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NCERT Solutions for Class 11 Maths Chapter 15 Statistics

NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Edited By Ramraj Saini | Updated on Sep 25, 2023 10:18 PM IST

Statistics Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 15 Statistics are provided here. You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous Class NCERT syllabus. In this NCERT book chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In class 11 maths chapter 15 question answer, you will get questions related to all the above topics. These NCERT solutions are prepared by expert team at Careers360 keeping in my the latest syllabus of CBSE 2023-24, easy and simple language, step by step comprehensive coverage of the concepts.

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This Story also Contains
  1. Statistics Class 11 Questions And Answers
  2. Statistics Class 11 Questions And Answers PDF Free Download
  3. Statistics Class 11 Solutions - Important Formulae
  4. Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Summary And Formulae Of Statistics Maths Chapter 15 class 11
  6. NCERT Solutions For Class 11 Mathematics
  7. Key Features Of Statistics Class 11 NCERT Solutions
  8. NCERT Solutions For Class 11 - Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions  for Class 11 Maths Chapter 15 Statistics
NCERT Solutions for Class 11 Maths Chapter 15 Statistics

This statistics class 11 solutions is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc . As statistics deal with the collected data for specific purposes and in the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data, and make a conclusive decision based on the data. Class 11 maths chapter 15 NCERT solutions will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. Interested students can find all NCERT solutions for class 11 at one place which will help to understand the concepts in an easy way.

Also Read :

Statistics Class 11 Questions And Answers PDF Free Download

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Statistics Class 11 Solutions - Important Formulae

Measure of Dispersion:

Dispersion measures the degree of variation in the values of a variable. It quantifies how scattered observations are around the central value in a distribution.

Range:

Range is the simplest measure of dispersion. It is defined as the difference between the largest and smallest observations in a distribution.

Range of distribution = Largest observation – Smallest observation.

Mean Deviation:

Mean Deviation for Ungrouped Data:

  • For n observations x1, x2, x3, ..., xn, the mean deviation about their mean x¯ is calculated as:

    • MD(\bar x) = (Σ |xᵢ - x¯|) / n

  • The mean deviation about its median M is calculated as:

    • MD(M) = (Σ |xᵢ - M|) / n

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Mean Deviation for Discrete Frequency Distribution:

  • For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the mean deviation is calculated as:

    • MD(\bar x) = (Σ fᵢ |xᵢ - x¯|) / (Σ fᵢ) = (Σ fᵢ |xᵢ - x¯|) / N

Variance:

  • Variance is the average of the squared deviations from the mean x¯. If x, x, …, xₙ are n observations with mean x¯, the variance denoted by σ² is calculated as:

    • σ² = (Σ(xᵢ - x¯)²) / n

Standard Deviation:

  • Standard deviation, denoted as σ, is the square root of the variance σ². If σ² is the variance, then the standard deviation is given by:

    • σ = √(σ²)

  • For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the standard deviation is calculated as:

    • σ = √((Σ fᵢ(xᵢ - x¯)²) / N)

Coefficient of Variation:

  • The coefficient of variation (CV) is used to compare two or more frequency distributions. It is defined as:

    • Coefficient of Variation = (Standard Deviation / Mean) × 100

    • CV = (σ / x¯) × 100

Free download NCERT Solutions for Class 11 Maths Chapter 15 Statistics for CBSE Exam.

Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 15 question answer - Exercise: 15.1

Question:1 . Find the mean deviation about the mean for the data.

4,7,8,9,10,12,13,17

Answer:

Mean ( x ) of the given data:

x=18i=18xi=4+7+8+9+10+12+13+178=10

The respective absolute values of the deviations from mean, |xix| are

6, 3, 2, 1, 0, 2, 3, 7

i=18|xi10|=24

M.D.(x)=1ni=1n|xix|

=248=3

Hence, the mean deviation about the mean is 3.

Question:2. Find the mean deviation about the mean for the data.

38,70,48,40,42,55,63,46,54,44

Answer:

Mean ( x ) of the given data:

x=18i=18xi=38+70+48+40+42+55+63+46+54+4410=50010=50

The respective absolute values of the deviations from mean, |xix| are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

i=18|xi50|=84

M.D.(x)=1ni=1n|xix|

=8410=8.4

Hence, the mean deviation about the mean is 8.4.

Question:3. Find the mean deviation about the median.

13,17,16,14,11,13,10,16,11,18,12,17

Answer:

Number of observations, n = 12, which is even.

Arranging the values in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median (M)

=(122)thobservation+(122+1)thobservation2=13+142=272=13.5

The respective absolute values of the deviations from median, |xiM| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

i=18|xi13.5|=28

M.D.(M)=112i=1n|xiM|

=2812=2.33

Hence, the mean deviation about the median is 2.33.

Question:4. Find the mean deviation about the median.

36,72,46,42,60,45,53,46,51,49

Answer:

Number of observations, n = 10, which is even.

Arranging the values in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Now, Median (M)

=(102)thobservation+(102+1)thobservation2=46+492=952=47.5

The respective absolute values of the deviations from median, |xiM| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

i=18|xi47.5|=70

M.D.(M)=110i=1n|xiM|

=7010=7

Hence, the mean deviation about the median is 7.

Question:5 Find the mean deviation about the mean.

xi510152025fi74635

Answer:

xi
fi
fixi
|xix|
fi|xix|
5
7
35
9
63
10
4
40
4
16
15
6
90
1
6
20
3
60
6
18
25
5
125
11
55

fi
= 25
fixi
= 350

fi|xix|
=158

N=i=15fi=25;i=15fixi=350

x=1Ni=1nfixi=35012=14

Now, we calculate the absolute values of the deviations from mean, |xix| and

fi|xix| = 158

M.D.(x)=125i=1nfi|xix|

=15825=6.32

Hence, the mean deviation about the mean is 6.32

Question:6. Find the mean deviation about the mean.

xi1030507090fi42428168

Answer:

xi
fi
fixi
|xix|
fi|xix|
10
4
40
40
160
30
24
720
20
480
50
28
1400
0
0
70
16
1120
20
320
90
8
720
40
320

fi
= 80
fixi
= 4000

fi|xix|
=1280

N=i=15fi=80;i=15fixi=4000

x=1Ni=1nfixi=400080=50

Now, we calculate the absolute values of the deviations from mean, |xix| and

fi|xix| = 1280

M.D.(x)=180i=15fi|xix|

=128080=16

Hence, the mean deviation about the mean is 16

Question:7. Find the mean deviation about the median.

xi 5 7 9 10 12 15

fi 8 6 2 2 2 6

Answer:

xi
fi
c.f.
|xiM|
fi|xiM|
5
8
8
2
16
7
6
14
0
0
9
2
16
2
4
10
2
18
3
6
12
2
20
5
10
15
6
26
8
48

Now, N = 26 which is even.

Median is the mean of 13th and 14th observations.

Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median, M =13thobservation+14thobservation2=7+72=142=7

Now, we calculate the absolute values of the deviations from median, |xiM| and

fi|xiM| = 84

M.D.(M)=126i=16|xiM|

=8426=3.23

Hence, the mean deviation about the median is 3.23

Question:8 Find the mean deviation about the median.

xi 15 21 27 30 35

fi 3 5 6 7 8

Answer:

xi
fi
c.f.
|xiM|
fi|xiM|
15
3
3
13.5
40.5
21
5
8
7.5
37.5
27
6
14
1.5
9
30
7
21
1.5
10.5
35
8
29
6.5
52

Now, N = 30, which is even.

Median is the mean of 15th and 16th observations.

Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore, Median, M =15thobservation+16thobservation2=30+302=30

Now, we calculate the absolute values of the deviations from median, |xiM| and

fi|xiM| = 149.5

M.D.(M)=129i=15|xiM|

=149.529=5.1

Hence, the mean deviation about the median is 5.1

Question:9. Find the mean deviation about the mean.

Income per day in Rs
Number of persons
4
8
9
10
7
5
4
3






Answer:

Income
per day
Number of
Persons fi
Mid
Points xi
fixi
|xix|
fi|xix|
0 -100
4
50
200
308
1232
100 -200
8
150
1200
208
1664
200-300
9
250
2250
108
972
300-400
10
350
3500
8
80
400-500
7
450
3150
92
644
500-600
5
550
2750
192
960
600-700
4
650
2600
292
1168
700-800
3
750
2250
392
1176

fi
=50

fixi
=17900

fi|xix|
=7896


N=i=18fi=50;i=18fixi=17900

x=1Ni=18fixi=1790050=358

Now, we calculate the absolute values of the deviations from mean, |xix| and

fi|xix| = 7896

M.D.(x)=150i=18fi|xix|

=789650=157.92

Hence, the mean deviation about the mean is 157.92

Question:10. Find the mean deviation about the mean.

Height in cms
Number of persond
9
13
26
30
12
10

Answer:

Height
in cms
Number of
Persons fi
Mid
Points xi
fixi
|xix|
fi|xix|
95 -105
9
100
900
25.3
227.7
105 -115
13
110
1430
15.3
198.9
115-125
26
120
3120
5.3
137.8
125-135
30
130
3900
4.7
141
135-145
12
140
1680
14.7
176.4
145-155
10
150
1500
24.7
247

fi
=100

fixi
=12530

fi|xix|
=1128.8


N=i=16fi=100;i=16fixi=12530

x=1Ni=16fixi=12530100=125.3

Now, we calculate the absolute values of the deviations from mean, |xix| and

fi|xix| = 1128.8

M.D.(x)=1100i=16fi|xix|

=1128.8100=11.29

Hence, the mean deviation about the mean is 11.29

Question:11. Find the mean deviation about median for the following data :

Marks
Number of girls
6
8
14
16
4
2

Answer:

Marks
Number of
Girls fi
Cumulative
Frequency c.f.
Mid
Points xi
|xiM|
fi|xiM|
0-10
6
6
5
22.85
137.1
10-20
8
14
15
12.85
102.8
20-30
14
28
25
2.85
39.9
30-40
16
44
35
7.15
114.4
40-50
4
48
45
17.15
68.6
50-60
2
50
55
27.15
54.3





fi|xiM|
=517.1

Now, N = 50, which is even.

The class interval containing (N2)th or 25th item is 20-30. Therefore, 20-30 is the median class.

We know,

Median =l+N2Cf×h

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median =20+251414×10=20+7.85=27.85

Now, we calculate the absolute values of the deviations from median, |xiM| and

fi|xiM| = 517.1

M.D.(M)=150i=16fi|xiM|

=517.150=10.34

Hence, the mean deviation about the median is 10.34

Question:12 Calculate the mean deviation about median age for the age distribution of 100 persons given below:

[ Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

Answer:

Age
(in years)
Number
fi
Cumulative
Frequency c.f.
Mid
Points xi
|xiM|
fi|xiM|
15.5-20.5
5
5
18
20
100
20.5-25.5
6
11
23
15
90
25.5-30.5
12
23
28
10
120
30.5-35.5
14
37
33
5
70
35.5-40.5
26
63
38
0
0
40.5-45.5
12
75
43
5
60
45.5-50.5
16
91
48
10
160
50.5-55.5
9
100
53
15
135





fi|xiM|
=735

Now, N = 100, which is even.

The class interval containing (N2)th or 50th item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

We know,

Median =l+N2Cf×h

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median =35.5+503726×5=35.5+2.5=38

Now, we calculate the absolute values of the deviations from median, |xiM| and

fi|xiM| = 735

M.D.(M)=1100i=18fi|xiM|

=735100=7.35

Hence, the mean deviation about the median is 7.35

Class 11 maths chapter 15 ncert solutions - Exercise: 15.2

Question:1. Find the mean and variance for each of the data.

6,7,10,12,13,4,8,12

Answer:

Mean ( x ) of the given data:

x=18i=18xi=6+7+10+12+13+4+8+128=728=9

The respective values of the deviations from mean, (xix) are

-3, -2, 1 3 4 -5 -1 3

i=18(xi10)2=74

σ2=1ni=1n(xix)2

18i=18(xix)2=748=9.25

Hence, Mean = 9 and Variance = 9.25

Question:2. Find the mean and variance for each of the data.

First n natural numbers.

Answer:

Mean ( x ) of first n natural numbers:

x=1ni=1nxi=n(n+1)2n=n+12

We know, Variance σ2=1ni=1n(xix)2

σ2=1ni=1n(xin+12)2

We know that (ab)2=a22ab+b2

nσ2=i=1nxi2+i=1n(n+12)22i=1nxin+12=n(n+1)(2n+1)6+(n+1)24×n2.(n+1)2.n(n+1)2

σ2=(n+1)(2n+1)6+(n+1)24(n+1)22

=(n+1)(2n+1)6(n+1)24=(n+1)[4n+23n312]=(n+1).(n1)12=n2112

Hence, Mean = n+12 and Variance = n2112

Question:3. Find the mean and variance for each of the data

First 10 multiples of 3

Answer:

First 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean ( x ) of the above values:

x=110i=110xi=3+6+9+12+15+18+21+24+27+3010=3.10(10+1)210=16.5

The respective values of the deviations from mean, (xix) are

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

i=110(xi16.5)2=742.5

σ2=1ni=1n(xix)2

110i=110(xix)2=742.510=74.25

Hence, Mean = 16.5 and Variance = 74.25

Question:4. Find the mean and variance for each of the data.

xi
6
10
14
18
24
28
30
fi
2
4
7
12
8
4
3


Answer:

xi
fi
fixi
(xix)
(xix)2
fi(xix)2
6
2
12
-13
169
338
10
4
40
-9
81
324
14
7
98
-5
25
175
18
12
216
-1
1
12
24
8
192
5
25
200
28
4
112
9
81
324
30
3
90
13
169
363

fi
= 40
fixi
= 760


fi(xix)2
=1736

N=i=17fi=40;i=17fixi=760

x=1Ni=1nfixi=76040=19

We know, Variance, σ2=1Ni=1n(xix)2

σ2=173640=43.4

Hence, Mean = 19 and Variance = 43.4

Question:5. Find the mean and variance for each of the data.

xi
92
93
97
98
102
104
109
fi
3
2
3
2
6
3
3


Answer:

xi
fi
fixi
(xix)
(xix)2
fi(xix)2
92
3
276
-8
64
192
93
2
186
-7
49
98
97
3
291
-3
9
27
98
2
196
-2
4
8
102
6
612
2
4
24
104
3
312
4
16
48
109
3
327
9
81
243

fi
= 22
fixi
= 2200


fi(xix)2
=640

N=i=17fi=22;i=17fixi=2200

x=1Ni=1nfixi=220022=100

We know, Variance, σ2=1Ni=1n(xix)2

σ2=64022=29.09

Hence, Mean = 100 and Variance = 29.09

Question:6 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


Answer:

Let the assumed mean, A = 64 and h = 1

xi
fi
yi=xiAh
yi2
fiyi
fiyi2
60
2
-4
16
-8
32
61
1
-3
9
-3
9
62
12
-2
4
-24
48
63
29
-1
1
-29
29
64
25
0
0
0
0
65
12
1
1
12
12
66
10
2
4
20
40
67
4
3
9
12
36
68
5
4
16
20
80

fi
=100


fiyi
= 0
fiyi2
=286

N=i=19fi=100;i=19fiyi=0

Mean,

x=A+1Ni=1nfiyi×h=64+0100=64

We know, Variance, σ2=1N2[Nfiyi2(fiyi)2]×h2

σ2=1(100)2[100(286)(0)2]=2860010000=2.86

We know, Standard Deviation = σ=Variance

σ=2.86=1.691

Hence, Mean = 64 and Standard Deviation = 1.691

Question:​​​​​​​7. Find the mean and variance for the following frequency distributions.

Classes
0-30
30-60
60-90
90-120
120-150
150-180
180-210
Frequencies
2
3
5
10
3
5
2


Answer:

Classes
Frequency
fi
Mid point
xi
fixi
(xix)
(xix)2
fi(xix)2
0-30
2
15
30
-92
8464
16928
30-60
3
45
135
-62
3844
11532
60-90
5
75
375
-32
1024
5120
90-120
10
105
1050
2
4
40
120-150
3
135
405
28
784
2352
150-180
5
165
825
58
3364
16820
180-210
2
195
390
88
7744
15488

fi = N
= 30

fixi
= 3210


fi(xix)2
=68280


x=1Ni=1nfixi=321030=107

We know, Variance, σ2=1Ni=1n(xix)2

σ2=6828030=2276

Hence, Mean = 107 and Variance = 2276

Question:​​​​​​​8. Find the mean and variance for the following frequency distributions.

Classes
0-10
10-20
20-30
30-40
40-50
Frequencies
5
8
15
16
6


Answer:

Classes
Frequency
fi
Mid-point
xi
fixi
(xix)
(xix)2
fi(xix)2
0-10
5
5
25
-22
484
2420
10-20
8
15
120
-12
144
1152
20-30
15
25
375
-2
4
60
30-40
16
35
560
8
64
1024
40-50
6
45
270
18
324
1944

fi = N
= 50

fixi
= 1350


fi(xix)2
=6600


x=1Ni=1nfixi=135050=27

We know, Variance, σ2=1Ni=1n(xix)2

σ2=660050=132

Hence, Mean = 27 and Variance = 132

Question:9. 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 students
3
4
7
7
15
9
6
6
3






Answer:

Let the assumed mean, A = 92.5 and h = 5

Height
in cms
Frequency
fi
Midpoint
xi
\dpi100yi=xiAh
yi2
fiyi
fiyi2
70-75
3
72.5
-4
16
-12
48
75-80
4
77.5
-3
9
-12
36
80-85
7
82.5
-2
4
-14
28
85-90
7
87.5
-1
1
-7
7
90-95
15
92.5
0
0
0
0
95-100
9
97.5
1
1
9
9
100-105
6
102.5
2
4
12
24
105-110
6
107.5
3
9
18
54
110-115
3
112.5
4
16
12
48

fi =N = 60



fiyi
= 6
fiyi2
=254

Mean,

y=A+1Ni=1nfiyi×h=92.5+660×5=93

We know, Variance, σ2=1N2[Nfiyi2(fiyi)2]×h2

σ2=1(60)2[60(254)(6)2]=1(60)2[1524036]=15204144=105.583

We know, Standard Deviation = σ=Variance

σ=105.583=10.275

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Question:10. 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.
[ Hint First make the data continuous by making the classes as 32.536.5,36.540.4,40.544.5,44.548.5,48.552.5 and then proceed.]

Answer:

Let the assumed mean, A = 92.5 and h = 5

Diameters
No. of
circles fi
Midpoint
xi
\dpi100yi=xiAh
yi2
fiyi
fiyi2
32.5-36.5
15
34.5
-2
4
-30
60
36.5-40.5
17
38.5
-1
1
-17
17
40.5-44.5
21
42.5
0
0
0
0
44.5-48.5
22
46.5
1
1
22
22
48.5-52.5
25
50.5
2
4
50
100

fi =N = 100



fiyi
= 25
fiyi2
=199

Mean,

x=A+1Ni=1nfiyi×h=42.5+25100×4=43.5

We know, Variance, σ2=1N2[Nfiyi2(fiyi)2]×h2

\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84

We know, Standard Deviation = σ=Variance

σ=30.84=5.553

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

Statistics class 11 NCERT solutions - Exercise: 15.3

Question:1. 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


Answer:

The group having a higher coefficient of variation will be more variable.

Let the assumed mean, A = 45 and h = 10

For Group A

Marks
Group A
fi
Midpoint
xi
\dpi100yi=xiAh
=xi4510
yi2
fiyi
fiyi2
10-20
9
15
-3
9
-27
81
20-30
17
25
-2
4
-34
68
30-40
32
35
-1
1
-32
32
40-50
33
45
0
0
0
0
50-60
40
55
1
1
40
40
60-70
10
65
2
4
20
40
70-80
9
75
3
9
27
81

fi =N = 150



fiyi
= -6
fiyi2
=342

Mean,

x=A+1Ni=1nfiyi×h=45+6150×10=44.6

We know, Variance, σ2=1N2[Nfiyi2(fiyi)2]×h2

σ2=1(150)2[150(342)(6)2]×102=1152[51264]=227.84

We know, Standard Deviation = σ=Variance

σ=227.84=15.09

Coefficient of variation = σx×100

C.V.(A) = 15.0944.6×100=33.83

Similarly,

For Group B

Marks
Group A
fi
Midpoint
xi
\dpi100yi=xiAh
=xi4510
yi2
fiyi
fiyi2
10-20
10
15
-3
9
-30
90
20-30
20
25
-2
4
-40
80
30-40
30
35
-1
1
-30
30
40-50
25
45
0
0
0
0
50-60
43
55
1
1
43
43
60-70
15
65
2
4
30
60
70-80
7
75
3
9
21
72

fi =N = 150



fiyi
= -6
fiyi2
=375

Mean,

x=A+1Ni=1nfiyi×h=45+6150×10=44.6

We know, Variance, σ2=1N2[Nfiyi2(fiyi)2]×h2

σ2=1(150)2[150(375)(6)2]×102=1152[56214]=249.84

We know, Standard Deviation = σ=Variance

σ=249.84=15.80

Coefficient of variation = σx×100

C.V.(B) = 15.8044.6×100=35.42

Since C.V.(B) > C.V.(A)

Therefore, Group B is more variable.

Question:2 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


Answer:

X( xi )
Y( yi )
xi2
yi2
35
108
1225
11664
54
107
2916
11449
52
105
2704
11025
53
105
2809
11025
56
106
8136
11236
58
107
3364
11449
52
104
2704
10816
50
103
2500
10609
51
104
2601
10816
49
101
2401
10201
=510
= 1050
=26360
=110290

For X,

Mean , x=1ni=1nxi=51010=51

Variance, σ2=1n2[nxi2(xi)2]

σ2=1(10)2[10(26360)(510)2]=1100.[263600260100]=35

We know, Standard Deviation = σ=Variance=35=5.91

C.V.(X) = σx×100=5.9151×100=11.58

Similarly, For Y,

Mean , y=1ni=1nyi=105010=105

Variance, σ2=1n2[nyi2(yi)2]

σ2=1(10)2[10(110290)(1050)2]=1100.[11029001102500]=4

We know, Standard Deviation = σ=Variance=4=2

C.V.(Y) = σy×100=2105×100=1.904

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.

Question:3(i) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


No. of wage earners
586
648
Mean of monthly wages
Variance of the distribution of wages
100
121

Which firm A or B pays larger amount as monthly wages?

Answer:

Given, Mean of monthly wages of firm A = 5253

Number of wage earners = 586

Total amount paid = 586 x 5253 = 30,78,258

Again, Mean of monthly wages of firm B = 5253

Number of wage earners = 648

Total amount paid = 648 x 5253 = 34,03,944

Hence firm B pays larger amount as monthly wages.

Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


No. of wage earners
586
648
Mean of monthly wages
Variance of the distribution of wages
100
121

Which firm, A or B, shows greater variability in individual wages?

Answer:

Given, Variance of firm A = 100

Standard Deviation = σA=Variance=100=10

Again, Variance of firm B = 121

Standard Deviation = σB=Variance=121=11

Since σB>σA , firm B has greater variability in individual wages.

Question:4 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 matches
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?

Answer:

No. of goals
scored xi
Frequency
fi
xi2
fixi
fixi2
0
1
0
0
0
1
9
1
9
9
2
7
4
14
28
3
5
9
15
45
4
3
16
12
48

fi =N = 25

fixi
= 50
fixi2
=130

For Team A,

Mean,

x=1Ni=1nfixi=5025=2

We know, Variance, σ2=1N2[Nfixi2(fixi)2]

σ2=1(25)2[25(130)(50)2]=750625=1.2

We know, Standard Deviation = σ=Variance

σ=1.2=1.09

C.V.(A) = σx×100=1.092×100=54.5

For Team B,

Mean = 2

Standard deviation, σ = 1.25

C.V.(B) = σx×100=1.252×100=62.5

Since C.V. of firm B is more than C.V. of A.

Therefore, Team A is more consistent.

Question:5 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=1457.6
Which is more varying, the length or weight?

Answer:

For lenght x,

Mean, x=1ni=1nxi=21250=4.24

We know, Variance, σ2=1n2[nfixi2(fixi)2]

σ2=1(50)2[50(902.8)(212)2]=1962500=0.0784

We know, Standard Deviation = σ=Variance=0.0784=0.28

C.V.(x) = σx×100=0.284.24×100=6.603

For weight y,

Mean,

Mean, y=1ni=1nyi=26150=5.22

We know, Variance, σ2=1n2[nfiyi2(fiyi)2]

σ2=1(50)2[50(1457.6)(261)2]=47592500=1.9036

We know, Standard Deviation = σ=Variance=1.9036=1.37

C.V.(y) = σy×100=1.375.22×100=26.24

Since C.V.(y) > C.V.(x)

Therefore, weight is more varying.

Statistics maths chapter 15 class 11 - Miscellaneous Exercise

Question:1 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.

Answer:

Given,

The mean and variance of 8 observations are 9 and 9.25, respectively

Let the remaining two observations be x and y,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean, X=6+7+10+12+12+13+x+y8=9

60 + x + y = 72

x + y = 12 -(i)

Now, Variance

=1ni=18(xix)2=9.25

9.25=18[(3)2+(2)2+12+32+42+x2+y218(x+y)+2.92]

9.25=18[(3)2+(2)2+12+32+42+x2+y2+18(12)+2.92] (Using (i))

9.25=18[48+x2+y2216+162]=18[x2+y26]

x2+y2=80 -(ii)

Squaring (i), we get

x2+y2+2xy=144 (iii)

(iii) - (ii) :

2xy = 64 (iv)

Now, (ii) - (iv):

x2+y22xy=8064(xy)2=16xy=±4 -(v)

Hence, From (i) and (v):

x – y = 4 x = 8 and y = 4

x – y = -4 x = 4 and y = 8

Therefore, The remaining observations are 4 and 8. (in no order)

Question:2 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.

Answer:

Given,

The mean and variance of 7 observations are 8 and 16, respectively

Let the remaining two observations be x and y,

Observations: 2, 4, 10, 12, 14, x, y

∴ Mean, X=2+4+10+12+14+x+y7=8

42 + x + y = 56

x + y = 14 -(i)

Now, Variance

=1ni=18(xix)2=16

16=17[(6)2+(4)2+22+42+62+x2+y216(x+y)+2.82]

16=17[36+16+4+16+36+x2+y216(14)+2(64)] (Using (i))

16=17[108+x2+y296]=17[x2+y2+12]

x2+y2=11212=100 -(ii)

Squaring (i), we get

x2+y2+2xy=196 (iii)

(iii) - (ii) :

2xy = 96 (iv)

Now, (ii) - (iv):

x2+y22xy=10096(xy)2=4xy=±2 -(v)

Hence, From (i) and (v):

x – y = 2 x = 8 and y = 6

x – y = -2 x = 6 and y = 8

Therefore, The remaining observations are 6 and 8. (in no order)

Question:3 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.

Answer:

Given,

Mean = 8 and Standard deviation = 4

Let the observations be x1,x2,x3,x4,x5 and x6

Mean, x=x1+x2+x3+x4+x5+x66=8

Now, Let yi be the the resulting observations if each observation is multiplied by 3:

yi=3xixi=yi3

New mean, y=y1+y2+y3+y4+y5+y66

=3[x1+x2+x3+x4+x5+x66]=3×8

= 24

We know that,

Standard Deviation = σ=Variance

\dpi100=1ni=1n(xix)2

\dpi10042=16i=16(xix)2i=16(xix)2=6×16=96 -(i)

Now, Substituting the values of xi and x in (i):

\dpi100i=16(yi3y3)2=96i=16(yiy)2=96×9=864

Hence, the variance of the new observations = 16×864=144

Therefore, Standard Deviation = σ=Variance = 144 = 12

Question:4 Given that x¯ is the mean and σ2 is the variance of n observations .Prove that the mean and variance of the observations ax1,ax2,ax3,....,axn , are ax¯ and a2σ2 respectively, (a0) .

Answer:

Given, Mean = x¯ and variance = σ2

Now, Let yi be the the resulting observations if each observation is multiplied by a:

yi=axixi=yia

y=1ni=1nyi=1ni=1naxi

y=a[1ni=1nxi]=ax

Hence the mean of the new observations ax1,ax2,ax3,....,axn is ax¯

We know,

\dpi100σ2=1ni=1n(xix)2

Now, Substituting the values of xi and x :

σ2=1ni=1n(yiaya)2a2σ2=1ni=1n(yiy)2

Hence the variance of the new observations ax1,ax2,ax3,....,axn is a2σ2

Hence proved.

Question:5(i) 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:

If wrong item is omitted.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

x=1ni=1nxi

10=120i=120xi i=120xi=200

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 = 192

Therefore, Correct Mean = (Correct Sum)/19

=19219

= 10.1

Now, Standard Deviation,

σ=1ni=1nxi2(1ni=1nx)2

22=1ni=1nxi2(x)21ni=1nxi2=4+(x)2120i=1nxi2=4+100=104i=1nxi2=2080 ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

σ=1n(i=1nxi2)(x)2=201619(10.1)2

\dpi100=106.1102.01=4.09

=2.02

Question:5(ii) 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:

If it is replaced by 12.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

x=1ni=1nxi

10=120i=120xi i=120xi=200

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 + 12 = 204

Therefore, Correct Mean = (Correct Sum)/20

\dpi100=20420

= 10.2

Now, Standard Deviation,

σ=1ni=1nxi2(1ni=1nx)2

22=1ni=1nxi2(x)21ni=1nxi2=4+(x)2120i=1nxi2=4+100=104i=1nxi2=2080 ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8) + (12x12)

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation =

σ=1n(i=1nxi2)(x)2=216020(10.2)2

=108104.04=3.96

= 1.98

Question:6 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
Standard deviation
12
15
20

Which of the three subjects shows highest variability in marks and which shows the lowest?

Answer:

Given,

Standard deviation of physics = 15

Standard deviation of chemistry = 20

Standard deviation of mathematics = 12

We know ,

C.V. =Standard DeviationMean×100

The subject with greater C.V will be more variable than others.

C.V.P=1532×100=46.87

C.V.C=2040.9×100=48.89

C.V.M=1242×100=28.57

Hence, Mathematics has lowest variability and Chemistry has highest variability.

Question:7 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.

Answer:

Given,

Initial Number of observations, n = 100

x=1ni=1nxi

\dpi10020=1100i=1100xi i=1100xi=2000

Thus, incorrect sum = 2000

Hence, New sum of observations = 2000 - 21-21-18 = 1940

New number of observation, n' = 100-3 =97

Therefore, New Mean = New Sum)/100

=194097

= 20

Now, Standard Deviation,

σ=1ni=1nxi2(1ni=1nx)2

32=1ni=1nxi2(x)21ni=1nxi2=9+(x)21100i=1nxi2=9+400=409i=1nxi2=40900 ,which is the incorrect sum.

Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

σ=1n(i=1nxi2)(x)2=3969497(20)2

=108104.04=3.96

= 3.036

Summary And Formulae Of Statistics Maths Chapter 15 class 11

  • Measures of dispersion- Range, mean deviation, Quartile deviation, standard deviation, variance are measures of dispersion.
  • Range- The difference between the maximum and the minimum values of each series is called the ‘Range’ of the data. Range = Maximum Value – Minimum Value.
  • Mean- The arithmetic average of a range of values or quantities, found by adding all values and dividing by the number of values. For example, the mean of 4, 10, and 1 is (4+10+1)/3=15/3=5.

X¯=ΣXN
X represents values or scores,
N represents the number of values or scores.

  • Median- If the total number of values(n) is an odd number, then the formula is-

Median=(n+12)thterm

  • Median- If the total number of values(n) is an even number, then the formula is-

Median=(n2)thterm+(n+12)thterm2

  • Mode- It is the most frequently occurring value or number.
  • Mean deviation for ungrouped data-

M.D.(x¯)=Σ|xix¯|n,M.D.(M)=Σ|xiM|n

  • Mean deviation for grouped data-

M.D.(x¯)=Σfi|xix¯|N,M.D.(M)=fiΣ|xiM|N,whereN=Σfi

interested students can practice class 11 maths ch 15 question answer using the following exercises.

NCERT Solutions For Class 11 Mathematics

Key Features Of Statistics Class 11 NCERT Solutions

Easy to Understand: The ch 15 maths class 11 solutions are written in simple language and are easy to understand, which makes them accessible for students who may not have a strong background in statistics.

Comprehensive Coverage: The statistics NCERT solutions cover all the important concepts and topics in Statistics Class 11 NCERT curriculum. This ensures that students are well-prepared and have a thorough understanding of the subject.

Structured Format: The ch 15 maths class 11 solutions are presented in a structured format that helps students to organize their thoughts and approach problems in a systematic manner.

NCERT Solutions For Class 11 - Subject Wise

This chapter seems very easy but it very useful in the field of science, research reliable predictions, or forecasts for future use, decision making, etc. There are 7 questions are given in the miscellaneous exercise. Try to solve them also to get command in this chapter. In NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. Explain the term standard deviation, which is discussed in NCERT solutions for class 11 maths chapter 15.

The concept of standard deviation involves measuring the amount of variation or deviation present in a given set of values, and the range is determined by considering the level of standard deviation. For a clearer understanding of this term, students are encouraged to download the PDF solutions provided by Careers360, which are available in both chapter-wise and exercise-wise formats to cater to their specific needs.

2. What are the topics discussed in Chapter 15 of NCERT Solutions for Class 11 Maths?

The topics discussed in class 11 maths statistics are
1. Introduction
2. Methods of Dispersion
3. Range
4. Mean Deviation
5. Variance and Standard Deviation
6. Analysis of Frequency Distributions

3. Are the NCERT Solutions for statistics class 11 maths helpful for the students?

The NCERT Solutions for class 11 chapter 15 maths offer precise and straightforward explanations that aid students in achieving good grades in their exams. The solutions' methodical approach to problem-solving provides students with a clear understanding of the marks allocation as per the latest CBSE Syllabus 2023. By using these class 11th statistics solutions , students can identify their weak areas and work towards improving them, resulting in better academic performance.

4. Where can I find the complete solutions of class 11 statistics solutions ?

Students can find a detailed NCERT solutions for class 11 maths  by clicking on the link. they can practice these solutions to get confidence in the concepts and in-depth understanding that ultimately lead to high score in the exam.

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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

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