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NCERT Solutions for Exercise 15.3 Class 11 Maths Chapter 15 Statistics

NCERT Solutions for Exercise 15.3 Class 11 Maths Chapter 15 Statistics

Edited By Sumit Saini | Updated on Jul 18, 2022 12:09 PM IST

NCERT solutions for Class 11 Maths chapter 15 exercise 15.3 has a total of 5 questions with some questions having subpart also. Questions from exercise 15.3 mainly deal with the comparison between two sets of data in which we have to compare the variation and consistency among the given sets. Exercise 15.3 Class 11 Maths uses the same concept, hence if the initial few questions are done properly, the whole exercise can be tackled easily.

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This Story also Contains
  1. Statistics Class 11 Chapter 15 -Exercise: 15.3
  2. More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

Solving NCERT Solutions for Class 11 Maths chapter 15 exercise 15.3 is highly recommended as it is easy as well as scoring from the exam point of view. Sometimes, direct questions are asked from this exercise as well in Boards. Also, students can refer to the following exercise of NCERT for further information about upcoming exercises.

  • Statistics Exercise 15.1

  • Statistics Exercise 15.2

  • Statistics Miscellaneous Exercise

    Statistics Class 11 Chapter 15 -Exercise: 15.3

    Question:1. From the data given below state which group is more variable, A or B?

    Marks
    10-20
    20-30
    30-40
    40-50
    50-60
    60-70
    70-80
    Group A
    9
    17
    32
    33
    40
    10
    9
    Group B
    10
    20
    30
    25
    43
    15
    7


    Answer:

    The group having a higher coefficient of variation will be more variable.

    Let the assumed mean, A = 45 and h = 10

    For Group A

    Marks
    Group A
    f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    = \frac{x_i-45}{10}
    y_i^2
    f_iy_i
    f_iy_i^2
    10-20
    9
    15
    -3
    9
    -27
    81
    20-30
    17
    25
    -2
    4
    -34
    68
    30-40
    32
    35
    -1
    1
    -32
    32
    40-50
    33
    45
    0
    0
    0
    0
    50-60
    40
    55
    1
    1
    40
    40
    60-70
    10
    65
    2
    4
    20
    40
    70-80
    9
    75
    3
    9
    27
    81

    \sum{f_i} =N = 150



    \sum f_iy_i
    = -6
    \sum f_iy_i ^2
    =342
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    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{227.84} = 15.09

    Coefficient of variation = \frac{\sigma}{\overline x}\times100

    C.V.(A) = \frac{15.09}{44.6}\times100 = 33.83

    Similarly,

    For Group B

    Marks
    Group A
    f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    = \frac{x_i-45}{10}
    y_i^2
    f_iy_i
    f_iy_i^2
    10-20
    10
    15
    -3
    9
    -30
    90
    20-30
    20
    25
    -2
    4
    -40
    80
    30-40
    30
    35
    -1
    1
    -30
    30
    40-50
    25
    45
    0
    0
    0
    0
    50-60
    43
    55
    1
    1
    43
    43
    60-70
    15
    65
    2
    4
    30
    60
    70-80
    7
    75
    3
    9
    21
    72

    \sum{f_i} =N = 150



    \sum f_iy_i
    = -6
    \sum f_iy_i ^2
    =375

    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{249.84} = 15.80

    Coefficient of variation = \frac{\sigma}{\overline x}\times100

    C.V.(B) = \frac{15.80}{44.6}\times100 = 35.42

    Since C.V.(B) > C.V.(A)

    Therefore, Group B is more variable.

    Question:2 From the prices of shares X and Y below, find out which is more stable in value:

    X
    35
    54
    52
    53
    56
    58
    52
    50
    51
    49
    Y
    108
    107
    105
    105
    106
    107
    104
    103
    104
    101


    Answer:

    X( x_i )
    Y( y_i )
    x_i^2
    y_i^2
    35
    108
    1225
    11664
    54
    107
    2916
    11449
    52
    105
    2704
    11025
    53
    105
    2809
    11025
    56
    106
    8136
    11236
    58
    107
    3364
    11449
    52
    104
    2704
    10816
    50
    103
    2500
    10609
    51
    104
    2601
    10816
    49
    101
    2401
    10201
    =510
    = 1050
    =26360
    =110290

    For X,

    Mean , \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51

    Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{35} = 5.91

    C.V.(X) = \frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58

    Similarly, For Y,

    Mean , \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105

    Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{4} = 2

    C.V.(Y) = \frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904

    Since C.V.(Y) < C.V.(X)

    Hence Y is more stable.

    Question:3(i) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


    Firm A
    Firm B
    No. of wage earners
    586
    648
    Mean of monthly wages
    Rs\hspace {1mm} 5253
    Rs\hspace {1mm} 5253
    Variance of the distribution of wages
    100
    121

    Which firm A or B pays larger amount as monthly wages?

    Answer:

    Given, Mean of monthly wages of firm A = 5253

    Number of wage earners = 586

    Total amount paid = 586 x 5253 = 30,78,258

    Again, Mean of monthly wages of firm B = 5253

    Number of wage earners = 648

    Total amount paid = 648 x 5253 = 34,03,944

    Hence firm B pays larger amount as monthly wages.

    Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


    Firm A
    Firm B
    No. of wage earners
    586
    648
    Mean of monthly wages
    Rs\hspace {1mm} 5253
    Rs\hspace {1mm} 5253
    Variance of the distribution of wages
    100
    121

    Which firm, A or B, shows greater variability in individual wages?

    Answer:

    Given, Variance of firm A = 100

    Standard Deviation = \sigma_A = \sqrt{Variance}= \sqrt{100} = 10

    Again, Variance of firm B = 121

    Standard Deviation = \sigma_B = \sqrt{Variance}= \sqrt{121} = 11

    Since \sigma_B>\sigma_A , firm B has greater variability in individual wages.

    Question:4 The following is the record of goals scored by team A in a football session:

    No. of goals scored
    0
    1
    2
    3
    4
    No. of matches
    1
    9
    7
    5
    3

    For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

    Answer:

    No. of goals
    scored x_i
    Frequency
    f_i
    x_i^2
    f_ix_i
    f_ix_i^2
    0
    1
    0
    0
    0
    1
    9
    1
    9
    9
    2
    7
    4
    14
    28
    3
    5
    9
    15
    45
    4
    3
    16
    12
    48

    \sum{f_i} =N = 25

    \sum f_ix_i
    = 50
    \sum f_ix_i ^2
    =130

    For Team A,

    Mean,

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i =\frac{50}{25}= 2

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(25)^2}\left [25(130) - (50)^2 \right ] \\ \\ = \frac{750}{625} =1.2

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{1.2} = 1.09

    C.V.(A) = \frac{\sigma}{\overline x}\times100 = \frac{1.09}{2}\times100 = 54.5

    For Team B,

    Mean = 2

    Standard deviation, \sigma = 1.25

    C.V.(B) = \frac{\sigma}{\overline x}\times100 = \frac{1.25}{2}\times100 = 62.5

    Since C.V. of firm B is more than C.V. of A.

    Therefore, Team A is more consistent.

    Question:5 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

    \sum_{i=1}^{50}x_i=212, \sum _{i=1}^{50}x_i^2=902.8,\sum _{i=1}^{50}y_i=261,\sum _{i=1}^{50}y_i^2=1457.6
    Which is more varying, the length or weight?

    Answer:

    For lenght x,

    Mean, \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24

    We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28

    C.V.(x) = \frac{\sigma}{\overline x}\times100 = \frac{0.28}{4.24}\times100 = 6.603

    For weight y,

    Mean,

    Mean, \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22

    We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37

    C.V.(y) = \frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24

    Since C.V.(y) > C.V.(x)

    Therefore, weight is more varying.


    More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3

    The NCERT Class 11 Maths chapter Statistics is scoring and easier than other chapters in Class 11 Maths. Students need to remember the exact method to solve the questions and in order to get good marks. Exercise 15.3 Class 11 Maths deals with the questions in which comparison is done between two sets of given data. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.3 provides a good opportunity to boost marks with less effort in the final examination.

    Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3

  • The Class 11 Maths chapter 15 exercise is prepared by expert faculties having wide experience.
  • Exercise 15.3 Class 11 Maths can help students who are struggling with other chapters as this is a relatively easier chapter.
  • Class 11 Maths chapter 15 exercise 15.3 solutions provided here are of good quality and can be referred for revision also.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exemplar Solutions

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