NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

NCERT Solutions for Class 11 Maths Chapter 15 Statistics, contains solutions for all exercise 15.2 questions. By practising these questions, students can build their confidence level. NCERT Solutions also help the students to excel their skills in statistics. Download the NCERT Solutions of Class 11 maths now and start practising.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

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Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Solutions : 12 Questions

Exercise 15.3 Solutions : 5 Questions

Miscellaneous Exercise Solutions: 7 Questions

Access NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2

Find the mean and variance for each of the data in Exercise 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

Solution:-

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 19

So, x̅ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

Let us make the table of the given data and append other columns after calculations.

Xi Deviations from mean

(xi – x̅)

(xi – x̅)2
6 6 – 9 = -3 9
7 7 – 9 = -2 4
10 10 – 9 = 1 1
12 12 – 9 = 3 9
13 13 – 9 = 4 16
4 4 – 9 = – 5 25
8 8 – 9 = – 1 1
12 12 – 9 = 3 9
74

We know that Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 20

σ2 = (1/8) × 74

= 9.2

∴Mean = 9 and Variance = 9.25

2. First n natural numbers

Solution:-

We know that Mean = Sum of all observations/Number of observations

∴Mean, x̅ = ((n(n + 1))2)/n

= (n + 1)/2

and also WKT Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 21

By substitute that value of x̅ we get,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 22

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 23

WKT, (a + b)(a – b) = a2 – b2

σ2 = (n2 – 1)/12

∴Mean = (n + 1)/2 and Variance = (n2 – 1)/12

3. First 10 multiples of 3

Solution:-

First we have to write the first 10 multiples of 3,

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

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 24

So, x̅ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

Let us make the table of the data and append other columns after calculations.

Xi Deviations from mean

(xi – x̅)

(xi – x̅)2
3 3 – 16.5 = -13.5 182.25
6 6 – 16.5 = -10.5 110.25
9 9 – 16.5 = -7.5 56.25
12 12 – 16.5 = -4.5 20.25
15 15 – 16.5 = -1.5 2.25
18 18 – 16.5 = 1.5 2.25
21 21 – 16.5 = – 4.5 20.25
24 24 – 16.5 = 7.5 56.25
27 27 – 16.5 = 10.5 110.25
30 30 – 16.5 = 13.5 182.25
742.5

Then, Variance

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 25

= (1/10) × 742.5

= 74.25

∴Mean = 16.5 and Variance = 74.25

4.

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

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi fixi Deviations from mean

(xi – x̅)

(xi – x̅)2 fi(xi – x̅)2
6 2 12 6 – 19 = 13 169 338
10 4 40 10 – 19 = -9 81 324
14 7 98 14 – 19 = -5 25 175
18 12 216 18 – 19 = -1 1 12
24 8 192 24 – 19 = 5 25 200
28 4 112 28 – 19 = 9 81 324
30 3 90 30 – 19 = 11 121 363
N = 40 760 1736

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 26

5.

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

Solution:-

Let us make the table of the given data and append other columns after calculations.

Xi fi fixi Deviations from mean

(xi – x̅)

(xi – x̅)2 fi(xi – x̅)2
92 3 276 92 – 100 = -8 64 192
93 2 186 93 – 100 = -7 49 98
97 3 291 97 – 100 = -3 9 27
98 2 196 98 – 100 = -2 4 8
102 6 612 102 – 100 = 2 4 24
104 3 312 104 – 100 = 4 16 48
109 3 327 109 – 100 = 9 81 243
N = 22 2200 640

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 27

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

Solution:-

Let the assumed mean A = 64. Here h = 1

We obtain the following table from the given data.

Xi Frequency fi Yi = (xi – A)/h 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
0 286

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 28

Where A = 64, h = 1

So, x̅ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 29

σ2 = (12/1002) [100(286) – 02]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

∴ Mean = 64 and Standard Deviation = 1.691

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

7.

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

Solution:-

Let us make the table of the given data and append other columns after calculations.

Classes Frequency fi Mid – points

xi

fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)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
N = 30 3210 68280

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 30

8.

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

Solution:-

Let us make the table of the given data and append other columns after calculations.

Classes Frequency fi Mid – points

xi

fixi (xi – x̅) (xi – x̅)2 fi(xi – x̅)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
N = 50 1350 6600

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 31

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 32

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
Frequencies 3 4 7 7 15 9 6 6 3

Solution:-

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

Let us make the table of the given data and append other columns after calculations.

Height (class) Number of children

Frequency fi

Midpoint

Xi

Yi = (xi – A)/h Yi2 fiyi fiyi2
70 – 75 2 72.5 -4 16 -12 48
75 – 80 1 77.5 -3 9 -12 36
80 – 85 12 82.5 -2 4 -14 28
85 – 90 29 87.5 -1 1 -7 7
90 – 95 25 92.5 0 0 0 0
95 – 100 12 97.5 1 1 9 9
100 – 105 10 102.5 2 4 12 24
105 – 110 4 107.5 3 9 18 54
110 – 115 5 112.5 4 16 12 48
N = 60 6 254

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 33

Where, A = 92.5, h = 5

So, x̅ = 92.5 + ((6/60) × 5)

= 92.5 + ½

= 92.5 + 0.5

= 93

Then, Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 34

σ2 = (52/602) [60(254) – 62]

= (1/144) [15240 – 36]

= 15204/144

= 1267/12

= 105.583

Hence, standard deviation = σ = √105.583

= 10.275

∴ Mean = 93, variance = 105.583 and Standard Deviation = 10.275

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 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.]

Solution:-

Let the assumed mean, A = 42.5 and h = 4

Let us make the table of the given data and append other columns after calculations.

Height (class) Number of children

(Frequency fi)

Midpoint

Xi

Yi = (xi – A)/h 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
N = 100 25 199

Mean,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 35

Where, A = 42.5, h = 4

So, x̅ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

NCERT Soluitons for Class 11 Maths Chapter 15 Statistics Image 34

σ2 = (42/1002)[100(199) – 252]

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553

∴ Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.

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