Math  /  Data & Statistics

QuestionSave \& Exil certlly Lesson: 2.1 Frequency Distributitans
Question 3 of 7 , Step 1 of 7 6/216 / 21 correct \longrightarrow
Consider the following data representing the pice of plasma televisions (in dollars). 1387,1135,1352,1220,1202,1217,1176,1268,1150,1182,1222,1391,1163,1168,1282,1250,1201,1225,1401,1175,13711387,1135,1352,1220,1202,1217,1176,1268,1150,1182,1222,1391,1163,1168,1282,1250,1201,1225,1401,1175,1371 \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{5}{|c|}{ Price of plasma Televisions (in Dollars) } \\ \hline Class & Frequenty & Class Boundaries & Midpoint & Relative Frequency & Cumulative Frequency \\ \hline 107711361077-1136 & & & & & \\ \hline 113711961137-1196 & & & & & \\ \hline 119712561197-1256 & & & & & \\ \hline 125713161257-1316 & & & & & \\ \hline 131713761317-1376 & & & & & \\ \hline 137714361377-1436 & & & & & \\ \hline \end{tabular}
Step 1 of 7: Determine the class width of the classes listed in the frequency table.
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Studdy Solution
Calculate the class width by subtracting the lower boundary of the first class from the lower boundary of the second class.
Lower boundary of the second class=1137 \text{Lower boundary of the second class} = 1137 Class width=11371077=60 \text{Class width} = 1137 - 1077 = 60
The class width of the classes listed in the frequency table is 6060.

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