IT Sample: Constructing a Frequency Distribution

Task:

Construct a frequency distribution with suitable class interval size of marks obtained by 50 students of a class are given below:

22, 51, 34, 39, 68, 69, 9, 37, 16, 48, 36, 27, 55, 59, 60, 59, 74, 80, 19, 29, 31, 69, 62, 60, 28, 67, 51, 73, 17, 24, 44, 45, 51, 66, 68, 72, 74, 41, 57, 41, 67, 51, 53, 23, 58, 34, 39, 48, 36, 79

Solution:

Arrange the marks in ascending order as

9,16,17,19,22,23,24,27,28,29,31,34,34,36,36,37,39,39,41,41,44,45,48,48,51,51,51, 51,53,55,57,58,59,59,60,60,62,62,66,67,67,68,68,69,69,72,73,74,74,79

Minimum \ Value = 9
Maximum \ Value = 79

Range \ = \ Maximum \ Value \ - \ Minimum \ Value = 79 - 9 = 70

Number \ of \ classes = 1 + 2.1*log(N) = 1 + 2.1 * log(50) = 1 + 2.1*1.70 = 4.57 \ or \ 5 \ approximate

Class interval size:

h = \frac{Range}{No.of \_ Classes} = \frac{70}{5} = 14

Marks Number of students, f Class Boundary Class mark, x
0 – 9 1 0 – 9,5 (0+9)/2 = 4,5
10 – 19 3 9,5 – 19,5 (10+19)/2 = 14,5
20 – 29 6 19,5 – 29,5 (20+29)/2 = 24,5
30 – 39 8 29,5 – 39,5 (30+39)/2 = 34,5
40 – 49 6 39,5 – 49,5 (40+49)/2 = 44,5
50 – 59 10 49,5 – 59,5 (50+59)/2 = 54,5
60 – 69 11 59,5 – 69,5 (60+69)/2 = 64,5
70 – 79 5 69,5 – 79,5 (70+79)/2 = 74,5
50

Note: For finding the class boundaries, we take half of the difference between lower class limit of the 2nd class and upper class limit of the 1st class (20−19)/2 = 0,5. This value is subtracted from lower class limit and added in upper class limit to get the required class boundaries.

You can also read our relative frequency distribution example published before as an examplary work of our experts.

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