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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,

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

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