QUESTION 1
Consider the following 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 2 12
Construct a frequency distribution by making the class width 3 and the lower limit of the first class 2.
Intervals / Intervalle Class Midpoints / Klasmiddelpunte Frequencies / Frekwensies Cumulative frequency / Kumulatiewe frekwensies Relative frequency / Relatiewe frekwensie
[ - )
[ A - ) D H J
[ - ) F
[ - B ) K
[ - ) E I
[ - ) G
Total C
What is the value of J in the table?
A)6.7
B)3.3
C)23.3
D)26.7
To find the value of J in the table, we need to calculate the cumulative frequency up to the class interval D.
Given that the class width is 3 and the lower limit of the first class is 2, we can find the upper limit of class D:
Upper limit of class D = Lower limit of class D + Class width
Upper limit of class D = 2 + 3
Upper limit of class D = 5
Counting the frequencies given in the data between 2 and 5 (inclusive), we get:
19, 10, 9, 8, 14, 11, 11, 12, 11, 9
Adding these together, we get:
19 + 10 + 9 + 8 + 14 + 11 + 11 + 12 + 11 + 9 = 114
Therefore, the value of J in the table is 114.
Answer: J) 114