Question

the data represent the time, in minutes, spent reading a political blog in a day. construct a frequency distribution using 5 classes. in the table, include the midpoints, relative frequencies, and cumulative frequencies. which class has the greatest frequency and which has the least frequency?
complete the table, starting with the lowest class limit.
(simplify your answers.)
which class has the greatest frequency?
the class with the greatest frequency is from □ to □
which class has the least frequency?
the class with the least frequency is from □ to □

Explanation:

Step1: Organize the data

First, list out all the data values: \(0,1,3,3,4,4,7,8,10,17,18,18,20,22,23,23,29\)

Step2: Calculate the class width

The range of the data \(R=\text{Max}-\text{Min}=29 - 0=29\). Since we want \(5\) classes, the class width \(w=\lceil\frac{R}{5}
ceil=\lceil\frac{29}{5}
ceil = 6\) (using ceiling function to get a whole - number class width)

Step3: Define the classes

The first class is \(0 - 5\), the second is \(6 - 11\), the third is \(12 - 17\), the fourth is \(18 - 23\), the fifth is \(24 - 29\)

Step4: Calculate the frequency

• For the class \(0 - 5\): The data values \(0,1,3,3,4,4\) give a frequency \(f_1 = 6\)
• For the class \(6 - 11\): The data values \(7,8,10\) give a frequency \(f_2=3\)
• For the class \(12 - 17\): The data value \(17\) gives a frequency \(f_3 = 1\)
• For the class \(18 - 23\): The data values \(18,18,20,22,23,23\) give a frequency \(f_4=6\)
• For the class \(24 - 29\): The data value \(29\) gives a frequency \(f_5 = 1\)

Step5: Calculate the mid - point

The mid - point formula is \(M=\frac{\text{Lower limit}+\text{Upper limit}}{2}\)

• For the class \(0 - 5\): \(M_1=\frac{0 + 5}{2}=2.5\)
• For the class \(6 - 11\): \(M_2=\frac{6+11}{2}=8.5\)
• For the class \(12 - 17\): \(M_3=\frac{12 + 17}{2}=14.5\)
• For the class \(18 - 23\): \(M_4=\frac{18+23}{2}=20.5\)
• For the class \(24 - 29\): \(M_5=\frac{24 + 29}{2}=26.5\)

Step6: Calculate the relative frequency

The relative frequency formula is \(rf=\frac{f}{n}\), where \(n=\sum f=6 + 3+1+6+1=17\)

• For the class \(0 - 5\): \(rf_1=\frac{6}{17}\approx0.353\)
• For the class \(6 - 11\): \(rf_2=\frac{3}{17}\approx0.176\)
• For the class \(12 - 17\): \(rf_3=\frac{1}{17}\approx0.059\)
• For the class \(18 - 23\): \(rf_4=\frac{6}{17}\approx0.353\)
• For the class \(24 - 29\): \(rf_5=\frac{1}{17}\approx0.059\)

Step7: Calculate the cumulative frequency

The cumulative frequency formula is \(CF=\sum_{i = 1}^{k}f_i\) (where \(k\) is the class number)

• For the class \(0 - 5\): \(CF_1 = 6\)
• For the class \(6 - 11\): \(CF_2=6 + 3=9\)
• For the class \(12 - 17\): \(CF_3=9+1 = 10\)
• For the class \(18 - 23\): \(CF_4=10 + 6=16\)
• For the class \(24 - 29\): \(CF_5=16+1=17\)

The completed table:

Class	Frequency	Mid - point	Relative Frequency	Cumulative Frequency
6 - 11	3	8.5	0.176	9
12 - 17	1	14.5	0.059	10
18 - 23	6	20.5	0.353	16
24 - 29	1	26.5	0.059	17

Step8: Determine the class with the greatest and least frequency

The class with the greatest frequency:
The frequencies are \(6,3,1,6,1\). The classes \(0 - 5\) and \(18 - 23\) have the greatest frequency.
The class with the least frequency:
The classes \(12 - 17\) and \(24 - 29\) have the least frequency.

Answer:

The class with the greatest frequency is from \(0\) to \(5\) and from \(18\) to \(23\).
The class with the least frequency is from \(12\) to \(17\) and from \(24\) to \(29\).