 NCERT Exemplar Solution for Class 10 Mathematics Chapter 13: Exercise 13.4

Here you will get the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 13, Statistics and Probability. Here you will get solutions to problems given in exercise 13.4 of class 10 Maths NCERT Exemplar Book. NCERT Exemplar Solution for Class 10 Mathematics

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 step-wise 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 2017-2018 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 0-10 and 9 students have scored marks below 20. So, marks of (9 – 5) = 4 students lie in the class interval 10-20. Proceeding in the same way, we get frequency of all class intervals as below:

 Marks Number of students (fi) 0-10 5 10-20 9 – 5 = 4 20-30 17 – 9 = 8 30-40 29 – 17 = 12 40-50 45 – 29 = 16 50-60 60 – 45 = 45 60-70 70 – 60 = 10 70-80 78 – 70 = 8 80-90 83 – 78 = 5 90-100 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 0-10 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 0-10. So, annual rainfall record for 66-22 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 X-axis meeting the X-axis at point (21.25, 0) which is the required median.

Hence, the median rainfall = 21.25 cm.

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