The marks obtained in the previous class test out of 20 were as follows:
19 10 9 8 14 11
11 12 11 9 19 19
17 10 13 18 16 17
5 14 11 15 18 12
8 7 10 16 4 12
Make the class width 3 and the lower limit of the first class 2. With the given data complete the following table:
Intervals / Intervalle Class Midpoints / Klasmiddelpunte Frequencies / Frekwensies Cumulative frequency / Kumulatiewe frekwensies Relative Cumulative frequency / Relatiewe Kumulatiewe frekwensies Relative frequency / Relatiewe frekwensie
[ - ) G
[ - ) C E I
[ A - ) B
[ - ) H
[ - ) F J
[ - ) D
Total
What is the value of H in the table?
30%
10%
0.333
0.6
To create a frequency distribution table, we first need to calculate the class boundaries and frequencies.
Given data:
19 10 9 8 14 11 11 12 11 9 19 19 17 10 13 18 16 17 5 14 11 15 18 12 8 7 10 16 4 12
Class width = 3
Lower limit of first class = 2
Class boundaries:
[2 - 5)
[5 - 8)
[8 - 11)
[11 - 14)
[14 - 17)
[17 - 20)
Frequency distribution table:
Intervals / Intervalle Class Midpoints / Klasmiddelpunte Frequencies / Frekwensies Cumulative frequency / Kumulatiewe frekwensies Relative Cumulative frequency / Relatiewe Kumulatiewe frekwensies Relative frequency / Relatiewe frekwensie
[2 - 5) 3.5 9 9 0.3 0.3
[5 - 8) 6.5 5 14 0.467 0.2333
[8 - 11) 9.5 10 24 0.8 0.333
[11 - 14) 12.5 8 32 1.067 0.2667
[14 - 17) 15.5 6 38 1.267 0.2
[17 - 20) 18.5 3 41 1.367 0.1
Total - - 41 - -
The value of H in the table is "0.6" which represents the relative cumulative frequency for the [8 - 11) class interval.