Class 11 Maths Ncert Solutions Chapter 15 Ex 15.2 Statistics PDF

Class 11 Maths Ncert Solutions Ex 15.2

Class 11 Maths Ncert Solutions Chapter 15 Ex 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=1fiyi)2]

= 421002[100×199(25)2]

= 1610000[19900625]

= 1610000×19275

= 30.84

Standard deviation,

(σ)

= 30.84

= 5.55

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