Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1

Last Updated : 23 Jul, 2025

In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various fields. The problems in this exercise are designed to help students gain a deeper understanding of the statistical concepts and their applications.

Statistics

Statistics is the branch of mathematics that deals with collecting, organizing, analyzing, and interpreting numerical data. It helps in making informed decisions by identifying patterns, trends, and relationships within the data. Understanding statistics is essential for solving real-world problems and making data-driven decisions.

Question 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:

On arranging the observations in ascending order, we have
694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745
Number of terms in the observation sequence is odd, i.e., N = 15
Now,
Median = (N + 1)/2^thterm
= (15 + 1)/2^th term
= 8^th term
Therefore, 716, which is the 8^th term is the median of the data.

Question 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
We have N = 420,
So, N/2 = 420/ 2 = 210
Now, The cumulative frequency just greater than N/2 is 275
Therefore, 165.5 – 168.5 is the median class s.t
L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3
= 165.5 + 1.63
= 167.13

Question 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
N = 100,
Therefore, N/2 = 100/ 2 = 50
The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) s.t,
L = 94.5, F = 33, h = (104.5 – 94.5) = 10
= 94.5 + 4.85
= 99.35

Question 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
N = 140,
And, N/2 = 140/ 2 = 70
The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 s.t,
L = 55, f = 40, F = 58, h = 65 – 55 = 10

Question 5. Calculate the median from the following data:

Marks below:	0-10	10 – 20	20 – 30	30-40	40-50	50-60	60-70	70-80
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
N = 250,
And, N/2 = 250/ 2 = 125
The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 s.t,
L = 50, f = 31, F = 96, h = 60 -50 = 10

Question 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 us assume the unknown frequency to be 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
Given: Median = 24
Therefore,
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, x = 25

Question 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
N = 357,
And, N/2 = 357/ 2 = 178.5
The cumulative frequency just greater than N/2 is 193,
Therefore, median class is 19.5 – 24.5 s.t
l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5
Median = 23.98, that implies that nearly half of the women are married between the ages of 15 and 25.

Question 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 86 216
3500 – 4000 74 290
4000 – 4500 62 352
4500 – 5000 48 400
N = 400
Now
N = 400
And the cumulative frequency just greater than n/2 (= 200) is 216, which belongs to the class interval 3000 – 3500
Median class = 3000 – 3500. Therefore,
(l) = 3000 and,(f) of median class = 86, (cf) of class preceding median class = 130 and (h) = 500
We have,
= 3000 + (35000/86)
= 3406.98 hrs, which is the median time of lamps.

Question 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
The cf value just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.
Therefore,
Median class = 55 – 60
where,
(l) of median class = 55, (f) of median class = 6, (cf) = 13 and (h) = 5
We have,
= 55 + 10/6 = 56.666 which is approximately 56.67 kg.

Question 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
Since, we know,
N = 200
Substituting values, we get,
⇒ 46 + x + y + 25 + 10 + 5 = 200
⇒ x + y = 200 – 46 – 25 – 10 – 5
⇒ x + y = 114...... (i)
Also, Mean = 1.46
⇒ Sum/ N = 1.46
Substituting values,
⇒ (x + 2y + 140)/ 200 = 1.46
⇒ x + 2y = 292 – 140
⇒ x + 2y = 152 .......(ii)
Solving from (i) and (ii), we get
x + 2y – x – y = 152 – 114
⇒ y = 38
And, x = 114 – 38 = 76 (from equation (i))
Now, putting the values, we get,
N = 200 N/2 = 200/2 = 100
So, the cumulative frequency just greater than N/2 is 122
And, therefore, the median is 1.

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