Finish times (to the nearest hour) for 10 dogsled teams are shown below. 310 271

Question

Finish times (to the nearest hour) for 10 dogsled teams are shown below. 310 271 278 295 283 236 281 266 269 Make a frequency table showing class limits, class boundaries, midpoints, frequency relative frequencies, and cumulative frequencies. Use three classes. (Round your answer for relative frequency to the nearest hundredth and for midpoint to the nearest tenth.) Class Limits 236-260 261-284 285-308 Boundaries 235.5-260.5 260.5 - 284.5 284.5 - 308.5 Midpoint 248.0 272.5 296.5 Freq 1 6 3 Relative Freq 0.10 0.60 0.30 Cumulative Freq. 1 7 10 Class Limits 236-260 261-285 286-310 Boundaries 2.5.5 - 260.5 260.5 - 285.5 285.5 - 310.5 Midpoint 248.0 273.0 298.0 Freq. 1 6 3 Relative Freq. 0.10 0.60 0.30 Cumulative Freq. 1 7 10 Class Limits 236-260 261-284 285-310 Boundaries 235.5 - 260.5 260.5 - 285.5 285.5 - 310.5 Midpoint 248.0 273.0 297.5 Freq. 2 6 2 Relative Freq. 0.20 0.60 0.20 Cumulative Freq. 2 8 10 Class Limits 236-260 261-284 285-308 Boundaries 235.5 - 260.5 260.5 - 284.5 284.5 - 308.5 Midpoint 248.0 272.5 296.5 Freq 2 6 2 Relative Freq. 0.20 0.60 0.20 Cumulative Freq. 2 8 10

Explanation / Answer

the number of values that fall between 236 and 260 is 1 and midpoint is ( 236+260)/2 =248

the number of values that fall between 261 and 284 are 6 and between 385 and 310 are 3

relative freq for part 1 will be 1/(1+6+3)=0.1

and second part will be 6/(1+6+3)=0.6 and last will be 0.3

also cummulative freq is sum of previous ones so for part 1 it is 0.1 , for part 2 it is 0.1+0.6=0.7 and for last part it will be 0.1+0.6+0.3=1

so answer is A

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