Find the mode of the distribution from the following data.
|
Class |
Frequcncy |
|
10-15 |
3 |
|
15-20 |
2 |
|
20-25 |
10 |
|
25-30 |
7 |
|
30-35 |
20 |
|
35-40 |
5 |
|
40-45 |
8 |
Here, the maximum class frequency is 20 and the class corresponding to this frequency is 30-35, so the modal class is 30-35.
Thus, we have
Modal class = 30-35
l (lower limit of modal class) 30
Class size (h) = 5
Frequency (f1) of the modal class = 20
Frequency (f0) of class preceding the modal class = 7
and Frequency (f2) of the class succeeding the modal class = 5
Henc, mode of the given data is 32.321.
Now, substitute the values in the formula of mode, we get
Calculate the median of the following distribution of incomes of employees of a company.
|
Income |
No. of Persons |
|
400-500 |
25 |
|
500-600 |
69 |
|
600-700 |
107 |
|
700-800 |
170 |
|
800-900 |
201 |
|
900-1000 |
142 |
|
1000-1100 |
64 |
Forming the cumulative frquency table, we get
|
Income |
No. of Persons |
c.f. |
|
400-500 |
25 |
25 |
|
500-600 |
69 |
94 |
|
600-700 |
107 |
201 |
|
700-800 |
170 |
371 |
|
800-900 |
201 |
572 |
|
900-1000 |
142 |
714 |
|
1000-1100 |
64 |
778 |
Determine the value of mode from the following frequency distribution table.
|
Murks (C.I.) |
No. of students (f) |
|
0-10 |
5 |
|
10-20 |
12 |
|
20-30 |
14 |
|
30-40 |
10 |
|
40-50 |
8 |
|
50-60 |
6 |
Here, the maximum class frequency is 14 and the class corresponding to frequency is 20-30.
So, the modal class is 20-30.
Thus, we have
Modal class = 20-30
l (lower limit of modal class) = 20
h (class size) = 10
f1 = (frequency of the modal class) = 14
f0 = (frequency of class preceding the modal class) = 12
f2 = (frequency of the class succeeding the modal class) = 10
Now, substituting these values in the formula of mode, we get
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.
|
Marks |
No. of students |
|
Less than 10 |
7 |
|
Less than 20 |
21 |
|
Less than 30 |
34 |
|
Less than 40 |
46 |
|
Less than 50 |
66 |
|
Less than 60 |
77 |
|
Less than 70 |
92 |
|
Less than 80 |
100 |
|
Class Marks (C.I.) |
No. of students (c.f.) |
f |
|
Less than 10 |
7 |
7 |
|
Less than 20 |
21 |
14 |
|
Less than 30 |
34 |
13 |
|
Less than 40 |
46 |
12 |
|
Less than 50 |
66 |
20 |
|
Less than 60 |
77 |
11 |
|
Less than 70 |
92 |
15 |
|
Less than 80 |
100 |
8 |
The class (40-50) has maximum frequency i.e. 20 therefore, modal class is 40-50.
f = 20
Now, l = 40, h = 10m, f1 = 20 and f2 = 11
f0 = 12 and modal class = 40 - 50 
The median of the following data is 52.5. Find the values of x and y if the total frequency is 100.
|
Class Interval |
Frequency |
|
0-10 |
2 |
|
10-20 |
5 |
|
20-30 |
x |
|
30-40 |
12 |
|
40-50 |
17 |
|
50-60 |
20 |
|
60-70 |
y |
|
70-80 |
9 |
|
80-90 |
7 |
|
90-100 |
4 |
|
Total 100 |
|
C.l. |
fi |
c.f. |
|
0-10 |
2 |
2 |
|
10-20 |
5 |
7 |
|
20-30 |
x |
7 + x |
|
30-40 |
12 |
19 + |
|
40-50 |
17 |
36 + x |
|
50-60 |
20 |
56 + x |
|
60-70 |
y |
56 + x + y |
|
70-80 |
9 |
65 + x + y |
|
80-90 |
7 |
72 + x + y |
|
90-100 |
4 |
76 + x + y |
It is given that n = 100
∴ 76 + x + y 100 ⇒ x + y = 24 ...(i)
The median is 52.5 which lies in the class 50-60.
∴ l = 50,f = 20, c.f. = 36 + x, h = 10
Using the formula.
52.5 - 50 = (14 - x) x 0.5
2.5 = 7 -0.5x 
25 = 70 - 5x
5x = 70 - 25 = 45
x = 9
Putting this value of x in (i), we get
9 + y = 24
y = 24 - 9 = 15
Hence, x = 9 and y= 15
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