In this chapter, we provide RD Sharma Class 10 Ex 7.4 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 7.4 Solutions Chapter 7 Statistics pdf, Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 7 |
| Chapter Name | Statistics |
| Exercise | 7.4 |
| Category | RD Sharma Solutions |
RD Sharma Solutions for Class 10 Chapter 7 Statistics Ex 7.4 Download PDF
Chapter 7: Statistics Exercise – 7.4
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:
Lives in hours of is pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719
Arrange the above data in ascending order = 694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745
N = 15 (odd)
terms = 8th terms = 716
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,
N/2 = 420/ 2 = 120
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 – 0.27
= 165.23
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 | 99 |
| 135 – 144 | 134.5 – 144.5 | 1 | 100 |
| N = 100 |
We have N = 100 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 + 4.88
= 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(f) |
| 45 – 55 | 25 | 58 |
| 55 – 65 | 40(f) | 28 |
| 65 – 75 | 20 | 38 |
| 75 – 85 | 15 | 183 |
| 85 – 95 | 7 | 140 |
| N = 140 |
We have N = 140 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
= 55 + 3 = 58
Question: 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 |
We have N = 250, 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
= 50 + 9.35
= 59.35
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:
| 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 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
Missing frequency = 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 | 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 (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 N/2 = 357/2 = 178.5 The cumulative frequency just greater than N/2 is 193,
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, Nearly half the women were 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: | 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:
| Life time | Number of lamps fi | Cumulative frequency (cf) |
| 1500 – 2000 | 14 | 14 |
| 2000 – 2500 | 56 | 70 |
| 2500 – 3000 | 60 | 130 |
| 3000 – 3500 | 86 | 216 |
| 3500 – 4000 | 74 | 290 |
| 4000 – 4500 | 62 | 352 |
| 4500 – 5000 | 48 | 400 |
| Total (n) | 400 |
Now we may observe that cumulative frequency just greater than n/2 (400/2 = 200) is 216 belongs to class interval 3000 – 3500 Median class = 3000 – 3500
Lower limits (l) of median class = 3000 Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130 Class size (h) = 500
= 3000 + (35000/86) = 3406.98 So, median life time of lamps is 3406.98 hours
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:
We may find cumulative frequency with their respective class intervals as below:
| 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 |
Cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belonging to class interval 55 – 60 Median class = 55 – 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cf) = 13 Class size (h) = 5
= 55 + 10/6 = 56.666
So, median weight is 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 |
Given N = 200 46 + x + y + 25 + 10 + 5 = 200 x + y = 200 – 46 – 25 – 10 – 5 x + y = 114 —- (1)
And, Mean = 1.46 Sum/ N = 1.46 (x + 2y + 140)/ 200 = 1.46 x + 2y = 292 – 140 x + 2y = 152 —- (2)
Subtract equation (1) from equation (2) x + 2y – x – y = 152 – 114 y = 38
Putting the value of y in equation (1), we have x = 114 – 38 = 76
| No of accidents | No of days | Cumulative frequency |
| 0 | 46 | 46 |
| 1 | 76 | 122 |
| 2 | 38 | 160 |
| 3 | 25 | 185 |
| 4 | 10 | 195 |
| 5 | 5 | 200 |
| N = 200 |
We have, N = 200 N/2 = 200/2 = 100 The cumulative frequency just more than N/2 is 122 then the median is 1
Question: 11
An incomplete distribution is given below:
| 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 the missing frequencies.
(ii) Calculate the AM of the completed distribution.
Solution:(i)
| Class interval | Frequency | Cumulative frequency |
| 10 – 20 | 12 | 12 |
| 20 – 30 | 30 | 42 |
| 30 – 40 | x | 42+ x (F) |
| 40 – 50 | 65 (f) | 107 + x |
| 50 – 60 | Y | 107 + x + y |
| 60 – 70 | 25 | 132 + x + y |
| 70 – 80 | 18 | 150 + x + y |
| N = 150 |
Given Median = 46 Then, median class = 40 – 50, L = 40, h = 50 – 40 = 10, f = 65, F = 42 + x
Given N = 230
12 + 30 + 34 + 65 + y + 25 + 18 = 230
184 + y = 230
Y = 230 – 184
Y = 46 (ii)
| Class interval | Mid value x | Frequency f | Fx |
| 10 -20 | 15 | 12 | 180 |
| 20 – 30 | 25 | 30 | 750 |
| 30 – 40 | 35 | 34 | 1190 |
| 40 – 50 | 45 | 65 | 2925 |
| 50 – 60 | 55 | 46 | 2530 |
| 60 – 70 | 65 | 25 | 1625 |
| 70 – 80 | 75 | 18 | 1350 |
| N = 230 | Σfx = 10550 |
= 45.87
Question: 12
If the median of the following frequency distribution is 28.5 find the missing frequencies:
| Class interval: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | Total |
| Frequency: | 5 | f1 | 20 | 15 | f2 | 5 | 60 |
Solution:
| Class interval | Frequency | Cumulative frequency |
| 0 – 10 | 5 | 5 |
| 10 – 20 | f1 | 5 + f1 (F) |
| 20 – 30 | 20 (f) | 25 + f1 |
| 30 – 40 | 15 | 40 + f1 |
| 40 – 50 | f2 | 40 + f1 + f2 |
| N = 60 |
Given Median = 28.5 Then, median class = 20 – 30
17 = 25 – f1
f1 = 25 – 17 = 8
Given Sum of frequencies = 60
5 + f1+ 20 + 15 + f2 + 5 = 60
5 + 8 + 20 + 15 + f2 + 5 = 60
f2 = 7 f1 = 8 and f2 = 7
Question: 13
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data.
| Class interval | Frequency | Class interval | Frequency |
| 0 – 100 | 2 | 500 – 600 | 20 |
| 100 – 200 | 5 | 600 – 700 | f2 |
| 200 – 300 | f1 | 700 – 800 | 9 |
| 300 – 400 | 12 | 800 – 900 | 7 |
| 400 – 500 | 17 | 900 – 1000 | 4 |
Solution:
| Class interval | Frequency | Cumulative frequency |
| 0 – 100 | 2 | 2 |
| 100 – 200 | 5 | 7 |
| 200 – 300 | f1 | 7 + f1 |
| 300 – 400 | 12 | 19 + f1 |
| 400 – 500 | 17 | 36 + f1 (F) |
| 500 – 600 | 20 (f) | 56 + f1 |
| 600 – 700 | f2 | 56 + f1 + f2 |
| 700 – 800 | 9 | 65 + f1 + f2 |
| 800 – 900 | 7 | 72 + f1 + f2 |
| 900 – 1000 | 4 | 76 + f1 + f2 |
| N = 100 |
Given Median = 525 Then, median class = 500 – 600 L = 500, f = 20, F = 36 + f1, h = 600- 500 = 100
25 = (14 – f1) x 5
5 = 14 – f1
f1 = 14 – 5
f1 = 9
Given Sum of frequencies = 100
2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
85 + f2 = 100
f2 = 100 – 85 = 15
f1 = 9 and f2 = 15
Question: 14
If the median of the following data is 32.5, find the missing frequencies.
| Class interval: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 -70 | Total |
| Frequency: | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Solution:
| Class interval | Frequency | Cumulative frequency |
| 0 – 10 | f1 | f1 |
| 10 – 20 | 5 | 5 + f1 |
| 20 – 30 | 9 | 14 + f1 |
| 30 – 40 | 12 (f) | 26 + f1 |
| 40 – 50 | f2 | 26 + f1 + f2 |
| 50 – 60 | 3 | 29 + f1 + f2 |
| 60 – 70 | 2 | 31 + f1 + f2 |
| N = 40 |
Given Median = 32.5. The median class = 90 – 40 L = 30, h = 40 – 30 = 10, f = 12, F = 14 + f1
2.5 (12) = (6 – f1) * 10
30 = (6 – f1) * 10
3 = 6 – f1
f1 = 6 – 3
f1 = 3
Given Sum of frequencies = 40
f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40
3 + 5 + 9 + 12 + f2 + 3 + 2 = 40
34 + f2 = 40
f2 = 40 – 34
= 6
f1 = 3 and f2 = 6
Question: 15
Compute the median for each of the following data
| (i) | (ii) | ||
| Marks | No of students | Marks | No of students |
| Less than 10 | 0 | More than 80 | 150 |
| Less than 30 | 10 | More than 90 | 141 |
| Less than 50 | 25 | More than 100 | 124 |
| Less than 70 | 43 | More than 110 | 105 |
| Less than 90 | 65 | More than 120 | 60 |
| Less than 110 | 87 | More than 130 | 27 |
| Less than 130 | 96 | More than 140 | 12 |
| Less than 150 | 100 | More than 150 | 0 |
Solution:(i)
| Marks | No of students | Class interval | Frequency | Cumulative frequency |
| Less than 10 | 0 | 0 – 10 | 0 | 0 |
| Less than 30 | 10 | 10 – 30 | 10 | 10 |
| Less than 50 | 25 | 30 – 50 | 15 | 25 |
| Less than 70 | 43 | 50 – 70 | 18 | 43 (F) |
| Less than 90 | 65 | 70 – 90 | 22 (f) | 65 |
| Less than 110 | 87 | 90 – 110 | 22 | 87 |
| Less than 130 | 96 | 110 – 130 | 9 | 96 |
| Less than 150 | 100 | 130 – 150 | 4 | 100 |
| N = 100 |
We have N = 100 N/2 = 100/2 = 50. The cumulative frequency just greater than N/2 is 65 then median class is 70 – 90 such that L = 70, h = 90 – 70 = 20, f = 22, F = 43
= 70 + 6.36
= 76.36
(ii)
| Marks | No of students | Class interval | Frequency | Cumulative frequency |
| More than 80 | 150 | 80 – 90 | 9 | 9 |
| More than 90 | 141 | 90 – 100 | 17 | 26 |
| More than 100 | 124 | 100 – 110 | 19 | 45 (F) |
| More than 110 | 105 | 110 – 120 | 45 (f) | 90 |
| More than 120 | 60 | 120 – 130 | 33 | 123 |
| More than 130 | 27 | 130 – 140 | 15 | 138 |
| More than 140 | 12 | 140 – 150 | 12 | 150 |
| More than 150 | 0 | 150 – 160 | 0 | 150 |
| N = 150 |
We have N = 150 N/ 2 = 150/ 2 = 75.
The cumulative frequency just more than N/ 2 is 90 then the median class is 110 – 120 such that L = 70,
h = 120 – 110 = 10, f = 45, F = 45
= 110 + 6.67
= 116.67
Question: 16
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
| Height in cm | No 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 the less than type, 140, 145, 150, 155, 160, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 160-165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e. the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 – 145 is 11 – 4 = 7. Similarly, the frequency of 145 – 150 is 29 – 11 = 18, for 150 – 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with given cumulative frequencies becomes:
| Class interval | frequency | Cumulative frequency |
| Below 140 | 4 | 4 |
| 140 – 145 | 7 | 11 |
| 145 – 150 | 18 | 29 |
| 150 – 155 | 11 | 40 |
| 155 – 160 | 6 | 46 |
| 160 – 165 | 5 | 51 |
Now n = 51. S0, n/ 2 = 51/ 2 =25.5 this observation lies in the class 145 – 150 Then, L (the lower limit) = 145 cf (the cumulative frequency of the class preceding 145 – 150) = 11 f (the frequency of the median class 145 – 150) = 18 h (the class size) = 5 Using the formula,
We have Median = 145 + 72.5/ 18 = 149.03 So, the median height of the girls is 149.03 cm This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height.
Question: 17
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 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Solution:
Here class width is not same. There is no need to adjust the frequencies according to class interval. Now given frequencies table is less 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 define class intervals with their respective cumulative frequency as below.
| Age (in years) | Number of policy holders fi | Cumulative frequency (cf) |
| 18 – 20 | 2 | 2 |
| 20 – 25 | 6 – 2 = 4 | 6 |
| 25 – 30 | 24 – 6 = 18 | 24 |
| 30 – 35 | 45 – 24 = 21 | 45 |
| 35 – 40 | 78 – 45 = 33 | 78 |
| 40 – 45 | 89 – 78 = 11 | 89 |
| 45 – 50 | 92 – 89 = 3 | 92 |
| 50 – 55 | 98 – 92 = 6 | 98 |
| 55 – 60 | 100 – 98 = 2 | 100 |
| Total |
Now from table we may observe that n = 100 Cumulative frequency (cf) just greater than n/ 2 (i.e. 100/ 2 = 50) is 78 belonging to interval 35 – 40 So median class = 35 – 40
Lower limit (l) of median class = 35 Class size (h) = 5 Frequency (f) of median class = 33 Cumulative frequency (cf) of class preceding median class = 45
So median age is 35.76 years
Question: 18
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) | No of leaves |
| 118 – 126 | 3 |
| 127 – 135 | 5 |
| 136 – 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Find the mean length of life
Solution:
The given data is not having continuous class intervals. We can observe the difference between two class intervals is 1. So we have to add and subtract 1/2 = 0.5 to upper class limits and lower class limits Now continuous class intervals with respective cumulative frequencies can be represented as below:
| Length (in mm) | Number of leaves fi | Cumulative frequency (cf) |
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | 8 |
| 135.5 – 144.5 | 9 | 17 |
| 144.5 – 153.5 | 12 | 29 |
| 153.5 – 162.5 | 5 | 34 |
| 162.5 – 171.5 | 4 | 38 |
| 171.5 – 180.5 | 2 | 40 |
From the table we may observe that cumulative frequency just greater then n/2 (i.e. 40/2 = 20) is 29, belongs to class interval 144.5 – 153.5
Median class = 144.5 – 153.5 Lower limit (l) = 144.5 Class size (h) = 9 Frequency (f) of median class = 12 Cumulative frequency (c f) of class preceding median class = 17
= 144.5 + 9/4
= 146.75
So median length of leaves is 146.75 mm
Question: 19
An incomplete distribution is given as follows:
| Variable: | 0 – 10 | 10 – 20 | 20 -30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 -70 |
| Frequency: | 10 | 20 | ? | 40 | ? | 25 | 15 |
You are given that the median value is 35 and sum is all the frequencies are 170. Using the median formula, fill up the missing frequencies
Solution:
| Class interval | Frequency | Cumulative frequency |
| 0 – 10 | 10 | 10 |
| 10 – 20 | 20 | 30 |
| 20 – 30 | f1 | 30 + f1 (F) |
| 30 – 40 | 40 (F) | 70 + f1 |
| 40 – 50 | f2 | 70 + f1 + f2 |
| 50 – 60 | 25 | 95 + f1 + f2 |
| 60 – 70 | 15 | 110 + f1 + f2 |
| N = 170 |
Given Median = 35 Then median class = 30 – 40 L = 30, h = 40 – 30 = 10, f = 40, F = 30 + f1
20 = 55 – f1
f1 = 55 – 20 = 35
Given Sum of frequencies = 170
10 + 20 + fi + 40 + f2 + 25 + 15 = 170
10 + 20 + 35 + 40 + f2 + 25 + 15 = 170
f2 = 25, f1 = 35 and f2 =25
All Chapter RD Sharma Solutions For Class10 Maths
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