NCERT Exemplar Solution for Class 10 Mathematics: Statistics, Exercise 13.1
Here you will get the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 13, Statistics. This part brings you the solutions to Exercise 13.1 that includes questions based on Statistics only. All these are the Multiple Choice Questions. All the questions are very important to prepare for CBSE Class 10 Board Exam 2017-2018.
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 d_{i }is the difference between a and each of the x_{i}’s (mid-value).
Thus, d_{i}’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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