Respuesta :
Answer:
The frequency table is shown below.
Step-by-step explanation:
The data set arranged ascending order is:
S = {33 , 34 , 39 , 48 , 49 , 50 , 53 , 54 , 55 , 56 , 58 , 58, 60 , 63 , 64 , 65 , 70 , 71 , 74 , 84}
It is asked to use the minimum value from the data set as the lower class limit for the first row.
So, the lower class limit for the first class interval is 33.
To determine the class width compute the range as follows:
[tex]\text{Range}=\text{Maximum}-\text{Minimum}[/tex]
[tex]=84-33\\=51[/tex]
The number of classes requires is 5.
The class width is:
[tex]\text{Class width}=\frac{Range}{5}=\frac{51}{2}=10.2\approx 10[/tex]
So, the class width is 10.
The classes are:
33 - 42
43 - 52
53 - 62
63 - 72
73 - 82
83 - 92
Compute the frequencies of each class as follows:
Class Interval Values Frequency
33 - 42 33 , 34 , 39 3
43 - 52 48 , 49 , 50 3
53 - 62 53 , 54 , 55 , 56 , 58 , 58, 60 7
63 - 72 63 , 64 , 65 , 70 , 71 5
73 - 82 74 1
83 - 92 84 1
TOTAL 20
Compute the relative frequencies as follows:
Class Interval Frequency Relative Frequency
33 - 42 3 [tex]\frac{3}{20}\times 100\%=15\%[/tex]
43 - 52 3 [tex]\frac{3}{20}\times 100\%=15\%[/tex]
53 - 62 7 [tex]\frac{7}{20}\times 100\%=35\%[/tex]
63 - 72 5 [tex]\frac{5}{20}\times 100\%=25\%[/tex]
73 - 82 1 [tex]\frac{1}{20}\times 100\%=5\%[/tex]
83 - 92 1 [tex]\frac{1}{20}\times 100\%=5\%[/tex]
TOTAL 20 100%
