Careers360 Logo
NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:21 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 15 includes solutions to questions of all types such as short answer questions, long answer questions, value-based questions, MCQ’s and HOTS. The solutions are given by experts in a simple manner and are solved step-by-step. If you who have doubts regarding the chapter Statistics, then refer to NCERT Exemplar Class 11 Maths chapter 15 solutions and follow the solutions for better understanding of the concept. Students can download the pdf for Class 11 Maths NCERT Exemplar Solutions Chapter 15 which consists of all the important questions. You can cover all the important topics and sub-topics using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function. You can also get your queries solved by referring to the solutions solved step by step in an easily comprehensible format.
Also, check - NCERT Class 11 Maths Solutions for Other Chapters

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:
  2. More about NCERT Exemplar Class 11 Maths Solutions Chapter 15
  3. NCERT Solutions for Class 11 Mathematics Chapters
  4. Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

Question:1

Find the mean deviation about the mean of the distribution:
 Size 2021222324 Frequency 64514

Answer:

We have to find the mean deviation about the mean of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

 Size (xi) Frequency (fi)fixi206120214842251102312324496 Total 20433
Here, mean X =ΣfixiΣfi=43320=21.65 So the above table can be rewritten as 
 Size (xi) Frequency (fi)fixidi=|xix¯|fidi2061201.659.90214840.652.602251100.351.75231231.351.35244962.359.40 Total 204336.3525.00

Hence, mean deviation becomes= Σfi(xixi¯)Σfi=2520=1.25 

Question:2

Find the mean deviation about the median of the following distribution:

 Marks obtained 1011121415 No. of students 23834

Answer:

The data distribution is given in the question.

We have to find the mean deviation about the median of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

 Marks obtained (xi) Number of students (fi) Cumulative frequency 10221135128131431615420 Total 20

Now here N=20 which is even.Here,median\(M=12[(N2)thobservation+(N2+1)th observation]  M=12[(202)thobservation+(202+1)th observation ]   M=12 [10thobservation+11thobservation ]   Both these observations lie in cumulative frequency 13, for which corresponding observation is 12 M=12[12+12]=12~ So the above updated table is as shown below 

 Marks obtained (xi) Number of  students (fi) Cumulative  frequency di=|xiM|fidi1022241135131281300143162615420312 Total 206.3525

Hence, mean deviation becomes=ΣfixiΣfi=2520=1.25 

Question:3

Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.

Answer:

Set of first n natural numbers when n is an odd number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is odd number.

So, mean is

x=1+2+3+..+nn=n(n+1)2n=n+12 ~~The deviations of numbers from the mean are as shown below,  1n+12, 2n+12,(n1)n12 , nn+12  Or,2(n+1)2 ,4(n+1)2.2(n1)(n+1)2, 2n(n+1)2 


~~~ Or, 1n2,3n2 , 5n2.n32 ,n12     (n1)2, (n3)2,(n5)2,,n52,n32,n12~ So the absolute values of deviation from the mean is |xix |=n12, n32 ,n52,.n52,n32,n12  ~ The sum of absolute values of deviations from the mean, is Σ|xix |=n12+n32+n52+n52+n32+n12      


Σ|xix |=2(1+2+3+.+n52+n32+n12) ~ That is 2 times the sum of n12terms,~so it can be written as Σ|xix |=2(n12(n12+1)2)=n(n12)(n+12)=(n214) ~~ Therefore, mean deviation about the mean is Σ|xix |n=(n214)n=(n214n)  

Question:4

Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.

Answer:

Set of first n natural numbers when n is an even number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is an even number.

So, mean is

x=1+2+3+..+nn=n(n+1)2n=n+12 ~~The deviations of numbers from the mean are as shown below,  1n+12, 2n+12,(n1)n12 , nn+12  Or,2(n+1)2 ,4(n+1)2.2(n1)(n+1)2, 2n(n+1)2 
~~~ Or, 1n2,3n2 , 5n2.n32 ,n12     (n1)2, (n3)2,(n5)2,,n52,n32,n12~ So the absolute values of deviation from the mean is |xix |=n12, n32 ,n52,.n52,n32,n12  ~ The sum of absolute values of deviations from the mean, is Σ|xix |=n12+n32+n52+n52+n32+n12      Σ|xix |=(12+32+.+n52+n32+n12) (n2)


We know that sum of first n natural numbers =\(n2\)\vspace\baselineskip Therefore, mean deviation about the mean is Σ|xix |n=(12+32+..+n12)(n2)n =(n2)2n=n24n=n4

Question:5

Find the standard deviation of the first n natural numbers.

Answer:

\(Set~of first n natural numbers is given \) We have to find the standard deviation  We can write in table the first n natutal numbers as 
xi12345..nxi21491625.n2

The sums are Σxi=1+2+3++n=n(n+1)2      Σxi2=12+22+32++n2=n(n+1)(2n+1)6    Therefore,thestandarddeviationcanbewrittenas,σ=Σxi2n(Σxin)2    ,σ=n(n+1)(2n+1)6n(n(n+1)2n)2=2n2+n+2n+16n2+2n+14 =4n2+6n+23n26n312=n2112

Question:6

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds.
Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have
\\ \sum_{i=1}^{15}{x_{i}}=279\: \: and\: \: \sum_{i=1}^{15} \mathrm{x}_{i}=5524.Calculate the standard derivation based on all 40 observations.

Answer:

Number ofobservations=25,mean=18.2seconds,standarddeviation=3.25seconds. There are another set of 15 observations x1,x2,..,x15~ also in seconds is i=115xi=279 and i=115xi2=5524  To find that the standard derivation based on all 40 observations  As per the given criteria in the first set Number of observations , n1=25


Mean=18.2 And standard deviation σ1=3.25  And in second set number of observations n2=15~ For the first set we have x1=18.2=Σxi25      Σxi=2518.2=455 
 Therefore the standard deviation becomes σ12=Σxi225(18.2)2  Substituting the values, we get (3.25)2=Σxi225331.24   10.5625+331.24=Σxi225=341.8025  Σxi2=25341.8025=8545.06For~the combined standard deviation of the 40 observation, n=40 and Σxi2=8545.06+5524=14069.69   Σxi=455+279=734Therefore the standard deviation can be written as,σ=Σxi2n(Σxin)2 
~~Substituting~the values, we get Therefore the standard deviation can be written as σ=14069.6940(73440)2=351.7265336.7225=15.004=3.87

Question:7

The mean and standard deviation of a set of n1 observations are X1¯ and s1, respectively while the mean and standard deviation of another set of n2 observations are X2¯ and s2, respectively. Show that the standard deviation of the combined set of (n1+n2) observations is given by
SD=n1(s1)2+n2(s2)2n1+n2+n1n2(x1x2)2(n1+n2)2

Answer:

It is given that mean and Standard deviation of a set of n1 observations are x1 and σ1 respectively while the mean and standard deviation of naother set of n2observations are x2 and σ2respectively\]x1=1n1i=1n1xix2=1n2j=1n2yjx=1(n1+n2)[i=1n1xi+j=1n2yj]=n1x1+n2x2n1+n2σ12=1n1i=1n1(xix)2


σ22=1n2j=1n2(yjx)2σ2=σ12+σ22=1n1+n2[i=1n1(xix)2+1n2j=1n2(yjx)2]i=1n1(xix)2=i=1n1(xixj+xjx)2=i=1n1(xixj)2+n1(xjx)2+2(xjx)i=1n1(xixj)i=1n1(xixj)=0


i=1n1(xix)2=n1s12+n1(x1x)2d1=x1x=x1n1x1+n2x2n1+n2=n2(x1x2)n1+n2i=1n1(xix)2=n1s12+n1n22(x1x2)2(n1+n2)2j=1n2(yjx)2=j=1n2(yjxi+xix)2
=j=1n2(yjxi)2+n2(xix)2+2(xix)j=1n2(yjxi)j=1n2(yjxi)=0j=1n2(yjx)2=n2s22+n2(x2x)2d2=x2x=x2n1x1+n2x2n1+n2=n1(x2x1)n1+n2
j=1n2(xix)2=n2s22+n12n2(x2x1)2(n1+n2)2σ2=σ12+σ22=1n1+n2[i=1n1(xix)2+j=1n2(yjx)2]=1n1+n2[n1s12+n1n22(x1x2)2(n1+n2)2+n2s22+n12n2(x2x1)2(n1+n2)2]=1n1+n2[n1s12+n2s22+n1n2(x1x2)2(n1+n2)2(n1+n2)]

=n1s12+n2s22n1+n2+n1n2(x1x2)2(n1+n2)2σ=n1s12+n2s22n1+n2+n1n2(x1x2)2(n1+n2)2

Question:8

Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

Answer:

Given that two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. We have to show that the standard deviation of the set obtained by combining the given two sets  As per given criteria, for first set Number of observations , n1=20 


Standard deviation s1=5 And mean x1=17~ For second set Number of observations, n2=20 Standard deviation s2=5 and mean x2=22We know the standard deviation for combined two series isσ=n1s12+n2s22n1+n2+(n1n2)(x1x2)2(n1+n2)2  Substituting the corresponding values we get σ=20(5)2+20(5)220+20+(2020)(1722)2(20+20)2=100040+100001600=25+254=100+254=1254=5.59

Question:9

The frequency distribution:
xA2A3A4A5A6Af211111
where A is a positive integer, has a variance of 160. Determine the value of A.

Answer:

A frequency distribution table is given where variance =160. We have to find the value of A, where A is a positive number. Let us make a table from the given data and fill out the other columns after calculation
 Size (xi) Frequency (fi)fixifixi2 A 22A2A22A12A4A23A13A9A24A14A16A25A15A25A26A16A36A2 Total N=722A92A2

And we know that variance is  σ2=Σfixi2n(Σfixin)2 


 Substituting the values from the above table and also given that variance=160we get 160=92A27(22A7)2=(92A27484A249)=792A2484A249=160A249     160=160A249   A2=49A=7 Hence, the value of A is 7

Question:10

For the frequency distribution:
x234567f491614116
Find the standard distribution.

Answer:

A frequency distribution table is given and we have to find the standard deviation

Let us make a table from the given data and fill out the other columns after calculation

 Size (xi) Frequency (fi)fixifixi2248163927814166425651470350611663967642294 Total N=602771393


 And we know that standard deviation is σ=Σfixi2n(Σfixin)2

Substituting the value from the above table

σ=139360(27760)2=23.2321.34=1.37

Question:11

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
 Marks 012345 Frequency x2xx2(x+1)22xx+1
where x is a positive integer. Determine the mean and standard deviation of the marks.

Answer:

It is given that there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

We have to find the mean and standard deviation of the marks.

It is given that there are 60 students in the class, so

Σfi=60 (x2)+x+x2+(x+1)2+2x+x+1=60 5x1+x2+x2+2x+1=602x2+7x=60On factorising 2x2+7x60=0we get (2x+15)(x4)=02x=15orx=4Given~that x is a positive number, so x can take 4 as the only value .And let assumed mean,a=3Now put x=4 and a=3 in the frequency distribution table and other columns aftercalculations, we get 

 Marks (xi) Frequency (fi)di=xiaifidifidi20x2=42=236181X=428162x2=16116163(x+1)2000=(4+1)2=2542x=24=81485X+1=4+1=521020 Total N=60277=1278

And we know that standard deviation is σ=Σfidi2n(Σfidin)2 


Substituting the values from the above table σ=7860(1260)2=1.3(0.2)2=1.30.04=1.12 Hence, the standard deviation is 1.12  Now mean is x=A+ΣfidiN =3+(1260)=315=145=2.8  Hence, the mean and standard deviation of the marks are 2.8 and 1.12 respectively

Question:12

The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

Answer:

The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. We have to find the overall standard deviation.

As per the given criteria. In the first set of samples number of sample bulbs , n1=60

 Standard deviation s1=8hrs~ Mean life, x1=650 And in second set of samples, number of sample bulbs n2=80 Standard deviation s2=7hrs~ Mean life x2=660 We know the standard deviation for combined two series is σ=n1s12+n2s22n1+n2+(n1n2)(x1x2)2(n1+n2)2Substituting the corresponding values we getσ=60(8)2+80(7)260+80+(6080)(650660)2(20+80)2=3887+120049==391649=8.9


Question:13

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all items and the sum of the squares of the items.

Answer:

Mean and standard deviation of 100 items are 50 and 4 respectively
We have to find the sum of all items and the sum of the squares of the items.
As per the question, number of items n =100
Mean of the given items, x=50 But we know,x=Σxin  Substituting the corresponding values we get50=Σxi100     Σxi=50100=5000Hence,~the sum of all the 100 items=5000Also given that the standard deviation of the 100 items is 4  σ=4

 But we know that Σxi2n(Σxin)2   Substituting the corresponding values, we get4=Σxi2100(5000100)2~~Now taking square on both sides, we get 42=Σxi2100(50)2   
 16=Σxi21002500   16+2500=Σxi2100    Σxi2100=2516 Σxi2=2516100=251600

Question:14

If for a distribution $$ \sum (x-5)=3, \sum (x-5)\textsuperscript2=43 and the total number of item is 18, find the mean and standard deviation.

Answer:

It is given that for a distribution Σ(x5)=3, Σ(x5)2=43 and the total number of item is 18We have to find the mean and standard deviationAs per given criteria, Number of items, n=18

~ Mean can be written as x=A+Σ(x5)n   Here assumed mean is 5, so substituting the corresponding values in above equation we get x=5+318=9318=5.17 And we know that the standard deviation can be written as,σ=Σ(x5)2n(Σ(x5)n)2 

Substituting the corresponding values we get

σ=4318(318)2=2.39(0.166)2=2.390.027=2.363=1.54

Thus the mean and standard deviation of given items are 5.17 and 1.54 respectively.

Question:15

Find the mean and variance of the frequency distribution given below:
x1x<33x<55x<77x<10f6451

Answer:

Frequency distribution is given. We have to find the mean and variance

Converting the ranges of x to groups the given table can be re written as

X (Class)  Size (xi) Frequency (fi)fixifixi213261224354416645765301807108.518.572.25 Total N=16=66.5340.25

And, we know that variance is  σ2=Σfixi2n(Σfixin)2 


~ Substituting values from above table, we get σ2=340.2516(66.516)2    σ2=21.265(4.16)2 =21.26517.305=3.96  mean X =ΣfixiΣfi=66.516=4.16  Hence, the mean and variance of the given frequency distribution is 4.16 and 3.96 respectively 

Question:16

Calculate the mean deviation about the mean for the following frequency distribution:
 Class interval 044881212161620 Frequency 46852

Answer:

A frequency distribution table is given, and we have to find the mean deviation about the mean
Let us make a table from the given data and fill out the other columns after calculation
 Class  Interval  Mid value (xi) Frequency (fi)fixi0424848663681210880121614570162018236 Total N=25=230
mean X =ΣfixiΣfi=23025=9.2~~ The above column can be re written as

 Class  Interval  Mid value (xi) Frequency (fi)fixidi=∣xix¯fidi042487.228.84866363.219.2812108800.86.41216145701.824.41620182368.817.6 Total N=25=230=96.0
meanDeviation=ΣfidiΣfi=9625=3.84

Question:17

Calculate the mean deviation from the median of the following data:
 Class interval 06612121818242430 Frequency 45362

Answer:

A frequency distribution table is given and we have to find the mean deviation about the median

Let us make a table from the given data and fill out the other columns after calculation

 Class  Interval  Mid value (xi) Frequency (fi) Cumulative  frequency (cff)06344612959121815312182421618243027220 Total N=20

Now here N=20, which is even

Here Median Class=N2=10th term 

This observation lie in the class interval 12-18, so median can be written asM=l+N2cffh Here l=12, cf=9, f=3, h=6 and N=20Substituting these values, the above equation becomes,M=12+202936M=12+10936

=12+163   M=12+2=14

 Class  Interval  Mid value (xi) Frequency (fi)di=|xiM|fidi0634114461295525121815313182421674224302721326 Total N=20=140

meanDeviation=ΣfidiΣfi=14020=7

Question:18

Determine the mean and standard deviation for the following distribution:
 Marks 2345678910111213141516 Frequency 166882230210001

Answer:

A frequency distribution table is given and we have to find the mean and standard deviation
Let us make a table from the given data and fill out the other columns after calculation

 Marks (xi) Frequency (fi)fixi21236184624584068487214821693271000112221211213001400150016116 Total N=40=239
mean X =ΣfixiΣfi=23940=5.9756
 Marks (xi) Frequency (fi)fixidi=xix¯fidifidi2212441636183185446242122458401886848000721412282162489327392710004001122251050121126636130070014008001500900161161010100 Total N=40=239=1=325
And we know that standard deviation is
σ=Σfidi2n(Σfidin)2=32540(140)2=8.125(0.025)2=2.8721

Question:19

The weights of coffee in 70 jars are shown in the following table:
 Weight (in grams)  Frequency 2002011320120227202203182032041020420512052061
Determine variance and standard deviation of the above distribution.

Answer:

A frequency distribution table for the weights of coffee in 70 jars is given and we have to find the variance and standard deviation of the distribution.
Let us make a table from the given data and fill out the other columns after calculation
 Weight (in  grams)  Mid-value (xi) Frequency (fi)fixi200201200.5132606.5201202201.5275440.5202203202.5183645203204203.5102035204205204.51204.5205206205.51205.5 total N=70=14137
mean X =ΣfixiN=1413770=201.9
 Weight (in  grams)  Mid-value (xi) Frequency (fi)di=xix¯fidifidi2200201200.5131.418.225.48201202201.5270.410.84.32202203202.5180.610.86.48203204203.5101.61625.6204205204.512.62.66.76205206205.513.63.612.96 total N=70=4=81.6
And we know that standard deviation is
σ=Σfidi2n(Σfidin)2=81.670(470)2=1.17(0.057)2=1.17=1.08  
Variance = 1.17

Question:20

Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Answer:

The first n terms of an A.P are given whose first term is a and common difference is d. We have to find mean and standard deviation.

The given AP in tabular form is as shown below,

\multicolumn1|c|xi\multicolumn1|c|di=xiadi2a00a+ddd2a+2d2d4d2a+3d3d9d2a+(n1)d(n1)d(n1)2d2
Here we have assumed a as mean ~ Given that AP has n terms. And we know the sum of all terms of AP can be written as 

xi=n2[2a+(n1)d]x=xin=2a+(n1)d2=a+(n1)2ddi=(xia)=d[1+2+3++(n1)]=d(n(n1)2)


di2=(xia)2=d2[12+22+32++(n1)2]=d2(n(n1)(2n1)6)σ=(xia)2 n((xia)n)2=d2(n(n1)(2n1)6)nd2n2(n1)24n2=d2(n1)(2n1)6d2(n1)24=d2(n1)2(2n13n12)=d2(n1)2(4n23n+36)=d2(n1)2(n+16)=dn2112

Question:21

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
 Ravi 25504530704236483560 Hashina 10705020955542604880
Who is more intelligent and who is more consistent?

Answer:

The marks obtained, out of 100, by 2 students Ravi and Hashina in 10 tests are given
We have to find who is more intelligent and who is more consistent
The marks of Ravi taken separately as follows
\multicolumn1|c|xi\multicolumn1|c|di=xi45\multicolumn1|c|di2252040050525450030152257025625423936981483935101006015225 TOTAL =441=9=1699
Here we have assumed 45 as mean
And we know that the standard deviation can be written as,
σ=Σ(xa)2n(Σ(xa)n)2  σ=169910(910)2=169.90.81=169.09 =13Now mean is x=A+ΣfidiN =45(910)=44.1   For Hashina
\multicolumn1|c|xi\multicolumn1|c|di=xi53\multicolumn1|c|di2104318497017289503920331089954217645524421112160749485258027729 TOTAL =530=0=5928
Here as 53010=53  So,53ismean

And we know that the standard deviation can be written as,
σ=Σ(xa)2n(Σ(xa)n)2  σ=592810(010)2=592.8=24.35 We have the mean and S.D of the Hashina and Ravi . For Ravi , C.V=σx 100=1344.1100=29.48
ForHasina,C.V=σx 100=24.3553100 =45.94  Now as CV (of Ravi )<CV of Hasina~Hence, Ravi is more consistent  Mean of Hasina>Mean of Ravi ~ Hence, Hasina is more intelligent

Question:22

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Answer:

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively.
If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27
respectively.
We have to find the correct standard deviation
As per the given criteria, Number of observations, n=100
Mean of the given observations before correction,X =40

But we know X =Σxin  

Substituting the corresponding values, we get 40=Σxi100  


  Σxi=40100=4000 It is given in the question that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively    So  Σxi=40003070+3+27=3930  So the correct mean after correction is X =Σxin=3930100=39.3 

Also given that standard deviation of the 100 observations is 10 before correction σ=10


But~we know that σ=Σxi2n(Σxin)2    Substituting the corresponding values we get 10=Σxi2100(4000100)2  ~Now taking square on both sides, we get 102=Σxi2100(40)2   100=Σxi21001600  100+1600=Σxi2100    Σxi2=170000 It is said that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively. 


So,~correction is Σxi2=170000(30)2(70)2+(3)3+(27)2     Σxi2=1700009004900+9+729=164938~~So, the correct standard deviation after correction is σ=164938100(3930100)2=1649.38(39.3)2=104.89=10.24

Question:23

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Answer:

If at the time of calculation of mean and variance of 10 readings, a student wrongly used the reading 52 instead of the correct reading 25. The mean and variance were obtained as 45 and 16 respectively. We have to find the correct mean and the variance.  As per the given criteria, Number of observations,n=10Mean of the given observations before correction,X =45~ But we know, X =Σxin  
  Substituting the corresponding values, we get 45=Σxi10    Σxi=4510=450 It~is~given in the question that one reading 25 was taken wrongly as 52 So  Σxi=45052+25=423  So the correct mean after correction is X =Σxin=42310=42.3  
~Also given that variance of the 10 observations is 16 before correction σ2=16 But~we know that σ2=Σxi2n(Σxin)2   Substituting the corresponding values we get16=Σxi210(45)2   Now taking square on both sides, we get 16=Σxi2102025   16+2025=Σxi210    Σxi2=20410 
It is said that one reading 25 was wrongly taken as 52. So,~correction is Σxi2=20410(52)2+(25)2  =204102704+625=18331~~So, the variance after correction is σ2=1833110(42310)2=1833.1(42.3)2=1833.11789.29 =43.81

Question:24

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
A. 2
B. 2.57
C. 3
D. 3.75

Answer:

Given data is 3,10, 10, 4, 7 ,10 ,5 . They are total 7.

Here mean, X =3+10+10+4+7+10+57=497=7 This can be written in table form as,

 Data (xi)\multicolumn1|c|di=|xix¯|34103103437010352 Total =18

Hence Mean Deviation becomes,M.D=Σdi7=187=2.57

Question:25

Mean deviation for n observations x1, x2, ..., xn from their mean X¯ is given by

\\A. \sum _{i=1}^{n}\left(x_{1}-\bar{x}\right)\\\\ B.\frac{1}{n} \sum _{i=1}^{n} \left |\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}} \right |\\\\C. \sum _{i=1}^{n} {\left(x_{i}-\bar{x}\right)^{2}}\\\\D \frac{1}{n}\sum _{i=1}^{n}(x-\bar{x})^{2}

Answer:

We know for n observations x1, x2,.,xn having X is given by M.D=Σdin  But we know that di=|xix|  So,~mean~deviation becomes M.D=Σ|xix|n  OrM.D=1ni=0n|xix |

Question:26

When tested, the lives (in hours) of 5 bulbs were noted as follows:
1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is

A. 178
B. 179
C. 220
D. 356

Answer:

Given~that the lives of 5 bulbs in hours is 1357, 1090, 1666, 1494 , 1623  Here mean X=1357+1090+1666+1494+16235=72307=1446  This can be written in table form as, 
 Lives (in hours )xi\multicolumn1|c|di=xix¯135789109035616662201494481623177 Total =890
Hence, mean deviation becomes M.D=Σdin=8905=178
M.D=Σdin=8905=178

Question:27

Following are the marks obtained by 9 students in a mathematics test:
50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is:

A. 9
B. 10.5
C. 12.67
D. 14.76

Answer:

Given that the marks obtained by 9 students in a mathematics test are 50, 69 , 20 ,33, 53, 39 ,40, 65, 59 As number of students =9 which is odd
So median will be9+12=5th term ~ Arranging these in ascending order, we get 20,33,39,40,50,53,59,65,69 ~ So the 5th term after arranging is 50 ~ So, median is 50 This~can~be written in table form as, 
 Marks (xi)\multicolumn1|c|di=∣xiMedian203033173911401050053359965156919 Total =114
Hence Mean deviation becomes M.D=Σdin=1149=12.67

Question:28

The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is

A.527
B.527
C.6
D.6

Answer:

Given data can be written in table form as,
 Data (xi)xi2636525981131691214486410100 Total =63619
But~we know that standard deviation can be written as σ=Σxi2n(Σxin)2=6197(637)2   σ=6197(9)2=619781 =527

Question:29

Let x1, x2, ..., xn be n observations and X¯ be their arithmetic mean. The formula for the standard deviation is given by

A.(xix)2B.(xix¯)2nC.(XiX)2nDxi2n+x¯2

Answer:

.We know standard deviation for x1, x2,..,xn~ observations can be written as σ=1n  i=0n(xix)2 

Question:30

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
A. 50000
B. 250000
C. 252500
D. 255000

Answer:

As per the question

Number of observation, n =100

Mean of the given observations, x=50 But~we know, X =Σxin Substituting the corresponding values, we get 50=Σxi100  Σxi=50100=5000It is also given that the standard deviation of the 100 observations is 5 σ=5σ=Σxi2n(Σxin)2Substituting the corresponding values, we get 5=Σxi2100(5000100)2 


 Now taking square on both sides we get 52=Σxi100(50)2 25=Σxi1002500 25+2500=Σxi2100      Σxi2=2525100 Σxi2=252500

Question:31

Let a, b, c, d, e be the observations with mean m and standard deviation s.

The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

A. s
B. ks
C. s + k
D. sk

Answer:

Given observations are a, b, c, d, e

So, the mean of the 5 observations is

m=a+b+c+d+e5=xi5xi=5m=a+b+c+d+e

And the standard deviation of the 5 observations is

σ=(xi)2 5(xi5)2=(xi)2 5m2

Now we will find the mean and standard deviation of the observations a+k , b+k, c+k, d+k, e+k we get

So, the mean of these 5 observations is

m1=(a+k)+(b+k)+(c+k)+(d+k)+(e+k)5=a+b+c+d+e5+k=m+kσ1=(xi+k)2 5((xi+k)5)2=(xi2+k2+2kxi)5(m+k)2=(xi2)5+5k25+2k(xi)5(m+k)2=(xi2)5+k2+2mkm2k22mk=(xi2)5m2σ1=σ

Question:32

Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s.
The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is

A. k + s
B.sk
C. ks
D. s

Answer:

Given observations are x1, x2, x3, x4, x5    So the mean of the 5 observations ism=x1+x2+x3+x4+x55=xi5xi=5m=x1+x2+x3+x4+x5σ=(xi)2 5(xi5)2=(xi)2 5m2m1=kx1+kx2+kx3+kx4+kx55=k(x1+x2+x3+x4+x5)5=km
σ1=(kxi)2 5(kxi5)2=k2(xi)2 5(m1)2=k2(xi)2 5k2m2=k2((xi)2 5m2)=k(xi)2 5m2σ1=kσ  Hence, the standard deviation of new set of observations kx1, kx2, kx3, kx4, kx5  is  kσ

Question:33

Let x\textsubscript1,x\textsubscript2,...x\textsubscriptn be n observations. w\textsubscripti=lx\textsubscripti+kfori=1,2,...n, where l and k are constants. If the mean of x\textsubscripti's is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviation of wi’s is 15, the values of l and k should be
A. l = 1.25, k = – 5
B. l = – 1.25, k = 5
C. l = 2.5, k = – 5
D. l = 2.5, k = 5

Answer:

Given x1, x2..xn be n observations And mean of these n observations, x=48 And their standard deviation SDx=12 Another series of n observations is given such that wi=lxi+k  fori=1,2,.nwhere L and k are constants And mean of these n observations, w=55 And their standard deviationSDw=15 Applying the given condition for mean we get wi=lxi+k Substituting~the corresponding given values of means, we get 55=l(48)+k(i)
 We know that if standard deviation of x series is s, then standard deviation of kx series is ks  So standard deviation of x1, x2,.xnis SDx~ And hence the SD of lx1, lx2..lxn is lSDx Similarly, If standard deviation of x series is s, then standard deviation of k+x series is s, So S.D of lx1, lx2.lxnis lSDxAnd hence the SD of lx1+k,lx2+k..lxn+kislSDx  So applying the given condition for standard deviation we get SDw=lSDx Substituting the given values we get 15=l(12) l=1512=1.25 Now substituting the value of l in equation (i), we get55=(1.25)(48)+k 55=60+k k=5

Question:34

Standard deviations for first 10 natural numbers is
A. 5.5
B. 3.87
C. 2.97
D. 2.87

Answer:

We know that the standard deviation of the first n natural numbers is n2112 Now for first 10 natural numbersn=10 substituting this in the equation of standard deviation we get σ=(10)2112=100112=9912=8.25=2.87

Question:35

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is
A. 6.5
B. 2.87
C. 3.87
D. 8.25

Answer:

We know that the standard deviation of the first n natural numbers is n2112 
Now for first 10 natural numbers n=10
substituting this in the equation of standard deviation we get σ=(10)2112=100112=9912=8.25=2.87
Now when 1 is added to each numbers of 1 , 2, 3, 4, 5, 6,7 ,8 ,9, 10 ; we get new series as 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1
Now we know if standard deviation of x series is s, then standard deviation of k+x series is s,
So the S.D of 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1 series is also same as the S.D of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 series.
σ=8.25 
Now for variance we will square on both the sides σ2=8.25

Question:36

Consider the first 10 positive integers. If we multiply each number by –1 and then add 1 to each number, the variance of the numbers so obtained is
A. 8.25
B. 6.5
C. 3.87
D. 2.87

Answer:

First 10 positive integers are 1,2, 3, 4, 5,6 ,7, 8,9, 10 On multiplying each number by
-1 we get-1, -2, -3, -4 , -5, -6, -7, -8, -9 , -10
On adding 1 to each of the number, we get 0, -1, -2, -3, -4, -5, -6, -7, -8, -9
Σxi=0123456789=45 and Σxi2=02+(1)2+(2)2++(9)2  But we know that Σxi2=9(9+1)(2(9)+1)6=285σ=Σxi2n(Σxin)2 Substituting the corresponding values, we get σ=28.5(4.5)2=28.520.25=8.25 Now for variance both the sides are squared σ2=8.25

Question:37

The following information relates to a sample of size 60:

$$ \sum x\textsuperscript2=18000, \sum x=960.
The variance is
A. 6.63
B. 16
C. 22
D. 44

Answer:

We know that S.D can be written as σ=Σxi2n(Σxin)2 But given Σx2=18000,Σx=960, N=60,substituting these corresponding values, we get σ=1800060(96060)2=300(16)2=300256=44 Now for variance we square both the sides σ2=44

Question:38

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is
A. 0
B. 1
C. 1.5
D. 2.5

Answer:

Given coefficient of variation of two distributions are CV1=50andCV2=60 And the arithmetic means are x1=30,x2=25 We know that coefficient of variation can be written as CV=σx 100 Now for first distribution, we have CV1=σ1x1 100Substituting corresponding values, we get 50=σ130100 σ1=15.(i)   Now for second distribution, we have CV2=σ2x2 100 
Substituting corresponding values, we get 60=σ225100 σ2=15.(ii)   So from both the equations we get that the difference of their standard deviation is 0 

Question:39

The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be
A. 81
B. 57
C. 36
D. 25

Answer:

Given that the standard deviation of some temperature data in degree celcius, σc=5We know that C=59(F32) F=9C5+32  Now we know that If S.D of x series is s, then S.D of kx series is ks.  So if SD of some temperature data is σc=5Then SD of some temperature data in9C5 ,9C5σc=595=9 Similarly, If SD of x series is s, then SD of k+x series is s,
 So SD of some temperature data in9C5 σc=9 And thus the SD of some temperature data in9C5+32willbeσF=9C5  σC=9Now for variance, we will square on both sides, we get σF=92=81

Question:40

Fill in the blanks
Coefficient of variation = ...mean100

Answer:

Standard Deviation
We know that the Coefficient of variation can be written as
C.V=σx 100whereσ=standard deviation 

Question:41

Fill in the blanks
If X¯is the mean of n values of x, then i=1n(xix) is always equal to _______.
If a has any value other than X¯ , then i=1n(xix)2 is _________ than (Xia)2

Answer:

Zero, less than Given x is mean of n values, then the sum of all n terms is denoted by i=1nxi,
so difference of both these is laways equal to zero i.e. Extra close brace or missing open brace
And square of the above equation is also equal to zero, so Extra close brace or missing open brace
Now if "a" has the value other than x then  i=1n(xix)2>0 
So, i=1n(xix)2< i=1n(xia)2  So, if a~ has any value other than x, then  i=1n(xix)2~is less than i=1n(xia)2

Question:42

Fill in the blanks:
If the variance of a data is 121, then the standard deviation of the data is _______.

Answer:

11
We know the square root of variance is standard deviation σ2=121
Taking square root on both sides, we get σ=121=11  
So, if the variance of a data is 121, then the Standard deviation of the data is 11.

Question:43

Fill in the blanks
The standard deviation of a data is ___________ of any change in origin but is _____ on the change of scale.

Answer:

Independent, dependent
Change of origin means some value has been added or subtracted in the observation. And we know the S.D does not change if any value is added or subtracted from the observations, So SD is independent of change in origin.
However, Standard deviation is only affected by a change in scale, that is, when some value is multiplied or divided to observations. Hence, the SD of any data is independent of any change in origin but dependent on any change of scale.

Question:44

The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

Answer:

Minimum
The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

Question:45

Fill in the blanks
The mean deviation of the data is _______ when measured from the median.

Answer:

Least
Mean deviation is sum of all deviations of a set of a data about the data’s mean. In addition, it is widely believed that the median is usually between the mean and the mode. So, the mean deviation of the data is least when measured from the median.

Question:46

Fill in the blanks
The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

Answer:

Greater than or equal to
SD is the difference between square of deviation of data about mean and square of mean. In addition, mean deviation is sum of all deviations of a set of data about the data’s mean. Hence, the SD is greater than or equal to the mean deviation taken from the arithmetic mean.

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

· Central Tendency

· Mean

· Median

· Mode

· Measure of Dispersion

· Range

· Quartile deviation

· Mean deviation

· Standard deviation

· Mean deviation for grouped data

· Mean deviation for ungrouped data

· Variance

· Analysis of Frequency Deviation

More about NCERT Exemplar Class 11 Maths Solutions Chapter 15

By referring to NCERT Exemplar Class 11 Maths chapter 15 solutions students can easily gather the various concepts of Statistics. Statistics carries a significant number of marks according to the CBSE paper pattern and the students should concentrate on important concepts like standard deviation and mean deviation, as they are frequently asked in the examination.

The step-wise and straightforward format of the NCERT exemplar solutions for Class 11 Maths chapter 15 makes it easy for the students to understand the most complex concepts of Statistics in a quick and easygoing manner. The pdf is exclusively created by our experts for students who find the concepts of Statistics difficult.

Students can easily score marks in this chapter by thoroughly studying a few topics which are commonly asked in the final examinations. The students will understand the concept in the simplest way possible by going through the NCERT Exemplar Class 11 Maths solutions chapter 15 Statistics.

As most students tend to skip the entire chapter due to doubts in Statistics, our experts have formulated the NCERT Exemplar Class 11 Maths chapter 15 solutions in the most student-friendly manner.

NCERT Solutions for Class 11 Mathematics Chapters

Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

Standard Deviation: Questions on Standard Deviation are frequently asked by the examiners. The students should go through the NCERT Exemplar Class 11 Maths chapter 15 solutions to solve all their doubts regarding standard deviation. The student should just follow the simple steps given in the solution and practice.

Mean Deviation: The students should not ignore solving the question of mean deviation as they are commonly asked in the exam. The students should memorize the formula for mean and standard deviation upon which the problem is based and solve the entire question by referring to the steps given in the Class 11 Maths NCERT Exemplar solutions chapter 15.

Mean, Median, Mode, and Range: These are the fundamental concepts of Statistics and can be asked in the examination for MCQ or short answer type questions.

Check Chapter-Wise NCERT Solutions of Book

NCERT Exemplar Class 11 Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Frequently Asked Questions (FAQs)

1. Which are the important topics one should cover in Statistics?

Standard Deviation, Mean Deviation and Central Tendency are the important topics one should cover in Statistics.

2. Where can we download the Class 11 Maths NCERT Exemplar Solutions Chapter 15?

Students can get PDF by using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function.

3. Are the solutions helpful for board examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 15 are helpful for board examinations.

4. Can these solutions be referred for cracking competitive examinations?

Yes, Class 11 Maths NCERT Exemplar solutions chapter 15 can be referred to when pursuing competitive examinations.   

Articles

Upcoming School Exams

Admit Card Date:30 December,2024 - 26 March,2025

Admit Card Date:08 January,2025 - 25 February,2025

Admit Card Date:20 January,2025 - 17 March,2025

Admit Card Date:25 January,2025 - 03 March,2025

View All School Exams
Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top