Skip to content
Related Articles

Related Articles

Improve Article

Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.2

  • Last Updated : 25 Feb, 2021

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

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

Solution:

We know,

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}



So, \bar{x} = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

xi

Deviations from mean 

(xi – x’)

(xi – x’)2
6 – 9 = -3 9
77 – 9 = -24
1010 – 9 = 11
1212 – 9 = 39
1313 – 9 = 416
44 – 9 = – 525
88 – 9 = – 11
1212 – 9 = 39
  74

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

σ2 = (1/8) × 74

= 9.2



Therefore, Mean = 9 and Variance = 9.25

Question 2. First n natural numbers

Solution:

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = ((n(n + 1))2)/n

= (n + 1)/2

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

On substituting the value of mean,

= \frac{1}{n}\sum_{i=1}^{n}(x_i - \frac{n+1}{2})^2 \\ = \frac{1}{n}\sum_{i=1}^{n}(x_i)^2 - \frac{1}{n}\sum_{i=1}^{n}2x_i(\frac{n+1}{2})+\frac{1}{n}\sum_{i=1}^{n}(\frac{n+1}{2})^2
Substituting values of Summation
\\ = \frac{1}{n}\frac{n(n+1)(2n+1)}{6}-\frac{n+1}{n}[\frac{n(n+1)}{2}]+\frac{(n+1)^2}{4n}×n

On extracting common values, we have, 

= (n+1)[\frac{4n+2-3n-3}{12}] \\ = \frac{(n+1)(n-1)}{12}



σ2 = (n2 – 1)/12

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

Question 3. First 10 multiples of 3

Solution:

The required multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

We know,

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

So, \bar{x} = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

xi

Deviations from mean



(xi – x’)

(xi – x’)2
33 – 16.5 = -13.5182.25
66 – 16.5 = -10.5110.25
99 – 16.5 = -7.556.25
1212 – 16.5 = -4.520.25
1515 – 16.5 = -1.52.25
1818 – 16.5 = 1.52.25
2121 – 16.5 = – 4.520.25
2424 – 16.5 = 7.556.25
2727 – 16.5 = 10.5110.25
3030 – 16.5 = 13.5182.25
  742.5

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/10) × 742.5

= 74.25

Therefore, Mean = 16.5 and Variance = 74.25

Question 4.

xi6101418242830
fi24712843

Solution:

xififixixi – x’(xi – x’)2fi(xi – x’)2
62126 – 19 = 13169338
1044010-19 = -981324
1479814-19 = -525175
181221618-19 = -1112
24819224-19 = 525200
28411228-19 = 981324
3039030-19 = 11121363
     1736

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

 \bar{x} = 760/40 

= 19

Also,



\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/40) × 1736

= 43.4

Question 5.

xi 92939798102104109
fi3232633

Solution:

xififixixi – x’(xi – x’)2fi(xi – x’)2
92327692-100 = -864192
93218693-100 = -74998
97329197-100 = -3927
98219698-100 = -248
1026612102-100 = 2424
1043312104-100 =4 1648
1093327109-100 = 981243
 N = 222200  640

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 2200/22

= 100

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/22) × 640

= 29.09



Therefore, Mean = 100 and Variance = 29.09

Question 6. Find the mean and standard deviation using short-cut method.

xi606162636465666768
fi21122925121045

Solution:

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where A = 64, h = 1

So, \bar{x} = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

σ2 = (12/1002) [100(286) – 02]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

Therefore,

Mean = 64 and Standard Deviation = 1.691

Question 7.

Classes0-3030-6060-9090-120120-150150-180180-210
Frequencies23510352

Solution:

Classesfixifixi(xi – x’)(xi – x’)2fi(xi – x’)2
0-3021530-92846416928
30-60345135-62384411532
60-90575375-3210245120
90-120101051050-2440
120-1503135405287842352
150-180516582558336416820
180-210219539088774415488
 N = 30 3210  68280

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

\bar{x} = 3210/30



= 107

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2

= (1/30) × 68280

= 2276

Therefore, Mean = 107 and Variance = 2276

Question 8.

Classes0-1010-2020-3030-4040-50
Frequencies5815166

Solution:

Classesfixifixi(xi-x’)(xi-x’)2fi(xi-x’)2
0-105525-224842420
10-20815120-121441152
20-301525375-2460
30-4016355608641024
40-50645270183241944
 N = 50 1350  6600

Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}

 \bar{x} = 1350/50

= 27

\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2



= (1/50) × 6600

= 132

Therefore, Mean = 27 and Variance = 132

Question 9. Find the mean, variance and standard deviation using short-cut method

Heights in cms70-7575-8080-8585-9090-9595-100100-105105-110110-115
Frequencies3477159663

Solution:

Height fiXiYi = (Xi-A)/hYi2fiyifiyi2
70-75272.5-419-1248
75-80177.5-39-1236
80-851282.5-24-1428
85-902987.5-11-77
90-952592.50000
95-1001297.51199
100-10510102.5241224
105-1104107.5391854
110-1155112.54161248
115-120N = 60   6254

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where, A = 92.5, h = 5

So, \bar{x}= 92.5 + ((6/60) × 5)

= 92.5 + 0.5

= 92.5 + 0.5

= 93

Then, Variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

Standard deviation = σ = √105.583

= 10.275

Question 10. The diameters of circles (in mm) drawn in a design are given below:

Diameters33-3637-4041-4445-4849-52
No. of circles1517212225

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:

HeightfixiYi = (Xi-A)/hYi2fiyifiyi2
32.5-36.51534.5-24-3060
36.5-40.51738.5-11-1717
40.5-44.52142.50000
44.5-48.52246.5112222
48.5-52.52550.52450100
 N=100   25199

\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h

Where, A = 42.5, h = 4

\bar{x} = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5



Then, Variance,

\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]

σ2 = (42/1002)[100(199) – 252]

On solving, we get,

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553




My Personal Notes arrow_drop_up
Recommended Articles
Page :