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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
Few questions from the Class 10 Maths NCERT Exemplar Exercise 13.4 are given below:
Q.┬аDetermine 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,┬аi.e. it lies between class interval 0-10 and 9 students have scored marks below 20. So, marks of (9 тАУ 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 тАУ 5 = 4 |
|
20-30 |
17 тАУ 9 = 8 |
|
30-40 |
29 тАУ 17 = 12 |
|
40-50 |
45 тАУ 29 = 16 |
|
50-60 |
60 тАУ 45 = 45 |
|
60-70 |
70 тАУ 60 = 10 |
|
70-80 |
78 тАУ 70 = 8 |
|
80-90 |
83 тАУ 78 = 5 |
|
90-100 |
85 тАУ 83 = 2 |
The complete frequency distribution table for given data can be obtained as follows:
Here, assumed mean,┬аa┬а= 45
And class width,┬аh┬а= 10
Thus, by step deviation method,
Q.┬аThe 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┬а(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┬а(in cm) |
Number of days |
|
More than or equal to 0 |
66 |
|
More than or equal to 10 |
66 тАУ 22 = 44 |
|
More than or┬аequal to 20 |
44 тАУ 10 = 34 |
|
More than or equal to 30 |
34 тАУ 8 = 26 |
|
More than or equal to 40 |
26 тАУ 15 =11 |
|
More than or equal to 50 |
11 тАУ 5 = 6 |
|
More than or equal to 60 |
6 тАУ 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.