Here we bring the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 13, Statistics and Prbobability. This part includes solutions to problems given in exercise 13.4 of class 10 Maths NCERT Exemplar Book. It contains only the long answer type questions framed from various important topics discussed in the chapter. Each question is provided with a detailed yet simple and stepwise solution to help students easily learn the concept and logic working behind that particular question.
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 20172018 as well as for other competitive exams.
CBSE Class 10 Maths NCERT Exemplar Solution: Exercise 13.3
Few questions from the Class 10 Maths NCERT Exemplar Exercise 13.4 are given below:
Q. Determine the mean of the following distribution
Marks 
Number of students 
Below 10 
5 
Below 20 
9 
Below 30 
17 
Below 40 
29 
Below 50 
45 
Below 60 
60 
Below 70 
70 
Below 80 
78 
Below 90 
83 
Below 100 
85 
Solution.
Given data is of less than type. First we have to convert it to the continuous type.
Here, 5 students have scored marks below 10, i.e. it lies between class interval 010 and 9 students have scored marks below 20. So, marks of (9 – 5) = 4 students lie in the class interval 1020. Proceeding in the same way, we get frequency of all class intervals as below:
Marks 
Number of students (f_{i}) 
010 
5 
1020 
9 – 5 = 4 
2030 
17 – 9 = 8 
3040 
29 – 17 = 12 
4050 
45 – 29 = 16 
5060 
60 – 45 = 45 
6070 
70 – 60 = 10 
7080 
78 – 70 = 8 
8090 
83 – 78 = 5 
90100 
85 – 83 = 2 
The complete frequency distribution table for given data can be obtained as follows:
Here, assumed mean, a = 45
And class width, h = 10
Thus, by step deviation method,
Q. The annual rainfall record of a city for 66 days is given in the following table.
Calculate the median rainfall using ogives (or more than type and of less than type)
Solution.
We observe that, number of days for which the rainfall was less than 0 cm is 0. Similarly, the number of days for which the annual rainfall was less than 10 cm is equal to the days having less than 0 cm rainfall plus the days having rainfall from 010 cm.
So, the number of days having rainfall less than 10 cm = 0 + 22 = 22 days.
Proceeding in the similar manner, we will get a less than type distribution as follows:
Less than type 

Rainfall (in cm) 
Number of days 
Less than 0 
0 
Less than 10 
0 + 22 = 22 
Less than 20 
22 + 10 = 32 
Less than 30 
32 + 8 = 40 
Less than 40 
40 + 15 = 55 
Less than 50 
55 + 5 = 60 
Less than 60 
60 + 6 = 66 
Also, we observe that the rainfall record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 010. So, annual rainfall record for 6622 days is more than or equal to 10 cm. Proceeding in the similar manner we will get the more than type distribution as follows:
More than type 

Rainfall (in cm) 
Number of days 
More than or equal to 0 
66 
More than or equal to 10 
66 – 22 = 44 
More than or equal to 20 
44 – 10 = 34 
More than or equal to 30 
34 – 8 = 26 
More than or equal to 40 
26 – 15 =11 
More than or equal to 50 
11 – 5 = 6 
More than or equal to 60 
6 – 6 = 0 
To draw less than type ogive we plot the points (0, 0), (10, 22), (20, 32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.
Also to draw the more than type we plot the points (0, 60), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the paper and join them by free hand.
The two ogives intersect at a point. From this point of intersection, we draw a line perpendicular to the Xaxis meeting the Xaxis at point (21.25, 0) which is the required median.
Hence, the median rainfall = 21.25 cm.
To get the complete exercise, visit the following link:
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