x1, x2, ...,xn. Prove that the mean and variance of the observations ax1, ax2, ax3, ., axn are ax¯ and a2σ2, respectively, (a¹0).


              Mean of n observations = x¯                      x1+x2+x3+...+xnn  = x¯ ...i      Variance of n observations  = σ2 x12+x22+x32+...+xn2nx1+x2+x3+...+xnn2 = σ2x12+x22+x32+...+xn2nx¯2 = σ2  x12+x22+x32+...+xn2n = σ2+x¯2 ...iiMean of ax1, ax2, ax3, ., axn = ax1+ ax2+ ax3+ .+ axnn      = ax1+ x2+ x3+ .+ xnnMean of ax1, ax2, ax3, ., axn = ax¯Variance of ax1, ax2, ax3, ., axn= a2x12+a2x22+a2x32+...+a2xn2n - ax1+ax2+ax3+...+axnn2= a2x12+x22+x32+...+xn2n - a2x1+x2+x3+...+xnn2= a2σ2+x¯2-a2x¯2= a2σ2+a2x¯2-a2x¯2  = a2σ2Thus, the mean and the variance of ax1,ax2,ax3,.,axn are ax¯ and a2σ2.

Q.22 The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.
(ii) If it is replaced by 12.


      Mean value of 20 observations=10         Total sum of 20 observations=200Standard deviation of 20 observations=2  Variance of 20 observations=4    x220(x20)2=4        x220(10)2=4                       x220=100+4                       x2=104×20      Incorrect observation=8(i)If wrong item is omitted, then Remaining observations=201 =19 New mean of remaining observations=200819=19219 =10.1New variance of remaining observations =x28219(x819)2 =104×208219(200819)2 =201619(10.1)2 =106.1102.01 =4.09New standard deviation of remaining observations =4.09 =2.02(ii) When 8 is replaced by 12,                New mean of 20 observations=2008+1220 =20420 =10.2New variance of 20 observations =x282+12220(x8+1220)2New variance of 20 observations=104×2082+12220(2008+1220)2 =208064+14420(10.2)2  =216020104.04  =108104.04New variance of 20 observations=3.96New standard deviation=1.9899   =1.99(Approx)

Q.23 The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard 12 15 20
Which of the three subjects shows the highest variability in marks and which shows the lowest?


C.V. of marks in Mathematics=σMean×100     =1242×100     =28.57            C.V. of marks in Physics=σMean×100      =1532×100     =46.87    C.V. of marks in Chemistry=σMean×100     =2040.9×100     =48.89Since, C.V. of Chemistry is the greatest and C. V. of Mathematics is the lowest.So, highest variability is in the marks of Chemistry and lowest variability is in the marks of Mathematics.

Q.24 The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.


Mean value of 100 observations=20Total sum of 100 observations=2000Standard deviation of 100 observations=3Variance of 100 observations=9      x2100(x100)2=9         x2100(20)2=9                        x2100=400+9                       x2=409×100   Incorrect observations=21,21,18If wrong observations are omitted, then Remaining observations=1003 =97New mean of remaining observations=200021211897 =194097 =20New variance of remaining observations =x221221218297(x21211897)2 =4090044144132497(20)2 =3969497400 =409.216400 =9.216      New standard deviation=9.216 =3.036

Thus, the new mean and new standard deviation are 20 and 3.036 respectively.

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FAQs (Frequently Asked Questions)

1. What are the key formulas in Statistics Class 11?

Mean, median, variance, and standard deviation formulae are important in Class 11. You’ll be able to develop a clear knowledge of the topics contained in the Chapter 15 Class 11 Mathematics once you’ve completed them. To do so, use the examples and activities in the NCERT Class 11 Mathematics CBSE book to solve and practise each sum.

2. Explain the concept of standard deviation, which is covered in Chapter 15 of the NCERT Solutions for Class 11 Mathematics.

The measurement of variation or deviation of a given set of values is dealt with by standard deviation. The range is determined by the standard deviation level. Students should obtain the solutions accessible at Extramarks to better grasp this term. Based on the needs of the students, they are presented in chapter and exercise format.

3. From where and how can I get the NCERT solutions for CBSE Mathematics Class 11 Chapter 15 Statistics?

 The NCERT Solutions for Class 11 Chapter 15, which can be found in this post, is created by subject experts and professionals affiliated with Extramarks who have considerable experience in the field of teaching and are well-versed in the entire course. The solutions are specifically created with NCERT recommendations in mind. Because they are written by subject-matter experts, these answers and notes save time and are reliable for preparation.

4. How much should you practise Mathematics for NCERT Class 11?

 While there is no set time for practising equations, the time required varies from person to person. Students are advised to select and practise challenging sums and equations from the Class 11 Mathematics NCERT Solutions Chapter 15. It’s also vital to set aside a few hours each week to practise those difficult equations, since this will help you study in the most efficient way possible.

5. Do I have to practise all of the questions in NCERT Solutions Class 11 Mathematics Statistics?

Constant practise is the key to mastering Ch 15 Mathematics Class 11 on Statistics. Students must memorise a number of formulas and conduct extensive calculations. A  simple error while solving these problems can cost the entire mark of that question. As a result, students should review these answers on a frequent basis to keep their skills sharp.

6. What is the total number of questions in NCERT Solutions Class 11 Mathematics Chapter 15 Statistics?

In Chapter 15 Class 11 Mathematics, there are 34 problems divided among three exercises and one miscellaneous exercise. These questions, handpicked by Mathematics professionals, provide children with a 360-degree picture of the entire chapter. They range from simple formula-based to complex word problems.