RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.4

The median is the middle of a distribution. In this exercise, students will practice finding the median of a discrete and grouped frequency distribution. For quick access to solutions, the RD Sharma Solutions Class 10 is the best place. It has well-structured solutions in simple language to match the level of all the students. For detailed studies regarding this exercise, download RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.4 PDF provided below.

RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.4 Download PDF

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Access RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.4

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

= 8th term

So, the 8th 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

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 1

= 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

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 2

= 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 median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10

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= 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 median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10

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= 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

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 5

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

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Median = 23.98

Which 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, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 7

= 3000 + (35000/86)

= 3406.98

Thus, the median life time 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, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 8

= 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), 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 become:

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.

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