I have data for cycles: 1.539, 3.4338, 5.4819, 9.5781,11.6262, 13.5719, 15.16, 1

Question

I have data for cycles: 1.539, 3.4338, 5.4819, 9.5781,11.6262, 13.5719, 15.16, 17.6681, 19.7162

Average period=10.5305?
Range=18.1772?
stand. dev. = 5.7776?

On part B when I had 3different run's with times. What I did wastake the (1st cycle - last)/ number of cycles
Ex. run #1 1.8978 - 20.0746 / 10 = 1.8177, does this equal theperiod?
And the average period is the 3 run's periods averagedtogether? I have data for cycles: 1.539, 3.4338, 5.4819, 9.5781,11.6262, 13.5719, 15.16, 17.6681, 19.7162

Average period=10.5305?
Range=18.1772?
stand. dev. = 5.7776?

On part B when I had 3different run's with times. What I did wastake the (1st cycle - last)/ number of cycles
Ex. run #1 1.8978 - 20.0746 / 10 = 1.8177, does this equal theperiod?
And the average period is the 3 run's periods averagedtogether?

Explanation / Answer

Average period is the avrage of each period.
Suppose for the cycles 1.539, 3.4338, 5.4819, 9.5781,11.6262, 13.5719, 15.16, 17.6681, 19.7162
Now the periods are 1st 1.539 - 0 = 1.539 2nd 3.4338 - 1.539 = 1.8948 3rd 5.4819 - 3.4338 = 2.0481 4th 9.5781 - 5.4819 = 4.0962 5th 11.6262 - 9.5781 = 2.0481 6th 13.5719 - 11.6262 = 1.9457 7th 15.16 - 13.5719 = 1.5881 8th 17.6681 - 15.16 = 2.5081 9th 19.7162 - 17.6681 = 2.0481 Then the average period is (1.539 + 1.8948 + 2.0481 + 4.0962 + 2.0481+ 1.9457 + 1.5881 + 2.5081 +  2.0481)/9 = 2.19

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