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.) class frequency midpoint relative frequency cumulative frequency

Explanation:

Step1: Find range

The data set is \(0,0,1,1,2,3,5,6,7,7,7,8,8,11,11,18,18,19\). The maximum value \(M = 19\) and the minimum value \(m=0\). The range \(R = M - m=19 - 0 = 19\).

Step2: Determine class - width

We want 5 classes. The class - width \(w=\frac{R}{n}\), where \(n = 5\). So \(w=\frac{19}{5}=3.8\). We round up to \(w = 4\).

Step3: Define classes

The first class starts at 0. The classes are \(0 - 3\), \(4 - 7\), \(8 - 11\), \(12 - 15\), \(16 - 19\).

Step4: Calculate frequencies

• For the class \(0 - 3\): The values in this class are \(0,0,1,1,2,3\), so the frequency \(f_1=6\).
• For the class \(4 - 7\): The values are \(5,6,7,7,7\), so the frequency \(f_2 = 5\).
• For the class \(8 - 11\): The values are \(8,8,11,11\), so the frequency \(f_3=4\).
• For the class \(12 - 15\): There are no values in this class, so the frequency \(f_4 = 0\).
• For the class \(16 - 19\): The values are \(18,18,19\), so the frequency \(f_5=3\).

Step5: Calculate mid - points

• For the class \(0 - 3\), the mid - point \(x_1=\frac{0 + 3}{2}=1.5\).
• For the class \(4 - 7\), the mid - point \(x_2=\frac{4+7}{2}=5.5\).
• For the class \(8 - 11\), the mid - point \(x_3=\frac{8 + 11}{2}=9.5\).
• For the class \(12 - 15\), the mid - point \(x_4=\frac{12+15}{2}=13.5\).
• For the class \(16 - 19\), the mid - point \(x_5=\frac{16 + 19}{2}=17.5\).

Step6: Calculate relative frequencies

The total frequency \(N=f_1 + f_2+f_3+f_4+f_5=6 + 5+4+0+3=18\).

• The relative frequency for the class \(0 - 3\) is \(rf_1=\frac{f_1}{N}=\frac{6}{18}=\frac{1}{3}\).
• The relative frequency for the class \(4 - 7\) is \(rf_2=\frac{f_2}{N}=\frac{5}{18}\).
• The relative frequency for the class \(8 - 11\) is \(rf_3=\frac{f_3}{N}=\frac{4}{18}=\frac{2}{9}\).
• The relative frequency for the class \(12 - 15\) is \(rf_4=\frac{f_4}{N}=0\).
• The relative frequency for the class \(16 - 19\) is \(rf_5=\frac{f_5}{N}=\frac{3}{18}=\frac{1}{6}\).

Step7: Calculate cumulative frequencies

• The cumulative frequency for the class \(0 - 3\) is \(cf_1=f_1 = 6\).
• The cumulative frequency for the class \(4 - 7\) is \(cf_2=f_1 + f_2=6 + 5 = 11\).
• The cumulative frequency for the class \(8 - 11\) is \(cf_3=f_1 + f_2+f_3=6 + 5+4 = 15\).
• The cumulative frequency for the class \(12 - 15\) is \(cf_4=f_1 + f_2+f_3+f_4=6 + 5+4+0 = 15\).
• The cumulative frequency for the class \(16 - 19\) is \(cf_5=f_1 + f_2+f_3+f_4+f_5=6 + 5+4+0+3 = 18\).

Class	Frequency	Mid - point	Relative Frequency	Cumulative Frequency
\(4 - 7\)	5	5.5	\(\frac{5}{18}\)	11
\(8 - 11\)	4	9.5	\(\frac{2}{9}\)	15
\(12 - 15\)	0	13.5	0	15
\(16 - 19\)	3	17.5	\(\frac{1}{6}\)	18

The class with the greatest frequency is \(0 - 3\) and the class with the least frequency is \(12 - 15\).

Answer:

Class	Frequency	Mid - point	Relative Frequency	Cumulative Frequency
\(4 - 7\)	5	5.5	\(\frac{5}{18}\)	11
\(8 - 11\)	4	9.5	\(\frac{2}{9}\)	15
\(12 - 15\)	0	13.5	0	15
\(16 - 19\)	3	17.5	\(\frac{1}{6}\)	18

The class with the greatest frequency is \(0 - 3\). The class with the least frequency is \(12 - 15\).