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NCERT Exemplar Solution for Class 10 Mathematics: Statistics, Exercise 13.1

Sep 8, 2017 12:17 IST
    Class 10 Maths NCERT Exemplar Solutions
    Class 10 Maths NCERT Exemplar Solutions

    Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions. This part includes solutions for questions 1 to 11 that based on Statistics from Exercise 13.1.  All these are the Multiple Choice Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

    CBSE Class 10 Mathematics Syllabus 2017-2018

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

    Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Statistics:

    Exercise 13.1

    Multiple Choice Questions:

    (a) lower limits of the classes

    (b) upper limits of the classes

    (c) mid-points of the classes

    (d) frequencies of the class marks

    Answer. (c)          

    Explanation:

    Here, a is assumed mean from class marks (xi) and di is the difference between a and each of the xi’s (mid-value).

    Thus, di’s are the deviation from a of mid-points of the classes.

    Question. 2 While computing mean of grouped data, we assume that the frequencies are

    (a) evenly distributed over all the classes

    (b) centred at the class marks of the classes

    (c) centred at the upper limits of the class

    (d) centred at the lower limits of the class

    Answer. (b)          

    Expalnation:

    While grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value is taken for any further calculation. In computing the mean of grouped data, it is assumed that the frequency of each class interval is centred around its mid value or class mark.

    Question. 5 The abscissa of the point of intersection of the Less than type and of the more than type cumulative frequency curves of a grouped data gives its

    (a) mean                                               

    (b) median           

    (c) mode                                               

    (d) All of these

    Answer. (b)

    Explanation:

    Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa because on X-axis we take the upper or lower limits respectively and on Y-axis we take cumulative frequency.

    Answer. (b)          

    Explanation:

    Commulative frequency for the given data is given below:

    Class

    Frequency

    Cumulative frequency

    0-5

    10

    10

    5-10

    15

    25

    10-15

    12

    37

    15-20

    20

    57

    20-25

    9

    66

    Now, the modal class is the class having the maximum frequency.

    The maximum frequency 20 belongs to class (15 - 20).

    So, lower limit of modal class is 15.

    And, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

    Here, n/2 = 66/2 = 33, which lies in the interval 10-15.

    So, lower limit of the median class is 10.

    Hence, required sum of lower limits of the median class and modal class is 10 + 15 = 25.

    (a) 17                                    

    (b) 17.5                                 

    (c) 18                                     

    (d) 18.5

    Answer. (b)

    Explanation:

    First make the classes continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

    Then find the Commulative frequency for the given data as below:

    Class

    Frequency

    Cumulative frequency

    0.5-5.5

    13

    13

    5.5-11.5

    10

    23

    11.5-17.5

    15

    38

    17.5-23.5

    8

    46

    23.5-29.5

    11

    57

    Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

    Here, n/2 = 57/2 = 28.5 which lies in the interval 11.5 - 17.5.

    Hence, the upper limit is 17.5.

    Question. 8 For the following distribution,

    Marks

    Number of students

    Below 10

    3

    Below 20

    12

    Below 30

    27

    Below 40

    57

    Below 50

    75

    Below 60

    80

    the modal class is

    (a) 10-20                                     

    (b) 20-30                                     

    (c) 30-40                                     

    (d) 50-60

    Answer. (c)

    Explanation:

    Commulative frequency for the given data is calculated as below:

    Marks

    Number of students

    Cumulative frequency

    Below 10

     3

    3

    10-20

     9

    12

    20-30

    15

    27

    30-40

    30

    57

    40-50

    18

    75

    50-60

    5

    80

    Now the class whose frequency is maximum is called the modal class. Here, we see the highest frequency is 30, which lies in the interval 30-40. Therefore, 30-40 is the modal class.

    Question.  9 Consider the data

    The difference of the upper limit of the median class and the lower limit of the modal class is

    (a) 0                                       

    (b) 19                                    

    (c) 20                                     

    (d) 38

    Answer.  (c)

    Explanation:

    Commulative frequency for the given data is calculated as below:

    Class

    Frequency

    Cumulative frequency

    65-85

    4

    4

    85-105

    5

    9

    105-125

    13

    22

    125-145

    20

    42

    145-165

    14

    56

    165-185

    7

    63

    185-205

    4

    67

    Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

    Here, n/2 = 67/2 = 33.5, which lies in the interval 125 - 145.

    Hence, upper limit of median class is 145.

    Again, the class whose frequency is maximum is called the modal class.

    Here, 20 is the highest frequency which lies in the interval 125 –145.

    Hence, the lower limit of modal class is 125.

    Thus, the required difference = Upper limit of median class - Lower limit of modal class

                                                    = 145 –125 = 20.

    Question. 10 The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below

    The number of atheletes who completed the race in less than 14.6 s is

    (a) 11    

    (b) 71    

    (c) 82     

    (d) 130

    Answer. (c)

    Exlanation:

     The number of atheletes who completed the race in less than 14.6

    = 2 + 4 + 5 + 71 = 82.

    Question.  11 Consider the following distribution

    Marks obtained

    Number of students

    More than or equal to 0

    63

    More than or equal to 10

    58

    More than or equal to 20

    55

    More than or equal to 30

    51

    More than or equal to 40

    48

    More than or equal to 50

    42

    the frequency of the class 30-40 is

    (a) 3       

    (b) 4       

    (c) 48     

    (d) 51

    Answer. (a)

    Explanation:

    The given frequency data can be represented in the form of class intervals as follows:

    Marks obtained

    Number of students

    Frequency

    0-10

    (63 – 58) = 5

    5

    10-20

    (58 – 55) = 3

    3

    20-30

    (55 – 51) = 4

    4

    30-40

    (51 – 48) = 3

    3

    40-50

    (48 – 42) = 6

    6

    50…

    42 − 0 = 42

    42

    Hence the frequency of 30 – 40 class interval is 3.

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