RS Aggarwal Class 9 Ex 14G Chapter 14
Q.1: Find the mode of the following items.
0,6,5,1,6,4,3,0,2,6,5,6
Sol:
Arrange the given data in ascending order, we have:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Observations (x) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 2 | 1 | 1 | 1 | 1 | 2 | 4 |
As 6 occurs the maximum number of times i.e. 4, mode = 6
Q.2: Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20
Sol:
Arranging the given data in ascending order, we have
15, 20, 22, 23, 25, 25, 25, 27, 40
The frequency table of the data is:
Observations (x) | 15 | 20 | 22 | 23 | 25 | 27 | 40 |
Frequency | 1 | 1 | 1 | 1 | 3 | 1 | 1 |
As 25 occurs the maximum number of times i.e. 3, mode= 25
Q.3: Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9
Sol:
Arranging the given data in ascending order, we have: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
The frequency table of the data is:
Observations (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency | 2 | 1 | 2 | 1 | 2 | 2 | 1 | 1 | 5 |
As 9, occurs the maximum number of times i.e. 5, mode = 9
Q.4: A cricket player scored the following runs in 12 one-day matches: 50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35. Find the modal score.
Sol:
Arranging the given data in ascending order, we have:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
The frequency table of the data is:
Observations (x) | 9 | 19 | 27 | 28 | 30 | 32 | 35 | 50 | 60 |
Frequency | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 4 | 1 |
As 50, occurs the maximum number of times i.e. 4, mode = 50
Thus, the modal score of the cricket player is 50.
Q.5: Calculate the mode of each of the following using the empirical formula: 17, 10, 12, 11, 10, 15, 14, 11, 12, 13
Sol:
Arranging the given data in ascending order, we have: 10, 10, 11, 11, 12, 12, 13, 14, 15, 17.
We prepare the table which is as given below:
Items (x) | Frequency (f) | Cumulative frequency | |
10 | 2 | 2 | 20 |
11 | 2 | 4 | 22 |
12 | 2 | 6 | 24 |
13 | 1 | 7 | 13 |
14 | 1 | 8 | 14 |
15 | 1 | 9 | 15 |
17 | 1 | 10 | 17 |
N=10 |
Here, n = 10, which is even.
Median =
=
=
=
Therefore, Median= 12
Now,
Therefore, Mean =
Mode = 3(Median) – 2(Mean) = 3(12) – 2(12.5) = 36 – 25
Thus, Mode = 11
Q.6:
Marks | 10 | 11 | 12 | 13 | 14 | 16 | 19 | 20 |
Number of students | 3 | 5 | 4 | 5 | 2 | 3 | 2 | 1 |
Sol:
We may prepare the table, given below:
Marks (x) | No. of students
(f) |
Cumulative Frequency | |
10 | 3 | 3 | 30 |
11 | 5 | 8 | 55 |
12 | 4 | 12 | 48 |
13 | 5 | 17 | 65 |
14 | 2 | 19 | 28 |
16 | 3 | 22 | 48 |
19 | 2 | 24 | 38 |
20 | 1 | 25 | 20 |
N=25 |
Here n= 25, which is odd.
Median =
=
Value of the 13th term = 13
Now,
So, mean =
Mode = 3(Median) – 2(Mean)
= 3(13) – 2(13.28) = 39 – 26.56 = 12.44
Thus, mode= 12.4
Q.7:
Items | 5 | 7 | 9 | 12 | 14 | 17 | 19 | 21 |
Frequency | 6 | 5 | 3 | 6 | 5 | 3 | 2 | 4 |
Sol:
We may prepare the table, given below:
Items (x) | Frequency (f) | Cumulative Frequency | |
5 | 6 | 6 | 30 |
7 | 5 | 11 | 35 |
9 | 3 | 14 | 27 |
12 | 6 | 20 | 72 |
14 | 5 | 25 | 70 |
17 | 3 | 28 | 51 |
19 | 2 | 30 | 38 |
21 | 4 | 34 | 84 |
N= |
Here, n = 34, which is even.
Median =
=
=
=
Therefore, Median = 12
Now,
Therefore, Mean =
Mode = 3(Median) – 2(Mean)
= 3(12) – 2(11.97)
= 36-23.94
Thus, Mode =12.06
Q.8:
x | 18 | 20 | 25 | 30 | 34 | 38 | 40 |
f | 6 | 7 | 3 | 7 | 7 | 5 | 5 |
Sol:
We may prepare the table, given below:
(x) | Frequency (f) | Cumulative Frequency | |
18 | 6 | 6 | 108 |
20 | 7 | 13 | 140 |
25 | 3 | 16 | 75 |
30 | 7 | 23 | 210 |
34 | 7 | 30 | 238 |
38 | 5 | 35 | 190 |
40 | 5 | 40 | 200 |
N= |
Here, n = 40, which is even.
Median =
=
=
=
Therefore, Median = 30
Now,
Therefore, Mean =
Mode = 3(Median) – 2(Mean)
= 3(30) – 2(29.025) = 31.95
Thus, Mode = 32
Q.9: The table given below shows the weight (in kg) of 50 persons:
Weight (in kg) | 42 | 47 | 52 | 57 | 62 | 67 | 72 |
No. of persons (f) | 3 | 8 | 6 | 8 | 11 | 5 | 9 |
Find the mean, median and mode.
Sol:
We may prepare the table, given below:
Weight (in kg) | No.of persons (f) | Cumulative frequency | |
42 | 3 | 3 | 126 |
47 | 8 | 11 | 376 |
52 | 6 | 17 | 312 |
57 | 8 | 25 | 456 |
62 | 11 | 36 | 682 |
67 | 5 | 41 | 335 |
72 | 9 | 50 | 648 |
N= |
Here,
Mean =
Therefore, mean weight = 58.7 kg
Here, N = 50 which is even.
Therefore, Median =
=
=
=
Therefore, Median weight = 59.5 kg
Mode = 3(Median) – 2(Mean)
= 3(59.5) – 2(58.7) = 178.5 – 117.4 = 61.1
Thus, Mode weight = 61.1 kg
Thus we have: Mean = 58.7 kg, Median = 59.5 kg and Mode = 61.1 kg
Q.10: The marks obtained by 80 students in a test are given below:
Marks | 4 | 12 | 20 | 28 | 36 | 44 |
No. of students | 8 | 10 | 16 | 24 | 15 | 7 |
Find the modal marks.
Sol:
We may prepare the table, given below:
Marks (x) | No. of students
(f) |
Cumulative frequency | |
4 | 8 | 8 | 32 |
12 | 10 | 18 | 120 |
20 | 16 | 34 | 320 |
28 | 24 | 58 | 672 |
36 | 15 | 73 | 540 |
44 | 7 | 80 | 308 |
N= |
Here, n = 80, which is even
Median =
=
=
=
Therefore, Median= 28
Now,
Therefore, Mean =
Mode = 3(Median) – 2(Mean)
= 3(28) – 2(24.9) = 84 – 49.8 = 34.2
Thus, Mode = 34.2
Q.11: The ages of the employees of a company are given below:
Age (in years) | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
No. of persons | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
Find the mean, median and mode of the above data.
Sol:
We may prepare the table, given below:
Age (in years)
(x) |
No. of persons
(f) |
Cumulative frequency | |
19 | 13 | 13 | 247 |
21 | 15 | 28 | 315 |
23 | 16 | 44 | 368 |
25 | 18 | 62 | 450 |
27 | 16 | 78 | 432 |
29 | 15 | 93 | 435 |
31 | 13 | 106 | 403 |
N= |
Here,
Mean =
Therefore, mean =25
Here, N = 106 which is even.
Therefore, Median =
=
=
=
Therefore, Median = 25
Mode = 3(Median) – 2(Mean)
= 3(25) – 2(25)
= 75 – 50 =25
Mode = 25
Thus we have, Mean = 25, Median = 25 and Mode = 25
Q.12: The following table shows the weight of 12 students:
Weight in kg | 47 | 50 | 53 | 56 | 60 |
No. of students | 4 | 3 | 2 | 2 | 4 |
Find the mean, median and mode for the above data.
Sol:
We may prepare the table, given as:
Weight in kg
(x) |
No. of students
(f) |
Cumulative frequency | |
47 | 4 | 4 | 188 |
50 | 3 | 7 | 150 |
53 | 2 | 9 | 106 |
56 | 2 | 11 | 112 |
60 | 4 | 15 | 240 |
N= |
Here,
Mean =
Therefore, mean = 53.06
Here, N= 15 which is odd.
Therefore, Median =
=
= value of 8th term = 53
Therefore, Median = 53
Mode = 3 (Median) -2 (Mean)
= 3(53) – 2(53.06)
= 159 – 106.12 = 52.88
Mode = 52.88
Thus we have: Mean = 53.06, Median = 53 and Mode = 52.88