NCERT Solutions For Class 11 Maths Chapter 15

NCERT Solutions Class 11 Maths Statistics Mean Median

NCERT solutions class 11 maths chapter 15 analysis of statistics is the perfect study material to learn statistics for class 11. The NCERT solutions for class 11 maths chapter 15 analysis of statistics is provided here to help students prepare in a better way. The NCERT solutions for class 11 maths chapter 15 is created according to the latest syllabus of the Central Board of Secondary Education. Check the NCERT solutions for class 11 maths chapter 15 pdf given below.

NCERT Solutions Class 11 Maths Chapter 15 Exercises

Exercise 15.1

 

Q1. Calculate the mean deviation about the mean for the given data

5, 8, 9, 10, 11, 13, 14, 18

 

Sol:

The given data is,

5, 8, 9, 10, 11, 13, 14, 18

 

Mean,

x¯=5+8+9+10+11+13+14+188 x¯=888 x¯=11

 

The deviations of the respective observations from the mean x¯,

xix¯ are -6, -3, -2, -1, 0, 2, 3, 7

 

The absolute values of the deviations,

|xix¯| are 6, 3, 2, 1, 0 , 2, 3, 7

 

The required mean deviation about the mean is,

M.D.(x¯)=8i=1|xix¯|8 M.D.(x¯)=6+3+2+1+0+2+3+78

M.D.(x¯)=248 = 3

 

 

Q2. Calculate the mean deviation about the mean for the given data

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

 

Sol:

The given data is,

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

 

Mean,

x¯=39+71+49+41+43+56+64+47+55+4510 x¯=51010 x¯=51

 

The deviations of the respective observations from the mean x¯,

xix¯ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.

 

The absolute values of the deviations,

|xix¯| are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

 

The required mean deviation about the mean is,

M.D.(x¯)=10i=1|xix¯|10 M.D.(x¯)=12+20+2+10+8+5+13+4+4+610

M.D.(x¯)=8410 = 8.4

 

Q3. Calculate the mean deviation about the median for the given data:

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

 

Sol:

The given data is,

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

There are 12 observations which is even number.

 

Arrange the data in ascending order:

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

 

Median,

M=(122)thobservation+(122+1)thobservation2

= 6thobservation+7thobservation2

= 13+142

= 272

= 13.5

 

The deviations of the respective observations from the median x¯,

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

 

The absolute values of the deviations,

|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

 

The required mean deviation about the median is,

M.D.(M)=12i=1|xiM|12

= 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.512

= 2812

= 2.33

 

Q4. Calculate the mean deviation about the median for the given data:

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

 

Sol:

The given data is,

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

There are 10 observations which is even number.

 

Arrange the data in ascending order:

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

 

Median,

M=(102)thobservation+(102+1)thobservation2

= 5thobservation+6thobservation2

= 46+492

= 952

= 47.5

 

The deviations of the respective observations from the median x¯,

xiM are –11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

 

The absolute values of the deviations,

|xiM| are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

 

The required mean deviation about the median is,

M.D.(M)=10i=1|xiM|10

= 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.510

= 7010

= 7

 

Q5. Calculate the mean deviation about the mean for the given data

 

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

 

Sol:

xi xi xi |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
25 350 158

 

N=5i=1fi=25

 

5i=1fixi=350

 

Therefore,

x¯=1N5i=1fixi

= 125×350

= 14

 

MD(x¯)=1N5i=1fi|xix¯|

= 125×158

= 6.32

 

Q6. Calculate the mean deviation about the mean for the given data

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

 

Sol:

xi xi xi |xix¯| fi|xix¯|
10 4 35 40 160
30 24 720 20 480
50 28 1400 0 0
70 16 1120 20 320
90 8 720 40 320
80 4000 1280

 

N=5i=1fi=80

 

5i=1fixi=4000

 

Therefore,

x¯=1N5i=1fixi

= 180×4000

= 50

 

MD(x¯)=1N5i=1fi|xix¯|

= 180×1280

= 16

 

Q7. Calculate the mean deviation about the median for the given data

xi 10 10 15 20 25
fi 7 4 6 3 5
xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

 

 Sol:

The observations given are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

xi fi c.f.
5 8 8
7 6 14
9 2 16
10 2 18
12 2 20
15 6 26

 

Here, N = 26, which is even.

 

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

 

Median=13thobservation+14thobservation2

= 7+72

= 7

 

The absolute values of the deviations from median,

|xiM| 2 0 2 3 5 8
fi 8 6 2 2 2 6
fi|xiM| 16 0 4 6 10 48

 

6i=1fi=26

 

6i=1fi|xiM|=84

 

Therefore,

MD(M)=1N6i=1fi|xiM|

= 126×84

= 3.23

 

Q8. Calculate the mean deviation about the median for the given data

xi 15 21 27 30 35
fi 3 5 6 7 8

 

Sol:

The observations given are in ascending order.

 

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

 

xi fi c.f.
15 3 3
21 5 8
27 6 14
30 7 21
35 8 29

 

Here, N = 29, which is odd.

 

Median=(29+12)thobservation

= 15th observation

 

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

 

Therefore,

Median = 30

 

The absolute values of the deviations from median,

|xiM| 15 9 3 0 5
fi 3 5 6 7 8
fi|xiM| 45 45 18 0 40

 

5i=1fi=29

 

5i=1fi|xiM|=148

 

Therefore,

MD(M)=1N5i=1fi|xiM|

= 129×148

= 5.1

 

Q9. Calculate the mean deviation about the mean for the given data:

Income per day Number of persons
0 – 100 4
100 – 200 8
200 – 300 9
300 – 400 10
400 – 500 7
500 – 600 5
600 – 700 4
700 – 800 3

 

 Sol:

The given table is formed.

Income per day Number of persons fi Mid – point xi fi xi |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
50 17900 7896

 

N=8i=1fi=50

 

8i=1fixi=17900

 

Therefore,

x¯=1N8i=1fixi

= 150×17900

= 358

 

MD(x¯)=1N8i=1fi|xix¯|

= 150×7896

= 175.92

 

Q10. Calculate the mean deviation about the mean for the given data:

Height in cm Number of boys
95 – 105 9
105 – 115 13
115 – 125 26
125 – 135 30
135 – 145 12
145 – 155 10

 

 

Sol:

The given table is obtained:

 

Height in cm Number of boys fi Mid – point xi fi xi |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

 

N=6i=1fi=100

 

6i=1fixi=12530

 

Therefore,

x¯=1N6i=1fixi

= 1100×12530

= 125.3

 

MD(x¯)=1N6i=1fi|xix¯|

= 1100×1128.8

= 11.28

 

Q11. Calculate the mean deviation about the median for the given data:

Marks Number of girls
0 – 10 6
10 – 20 8
20 – 30 14
30 – 40 16
40 – 50 4
50 – 60 2

 

 

Sol:

The given table is obtained:

 

Marks Number of girls fi Cumulative frequency (c.f.) Mid – point xi |xiM| fi|xix¯|
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
50 517.1

 

The class interval containing (N2)th or 25th item is 20 – 30.

Therefore, 20 -30 is the median class.

 

Now,

Median=l+N2Cf×h

Where, l = 20

C = 14

f = 14

h = 10

N = 50

 

Median=20+251414×10

= 20+11014

= 20 + 7.85

= 27.85

 

Mean deviation about the median is,

MD(M)=1N6i=1fi|xiM|

= 150×517.1

= 10.34

 

Q12.  Calculate the mean deviation about the median for the given data:

            

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

 

Sol:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

 

The table is as given below:

 

Age Number fi Cumulative frequency (c.f.) Mid – point xi |xiM| fi|xix¯|
16 – 20 5 5 18 20 100
21 – 25 6 11 23 15 90
26 – 30 12 23 28 10 120
31 – 35 14 37 33 5 70
36 – 40 26 63 38 0 0
41 – 45 12 75 43 5 60
46 – 50 16 91 48 10 160
51 – 55 9 100 53 15 135
100

 

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

 

Now,

Median=l+N2Cf×h

Where, l = 35.5

C = 37

f = 26

h = 5

N = 100

 

Median=35.5+503726×5

= 35.5+13×526

= 35.5 + 2.5

= 38

 

Mean deviation about the median is,

MD(M)=1N8i=1fi|xiM|

= 1100×735

= 7.35

 

Exercise 15.2

 

Q1. Calculate mean and variance for the given data 6, 7, 10, 12, 13, 4, 8, 12

 

Sol:

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

Mean,

x¯=8i=1xin

= 6+7+10+12+13+4+8+128

= 728

= 9

The given table is obtained,

xi (xix¯) (xix¯)2
6 -3 9
7 -2 4
10 -1 1
12 3 9
13 4 16
4 -5 25
8 -1 1
12 3 9
74

Variance,

(σ2)=1n8i=1(xix¯)2

= 18×74

= 9.25

 

Q2. Calculate the mean and variance for the first n natural numbers.

 

Sol:

The mean of first n natural number is,

Mean=SumofallobservationsNumberofobservations Mean=n(n+1)2n

= n+12

Variance,

(σ2)

= 1nni=1(xix¯)2

= 1nni=1[xi(n+12)]2

= 1nni=1x2i1nni=12(n+12)xi+1nni=1(n+12)2

= 1nn(n+1)(2n+1)6(n+1n)[n(n+1)2]+(n+1)24n×n

= (n+1)(2n+1)6(n+1)22+(n+1)24

= (n+1)(2n+1)6(n+1)24

= (n+1)[4n+23n312]

= (n+1)(n1)12

= n2112

 

Q3. Calculate the mean and variance for the first 10 multiples of 3.

Sol:

The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Number of observations,

n = 10

Mean,

x¯=10i=110

= 16510

= 16.5

The given table is obtained,

xi (xix¯) (xix¯)2
3 -13.5 182.25
6 -10.5 110.25
9 -7.5 56.25
12 -4.5 20.25
15 -1.5 2.25
18 1.5 2.25
21 4.5 20.25
24 7.5 56.25
27 10.5 110.25
30 13.5 182.25
742.5

Variance,

(σ2)

= 1n10i=1(xix¯)2

= 110×742.5

= 74.25

 

Q4. Calculate the mean and variance for the data

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

 

Sol:

The data obtained is given in the tabular form:

xi xi 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 11 121 363
40 760 1736

Here,

N = 40

7i=1fixi = 760

x¯=7i=1fixiN

= 76040

= 19

Variance,

(σ2)

= 1n7i=1(xix¯)2

= 140×1736

= 43.4

 

Q5. Calculate the mean and variance for the data

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

 

Sol:

The data obtained is given in the tabular form:

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
22 2200 640

Here,

N = 22

7i=1fixi = 2200

x¯=7i=1fixiN

= 220022

= 100

Variance,

(σ2)

= 1n7i=1(xix¯)2

= 122×640

= 29.09

 

Q6. Calculate 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

 

Sol:
The data is obtained in tabular form as follows.

xi fi fi=xi641 y2i fiyi fiy2i
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
100 220 0 286

Mean,

x¯=A9i=1fiyiN×h

= 64+0100×1

= 64 + 0

= 64

Variance,

(σ2)

= h2N2[N9i=1fiy2i(9i=1fiyi)2]

= 11002[100×2860]

= 2.86

Standard deviation,

σ=2.86

= 1.69

 

Q7. Calculate the mean and variance for the given frequency distribution.

 

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

 

Sol:

Class fi xi yi=xi10530 y2i fiyi fiy2i
0 – 30 2 15 -3 9 -6 18
30 – 60 3 45 -2 4 -6 12
60 – 90 5 75 -1 1 -5 5
90 – 120 10 105 0 0 0 0
120 – 150 3 135 1 1 3 3
150 – 180 5 165 2 4 10 20
180 – 210 2 195 3 9 6 18
30 2 76

Mean,

x¯

= A+7i=1fiyiN×h

= 105+230×30

= 105 + 2

= 107

Variance,

(σ2)

= h2N2[N7i=1fiy2i(7i=1fiyi)2]

= 302302[30×76(2)2]

= 2280 – 4

= 2276

 

Q8. Calculate the mean and variance for the following frequency distribution.

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

Sol:

Class Frequency fi Mid – point xi yi=xi2510 y2i fiyi fiy2i
0 – 10 5 5 -2 4 -10 20
10 – 20 8 15 -1 1 -8 8
20 – 30 15 25 0 0 0 0
30 – 40 16 35 1 1 16 16
40 – 50 6 45 2 4 12 24
50 10 68

Mean,

x¯

= A+5i=1fiyiN×h

= 25+1050×10

= 25 + 2

= 27

Variance,

(σ2)

= h2N2[N5i=1fiy2i(5i=1fiyi)2]

= 102502[50×68(10)2]

= 125[3400100]

= 330025

= 132

 

Q9. Calculate the mean, variance and standard deviation using short – cut method:

 

Height in cm Number of children
70 – 75 3
75 – 80 4
80 – 85 7
85 – 90 7
90 – 95 15
95 – 100 9
100 – 105 6
105 – 110 6
110 – 115 3

 

Sol:

Class fi xi yi=xi92.55 y2i fiyi fiy2i
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
60 6 254

Mean,

x¯

= A+9i=1fiyiN×h

= 92.5+660×5

= 92.5 + 0.5

= 93

Variance,

(σ2)

= h2N2[N9i=1fiy2i(9i=1fiyi)2]

= 52602[60×254(6)2]

= 253600[155204]

= 105.58

Standard deviation,

(σ)

= 105.58

= 10.27

 

Q10. The diameters of circles (in mm) drawn in a design are given below:

 

 

Diameters Number of Children
33 – 36 15
37 – 40 17
41 – 44 21
45 – 48 22
49 – 52 25

 

Sol:

Class fi xi yi=xi42.54 y2i fiyi fiy2i
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
100 25 199

N = 100

h = 4

Let us assume that mean A be 42.5

Mean,

x¯

= A+5i=1fiyiN×h

= 42.5+25100×4

= 43.5

Variance,

(σ2)

= h2N2[N5i=1fiy2i(5i=1fiy