chapter statistics ex13.1 full
Answers
Answer:
Step-by-step explanation:
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
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.
