RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4:
Chapter 7 of RD Sharma's Class 10 Maths book delves into Statistics, with Exercise 7.4 focusing on measures of central tendency, specifically the mean, median, and mode of grouped data. The exercise teaches students methods to compute these values efficiently using class intervals and frequencies.
Topics covered include calculating the mean using the direct, assumed mean, and step-deviation methods; finding the median by locating the cumulative frequency; and determining the mode by using the modal class. These concepts help students understand data summarization and interpretation, enhancing their skills in handling real-life statistical data.
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4 Overview
Chapter 7, Exercise 7.4 of RD Sharma's Class 10 Maths book focuses on Statistics, specifically on calculating measures of central tendency like mean, median, and mode. These concepts are essential for students as they form the basis for data interpretation, analysis, and real-life applications such as surveys and predictions.
This exercise helps build a strong foundation in understanding data distribution and variability, which is crucial for advanced studies in mathematics, economics, and science. By solving these questions, students gain proficiency in handling data systematically, preparing them for competitive exams and practical applications in various fields.
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4 PDF
Below, we have provided the PDF for RD Sharma Solutions for Class 10 Maths, Chapter 7, Exercise 7.4 on Statistics. This comprehensive solution guide will help you understand and solve problems related to statistics concepts covered in this exercise. With step-by-step explanations, students can easily grasp complex topics and enhance their problem-solving skills. Download the PDF to get detailed answers and clear explanations to help you prepare effectively for your exams.
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4 PDF
RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4 Statistics
Below is the RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4 Statistics -
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:
Arranging the given data in ascending order, we have
694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745
As the number of terms is an old number, i.e., N = 15
We use the following procedure to find the median.
Median = (N + 1)/2
th
term
= (15 + 1)/2
th
term
= 8
th
term
So, the 8
th
term in the arranged order of the given data should be the median.
Therefore, 716 is the median of the data.
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:
|
Class interval (exclusive)
|
Class interval (inclusive)
|
Class interval frequency
|
Cumulative frequency
|
|
160 – 162
|
159.5 – 162.5
|
15
|
15
|
|
163 – 165
|
162.5 – 165.5
|
118
|
133(F)
|
|
166 – 168
|
165.5 – 168.5
|
142(f)
|
275
|
|
169 – 171
|
168.5 – 171.5
|
127
|
402
|
|
172 – 174
|
171.5 – 174.5
|
18
|
420
|
|
|
|
N = 420
|
|
Here, we have N = 420,
So, N/2 = 420/ 2 = 210
The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such that
L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3
= 165.5 + 1.63
= 167.13
3. Following is the distribution of I.Q. of 100 students. Find the median I.Q.
|
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:
|
Class interval (exclusive)
|
Class interval (inclusive)
|
Class interval frequency
|
Cumulative frequency
|
|
55 – 64
|
54.5 – 64-5
|
1
|
1
|
|
65 – 74
|
64.5 – 74.5
|
2
|
3
|
|
75 – 84
|
74.5 – 84.5
|
9
|
12
|
|
85 – 94
|
84.5 – 94.5
|
22
|
34(F)
|
|
95 – 104
|
94.5 – 104.5
|
33(f)
|
67
|
|
105 – 114
|
104.5 – 114.5
|
22
|
89
|
|
115 – 124
|
114.5 – 124.5
|
8
|
97
|
|
125 – 134
|
124.5 – 134.5
|
2
|
98
|
|
135 – 144
|
134.5 – 144.5
|
1
|
100
|
|
|
|
N = 100
|
|
Here, we have N = 100,
So, N/2 = 100/ 2 = 50
The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) such that L = 94.5, F = 33, h = (104.5 – 94.5) = 10
= 94.5 + 4.85
= 99.35
4. Calculate the median from the following data:
|
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:
|
Class interval
|
Frequency
|
Cumulative frequency
|
|
15 – 25
|
8
|
8
|
|
25 – 35
|
10
|
18
|
|
35 – 45
|
15
|
33
|
|
45 – 55
|
25
|
58(F)
|
|
55 – 65
|
40(f)
|
98
|
|
65 – 75
|
20
|
118
|
|
75 – 85
|
15
|
133
|
|
85 – 95
|
7
|
140
|
|
|
N = 140
|
|
Here, we have N = 140,
So, N/2 = 140/ 2 = 70
The cumulative frequency just greater than N/ 2 is 98 then the median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10
= 55 + 3 = 58
5. Calculate the median from the following data:
|
Marks below:
|
10 – 20
|
20 – 30
|
30 – 40
|
40 – 50
|
50 – 60
|
60 – 70
|
70 – 80
|
85 – 95
|
|
No of students:
|
15
|
35
|
60
|
84
|
96
|
127
|
198
|
250
|
Solution:
|
Marks below
|
No. of students
|
Class interval
|
Frequency
|
Cumulative frequency
|
|
10
|
15
|
0 – 10
|
15
|
15
|
|
20
|
35
|
10 – 20
|
20
|
35
|
|
30
|
60
|
20 – 30
|
25
|
60
|
|
40
|
84
|
30 – 40
|
24
|
84
|
|
50
|
96
|
40 – 50
|
12
|
96(F)
|
|
60
|
127
|
50 – 60
|
31(f)
|
127
|
|
70
|
198
|
60 – 70
|
71
|
198
|
|
80
|
250
|
70 – 80
|
52
|
250
|
|
|
|
|
N = 250
|
|
Here, we have N = 250,
So, N/2 = 250/ 2 = 125
The cumulative frequency just greater than N/ 2 is 127, then the median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10
= 50 + 9.35
= 59.35
6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
|
Age in years:
|
0 – 10
|
10 – 20
|
20 – 30
|
30 – 40
|
40 – 50
|
|
No of persons:
|
5
|
25
|
?
|
18
|
7
|
Solution:
Let the unknown frequency be taken as x,
|
Class interval
|
Frequency
|
Cumulative frequency
|
|
0 – 10
|
5
|
5
|
|
10 – 20
|
25
|
30(F)
|
|
20 – 30
|
x (f)
|
30 + x
|
|
30 – 40
|
18
|
48 + x
|
|
40 – 50
|
7
|
55 + x
|
|
|
N = 170
|
|
It’s given that
Median = 24
Then, median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30
4x = 275 + 5x – 300
4x – 5x = – 25
– x = – 25
x = 25
Therefore, the Missing frequency = 25
7. The following table gives the frequency distribution of married women by age at marriage.
|
Age (in years)
|
Frequency
|
Age (in years)
|
Frequency
|
|
15 – 19
|
53
|
40 – 44
|
9
|
|
20 – 24
|
140
|
45 – 49
|
5
|
|
25 – 29
|
98
|
45 – 49
|
3
|
|
30 – 34
|
32
|
55 – 59
|
3
|
|
35 – 39
|
12
|
60 and above
|
2
|
Calculate the median and interpret the results.
Solution:
|
Class interval (exclusive)
|
Class interval (inclusive)
|
Frequency
|
Cumulative frequency
|
|
15 – 19
|
14.5 – 19.5
|
53
|
53 (F)
|
|
20 – 24
|
19.5 – 24.5
|
140 (f)
|
193
|
|
25 – 29
|
24.5 – 29.5
|
98
|
291
|
|
30 – 34
|
29.5 – 34.5
|
32
|
323
|
|
35 – 39
|
34.5 – 39.5
|
12
|
335
|
|
40 – 44
|
39.5 – 44.5
|
9
|
344
|
|
45 – 49
|
44.5 – 49.5
|
5
|
349
|
|
50 – 54
|
49.5 – 54.5
|
3
|
352
|
|
55 – 54
|
54.5 – 59.5
|
3
|
355
|
|
60 and above
|
59.5 and above
|
2
|
357
|
|
|
|
N =357
|
|
Here, we have N = 357,
So, N/2 = 357/ 2 = 178.5
The cumulative frequency just greater than N/2 is 193, so then the median class is (19.5 – 24.5) such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5
Median = 23.98
This means nearly half the women were married between the ages of 15 and 25
8. The following table gives the distribution of the life time of 400 neon lamps:
|
Life 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:
|
Life time
|
Number of lamps fi
|
Cumulative frequency (cf)
|
|
1500 – 2000
|
14
|
14
|
|
2000 – 2500
|
56
|
70
|
|
2500 – 3000
|
60
|
130(F)
|
|
3000 – 3500
|
86(f)
|
216
|
|
3500 – 4000
|
74
|
290
|
|
4000 – 4500
|
62
|
352
|
|
4500 – 5000
|
48
|
400
|
|
|
N = 400
|
|
It’s seen that the cumulative frequency just greater than n/2 (400/2 = 200) is 216, and it belongs to the class interval 3000 – 3500, which becomes the Median class = 3000 – 3500
Lower limits (l) of median class = 3000 and,
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
And the Class size (h) = 500
Thus, by calculating the median by the formula, we get
= 3000 + (35000/86)
= 3406.98
Thus, the median lifetime of lamps is 3406.98 hours
9. The distribution below gives the weight of 30 students in a class. Find the median weight of 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:
|
Weight (in kg)
|
Number of students fi
|
Cumulative frequency (cf)
|
|
40 – 45
|
2
|
2
|
|
45 – 50
|
3
|
5
|
|
50 – 55
|
8
|
13
|
|
55 – 60
|
6
|
19
|
|
60 – 65
|
6
|
25
|
|
65 – 70
|
3
|
28
|
|
70 – 75
|
2
|
30
|
It’s seen that the cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.
So, it’s chosen that
Median class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) = 13
And, Class size (h) = 5
Thus, by calculating the median by the formula, we get
= 55 + 10/6 = 56.666
So, the median weight is 56.67 kg.
10. Find the missing frequencies and the median for the following distribution if the mean is 1.46
|
No. of accidents:
|
0
|
1
|
2
|
3
|
4
|
5
|
Total
|
|
Frequencies (no. of days):
|
46
|
?
|
?
|
25
|
10
|
5
|
200
|
Solution:
|
No. of accidents (x)
|
No. of days (f)
|
fx
|
|
0
|
46
|
0
|
|
1
|
x
|
x
|
|
2
|
y
|
2y
|
|
3
|
25
|
75
|
|
4
|
10
|
40
|
|
5
|
5
|
25
|
|
|
N = 200
|
Sum = x + 2y + 140
|
It’s given that N = 200
⇒ 46 + x + y + 25 + 10 + 5 = 200
⇒ x + y = 200 – 46 – 25 – 10 – 5
⇒ x + y = 114 —- (i)
And also given, Mean = 1.46
⇒ Sum/ N = 1.46
⇒ (x + 2y + 140)/ 200 = 1.46
⇒ x + 2y = 292 – 140
⇒ x + 2y = 152 —- (ii)
Subtract equation (i) from equation (ii), and we get
x + 2y – x – y = 152 – 114
⇒ y = 38
Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76
Thus, the table becomes:
|
No. of accidents (x)
|
No. of days (f)
|
Cumulative frequency
|
|
0
|
46
|
46
|
|
1
|
76
|
122
|
|
2
|
38
|
160
|
|
3
|
25
|
185
|
|
4
|
10
|
195
|
|
5
|
5
|
200
|
|
|
N = 200
|
|
It’s seen that,
N = 200 N/2 = 200/2 = 100
So, the cumulative frequency just more than N/2 is 122
Therefore, the median is 1.
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.4
Solving RD Sharma Solutions for Class 10 Maths, Chapter 7 (Statistics), Exercise 7.4 offers multiple benefits:
In-Depth Understanding of Statistics Concepts
: This exercise covers key statistical concepts such as mean, median, and mode, helping students to understand their applications and calculations thoroughly.
Improves Problem-Solving Skills
: By working through these solutions, students enhance their analytical and problem-solving skills, learning different methods to approach statistical problems effectively.
Clarification of Step-by-Step Methods
: RD Sharma’s solutions provide detailed, step-by-step explanations for each question, which is particularly helpful for students in grasping the sequential approach to solving complex statistical problems.
Preparation for Exams
: These exercises closely align with CBSE exam patterns, helping students to get accustomed to the types of questions that may appear in their exams, ultimately boosting confidence and preparedness.
Practice in Data Interpretation
: The exercise focuses on data interpretation skills, which are crucial not only for exams but also for real-world applications where statistical data is essential.
Builds Speed and Accuracy
: Regular practice with RD Sharma solutions helps students improve their calculation speed and accuracy, both essential for handling time constraints in exams.
