RD Sharma Solutions Chapter 7 Statistics Exercise 7.4 Class 10 Maths
Chapter Name  RD Sharma Chapter 7 Statistics 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 7.4 Solutions
1. Following are the lives in hours of 15 pieces of the components of aircraft engine. find the median:
715,724,,725,710,729,745,694,699,696,712,734,728,716,705,719.
Solution
Lives in hours of is pieces are
= 715, 724, 725, 710, 729, 745, 694, 699, 696, 712,734,728,719,705,705,719
Arrange the above data in a sending order
694,696,699,705,710,712,715,716,719,721,725,728,729,734,745
N = 15(odd)
2. The following is the distribution of height of students of a certain class in a certain city.
Height (in cm): 
160 – 162 
163 – 165 
166 – 168 
169 – 171 
172 – 174 
No. of students: 
15 
118 
142 
127 
18 
Find the median height.
Solution
I.Q.: 
55 – 64 
65 – 74 
75 – 84 
85 – 94 
95 – 104 
105 – 114 
115 – 124 
125 – 134 
135 – 144 
No. of Students: 
1 
2 
9 
22 
33 
22 
8 
2 
1 
Solution
Rent (in Rs.): 
15 – 25 
25 – 35 
35 – 45 
45 – 55 
55 – 65 
65 – 75 
75 – 85 
85 – 95 
No. of houses: 
8 
10 
15 
25 
40 
20 
15 
7 
Solution
Marks below: 
10 
20 
30 
40 
50 
60 
70 
80 
No. of students: 
15 
35 
60 
84 
96 
127 
198 
250 
Solution
Variable: 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
Frequency: 
10 
20 
? 
40 
? 
25 
15 
Solution
Age in years: 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
No. of persons: 
5 
25 
? 
18 
7 
Solution
No. of accidents: 
0 
1 
2 
3 
4 
5 
Total 
Frequency (No. of days): 
46 
? 
? 
25 
10 
5 
200 
Solution
Variable: 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
70 – 80 
Frequency: 
12 
30 
 
65 
 
25 
18 
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Solution
Age(in years) 
Frequency 
Age(in years) 
Frequency 
15 – 19 
53 
40 – 44 
9 
20 – 24 
140 
45 – 49 
5 
25 – 29 
98 
50 – 54 
3 
30 – 34 
32 
55 – 59 
3 
35 – 39 
12 
60 and above 
2 
Calculate the median and interpret the results
Solution
Class interval: 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
Total 
Frequency: 
5 
F_{1} 
20 
15 
F_{2} 
5 
60 
Solution
Class interval 
Frequency 
Class interval 
Frequency 
0 – 100 
2 
500 – 600 
20 
100 – 200 
5 
600 – 700 
F_{2} 
200 – 300 
F_{1} 
700 – 800 
9 
300 – 400 
12 
800 – 900 
7 
400 – 500 
17 
900 – 1000 
4 
Solution
Class interval: 
0 – 10 
10 – 20 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
Total 
Frequency: 
F_{1} 
5 
9 
12 
F_{2} 
3 
2 
40 
Solution
(i) Marks 
No. of students 
(ii) Marks 
No. of students 
Less than 10 
0 
More than 150 
0 
Less than 30 
10 
More than 140 
12 
Less than 50 
25 
More than 130 
27 
Less than 70 
43 
More than 120 
60 
Less than 90 
65 
More than 110 
105 
Less than 110 
87 
More than 100 
124 
Less than 130 
96 
More than 90 
141 
Less than 150 
100 
More than 80 
150 
Solution
Height in cm 
Number of Girls 
Less than 140 
4 
Less than 145 
11 
Less than 150 
29 
Less than 155 
40 
Less than 160 
46 
Less than 165 
51 
Find the median height.
Solution
To calculate the median height, we need to find the class intervals and their corresponding frequencies
The given distribution being of thee less than type 140, 145, 150, ........, 165 give the upper limits of corresponding class intervals. So, the classes should be below 140 , 145, 150, .....160, 165 observe that from the given distribution , we find that there are 4  girls with height less than 140 is 4. Now there are 4 girls with heights less than 140. Therefore, the number of girls with height in the interval 140, 145 is 11  4 = 7, similarly. The frequencies of 145, 150 is 29  11 = 18, for 150  155 it is 40 29 =11, and so on so our frequencies distribution becomes.
16. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
Age in years 
Number of policy holders 
Below 20 
2 
Below 25 
6 
Below 30 
24 
Below 35 
45 
Below 40 
78 
Bwlow 45 
89 
Below 50 
90 
Below 55 
98 
Below 60 
100 
Solution
Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequencies table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can definite class intervals with their respective cumulative frequencies as below
17. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length( in mm): 
118 – 126 
127 – 135 
136 – 144 
145 – 153 
154 – 162 
163 – 171 
172 – 180 
No.of leaves: 
3 
5 
9 
12 
5 
4 
2 
Find the mean length of life.
Solution
The given data is not having continuous class intervals is 1. So, we have to add and subtract 1/2 = 0.5. upper class limits and lower class limits
Now continuous class intervals with respective cumulative frequencies can be presented as below
18. The following table gives the distribution of the life time fo 400 neon lamps:
Lite time(in hours) 
Number of lamps (1500 – 2000) 14 
2000 – 2500 
56 
2500 – 3000 
60 
3000 – 3500 
86 
3500 – 4000 
74 
4000 – 4500 
62 
4500 – 5000 
48 
Find the median life.
Solution
We can find cumulative frequencies with their respective class intervals as below
19. The distribution below gives the weight of 30 students in a class. Find the median weight f students:
Weight (in kg): 
40 – 45 
45 – 50 
50 – 55 
55 – 60 
60 – 65 
65 – 70 
70 – 75 
No. of students 
2 
3 
8 
6 
6 
3 
2 
Solution