A sample of athlete's workouts had a mean time of 132 minutes with a variance of...
A sample of athlete's workouts had a mean time of 132 minutes with a variance of 23 minutes. The CVar is ______ %. (round to 2 decimal places.)
Homework Answers
Sol:
variance=23
standard deviation=sqrt(23)= 4.795832
coefficient of variation=standard deviation/mean*100
=4.795832/132*100
=3.633206
=3.63%
ANSWER:
CVar=3.63%
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