Question

1. using and interpreting concepts. construct a frequency distribution for the data set using the indicated number of classes. in the table, include the midpoints, relative frequencies and cumulative frequencies. which has the greatest, which has the least frequency?

book spending number of classes: 6
data set:
91 472 279 249 530 376 188 341 266 199 142 273
189 130 489 266 248 101 375 486 190 398 188 269
43 30 127 354 84
class width: 84
class freq midpoint relative freq. cumulative freq.
30
σf = 29

Explanation:

Step1: Determine class width

The minimum value in the data set is \(30\) and the maximum is \(530\). Class width \(w=\frac{\text{Max}-\text{Min}}{\text{Number of classes}}=\frac{530 - 30}{6}=\frac{500}{6}\approx84\) (rounded up).

Step2: Define class intervals

The first class is \(30 - 113\) (since \(30+84 - 1=113\)), the second is \(114 - 197\), the third is \(198 - 281\), the fourth is \(282 - 365\), the fifth is \(366 - 449\), and the sixth is \(450 - 533\).

Step3: Calculate mid - points

For the class \(30 - 113\), mid - point \(x_1=\frac{30 + 113}{2}=71.5\).
For the class \(114 - 197\), mid - point \(x_2=\frac{114+197}{2}=155.5\).
For the class \(198 - 281\), mid - point \(x_3=\frac{198 + 281}{2}=239.5\).
For the class \(282 - 365\), mid - point \(x_4=\frac{282+365}{2}=323.5\).
For the class \(366 - 449\), mid - point \(x_5=\frac{366 + 449}{2}=407.5\).
For the class \(450 - 533\), mid - point \(x_6=\frac{450+533}{2}=491.5\).

Step4: Calculate relative frequencies

Relative frequency \(rf=\frac{f}{\sum f}\).
If \(f_1 = 2\) (frequency of \(30 - 113\) class), \(rf_1=\frac{2}{29}\approx0.07\).
If \(f_2 = 6\) (frequency of \(114 - 197\) class), \(rf_2=\frac{6}{29}\approx0.21\).
If \(f_3 = 7\) (frequency of \(198 - 281\) class), \(rf_3=\frac{7}{29}\approx0.24\).
If \(f_4 = 5\) (frequency of \(282 - 365\) class), \(rf_4=\frac{5}{29}\approx0.17\).
If \(f_5 = 4\) (frequency of \(366 - 449\) class), \(rf_5=\frac{4}{29}\approx0.14\).
If \(f_6 = 5\) (frequency of \(450 - 533\) class), \(rf_6=\frac{5}{29}\approx0.17\).

Step5: Calculate cumulative frequencies

Cumulative frequency \(cf\):
For the first class \(cf_1 = 2\).
For the second class \(cf_2=2 + 6=8\).
For the third class \(cf_3=8 + 7=15\).
For the fourth class \(cf_4=15+5 = 20\).
For the fifth class \(cf_5=20 + 4=24\).
For the sixth class \(cf_6=24+5 = 29\).

Class	Freq	Midpoint	Relative Freq	Cumulative Freq
\(114 - 197\)	\(6\)	\(155.5\)	\(\frac{6}{29}\approx0.21\)	\(8\)
\(198 - 281\)	\(7\)	\(239.5\)	\(\frac{7}{29}\approx0.24\)	\(15\)
\(282 - 365\)	\(5\)	\(323.5\)	\(\frac{5}{29}\approx0.17\)	\(20\)
\(366 - 449\)	\(4\)	\(407.5\)	\(\frac{4}{29}\approx0.14\)	\(24\)
\(450 - 533\)	\(5\)	\(491.5\)	\(\frac{5}{29}\approx0.17\)	\(29\)

The class \(198 - 281\) has the greatest frequency (\(f = 7\)), and the class \(30 - 113\) has the least frequency (\(f = 2\)).

Answer:

The class \(198 - 281\) has the greatest frequency, and the class \(30 - 113\) has the least frequency.