The median is the middle of a distribution. In this exercise, students will practise 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
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
= 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.
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