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Page No 193:
Question 1:
The table below classifies the number of days of a month according to the amount of rainfall received in a certain locality.
Find the mean daily rainfall during this month.
|
Rainfall (mm) |
Number of days |
||
|
54 56 58 55 50 47 44 41 |
3 5 6 3 2 4 5 2 |
Answer:
We can make the complete table as follows:
|
Rainfall (mm) |
Number of Days |
Total Rainfall (mm) |
|||
|
54 |
3 |
162 |
|||
|
56 |
5 |
280 |
|||
|
58 |
6 |
348 |
|||
|
55 |
3 |
165 |
|||
|
50 |
2 |
100 |
|||
|
47 |
4 |
188 |
|||
|
44 |
5 |
220 |
|||
|
41 |
2 |
82 |
|||
|
Total |
30 |
1545 |
Mean daily rainfall = 
Page No 194:
Question 1:
The table below classifies the members of a committee according to their ages.
|
Age |
Number of Members |
||
|
25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60 |
6 14 16 22 5 4 3 |
Calculate the mean age of the members of this committee.
Answer:
We can make the complete table as follows:
|
Age |
Number of Member |
Class Average |
Total Age |
||||
|
25 − 30 |
6 |
27.5 |
165 |
||||
|
30 − 35 |
14 |
32.5 |
455 |
||||
|
35 − 40 |
16 |
37.5 |
600 |
||||
|
40 − 45 |
22 |
42.5 |
935 |
||||
|
45 − 50 |
5 |
47.5 |
237.5 |
||||
|
50 − 55 |
4 |
52.5 |
210 |
||||
|
55 − 60 |
3 |
57.5 |
172.5 |
||||
|
Total |
70 |
|
2775 |
Mean age of the members of the committee =
Page No 194:
Question 2:
The table below shows the number of students in class 10 of a school, classified according to their heights:
|
Height (cm) |
Number of Students |
||
|
120 − 125 125 − 130 130 − 135 135 − 140 140 − 145 145 − 150 150 − 155 155 − 160 |
19 36 23 23 43 21 23 12 |
Calculate the mean height
Answer:
We can make the complete table as follows:
|
Height (cm) |
Number of Student |
Class Average |
Total Height |
||||
|
120 − 125 |
19 |
122.5 |
2327.5 |
||||
|
125 − 130 |
36 |
127.5 |
4590.0 |
||||
|
130 − 135 |
23 |
132.5 |
3047.5 |
||||
|
135 − 140 |
23 |
137.5 |
3162.5 |
||||
|
140 − 145 |
43 |
142.5 |
6127.5 |
||||
|
145 − 150 |
21 |
147.5 |
3097.5 |
||||
|
150 − 155 |
23 |
152.5 |
3507.5 |
||||
|
155 − 160 |
12 |
157.5 |
1890.0 |
||||
|
Total |
200 |
|
27750 |
Mean height =
Page No 199:
Question 1:
The table below classifies according to weight, the infants born during a week in a hospital:
|
Weight (kg) |
Number of Infants |
||
|
2,500 2,600 2,750 2,800 3,000 3,150 3,250 3,300 3,500 |
4 6 8 10 12 10 8 7 5 |
Find the median weight.
Answer:
The given data can be shown as:
|
Weight (kg) |
Number of Infant |
||
|
Up to 2.500 |
4 |
||
|
Up to 2.600 |
10 |
||
|
Up to 2.750 |
18 |
||
|
Up to 2.800 |
28 |
||
|
Up to 3.000 |
40 |
||
|
Up to 3.150 |
50 |
||
|
Up to 3.250 |
58 |
||
|
Up to 3.300 |
65 |
||
|
Up to 3.500 |
70 |
Here, we need to find the weight of infant.
From the table it can be inferred that the weight of the 29th infant to the 40th infant is 3.000 kg.
So, the weight of the 35th infant is 3.000 kg.
Thus, the median weight of the infants born during the week in the hospital is 3.000 kg.
Page No 200:
Question 1:
The table below shows the number of employees of an office, classified according to the income-tax paid by them.
|
Income Tax (Rupees) |
Number of Employees |
||
|
1000 − 2000 2000 − 3000 3000 − 4000 4000 − 5000 5000 − 6000 6000 − 7000 7000 − 8000 8000 − 9000 |
8 10 15 18 22 8 6 3 |
Compute the median income-tax.
Answer:
The given data can be shown as:
|
Income Tax (Rupees) |
Number of Employee |
||
|
Below 2000 |
8 |
||
|
Below 3000 |
18 |
||
|
Below 4000 |
33 |
||
|
Below 5000 |
51 |
||
|
Below 6000 |
73 |
||
|
Below 7000 |
81 |
||
|
Below 8000 |
87 |
||
|
Below 9000 |
90 |
Let us tabulate the numbers 2000, 3000, … in the first row and 8, 18, … in the second row.
|
x |
2000 |
3000 |
4000 |
5000 |
6000 |
7000 |
8000 |
9000 |
|||||||||
|
y |
8 |
18 |
33 |
51 |
73 |
81 |
87 |
90 |
We know that the change in y is proportional to the corresponding change in x.
We have to find that value of x for which y =
The number 45 is between the numbers y = 33 and y = 51.
The corresponding numbers are x = 4000 and x = 5000.
Thus, we must have:
Thus, the median income tax of the employees of the office is Rs.4666.67.
Page No 200:
Question 2:
The table below classifies the candidates who took and examination, according to the marks scored by them:
|
Marks |
Number of Candidates |
||
|
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 |
44 40 35 20 12 10 8 6 4 1 |
Find the median mark.
Answer:
Given data can be shown as:
|
Marks |
Number of candidate |
||
|
Below 10 |
44 |
||
|
Below 20 |
84 |
||
|
Below 30 |
119 |
||
|
Below 40 |
139 |
||
|
Below 50 |
151 |
||
|
Below 60 |
161 |
||
|
Below 70 |
169 |
||
|
Below 80 |
175 |
||
|
Below 90 |
179 |
||
|
Below 100 |
180 |
Let us tabulate the numbers 10, 20, … in the first row and 44, 84, … in the second row.
|
x |
10 |
20 |
30 |
40 |
50 |
60 |
70 |
80 |
90 |
100 |
|||||||||||
|
y |
44 |
84 |
119 |
139 |
151 |
161 |
169 |
175 |
179 |
180 |
We know that the change in y is proportional to the corresponding change in x.
We have to find that value of x for which y =
The number 90 is between the numbers y = 84 and y = 119.
The corresponding numbers are x = 20 and x = 30.
Thus, we must have:
Thus, the median marks of the candidates who took an examination is 21.71.
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