Question

A cyclist won a bicycle race for seven consecutive years. His "winning" times and "victory" margins (time difference of the second place finisher) are given in the figure below.

-Year------Time--------Margin
----------(h:m:s)------ (m:s)
1999-----91:32:17-----7:38
2000-----92:33:01-----6:01
2001-----86:17:28-----6:44
2002-----82:05:13-----7:16
2003-----83:41:13-----1:07
2004-----83:36:04-----6:17
2005-----86:15:04-----4:30
(a) Find the standard deviation of the cyclist's times. (Round your answers to the nearest second.)
: : h:m:s

(b) Find the standard deviation of the cyclist's margins. (Round your answers to the nearest second.)
: m:s

I have no idea what to do

Answer 1

a)
You will have to find the average time, which involves adding the 7 times and then dividing by 7
Too bad that "we" have not gone metric with units of time, so you will have to convert each of the times to a decimal number in terms of hours, (unless you use the
D"M'S key on your calculator and add them up that way.)

e.g.
91:32:17 = 91 + 32/60 + 17/3600 hrs = 91.5381
do this for the other six times
Add them up then divide by 7
this will be your mean.

now form a column of differences of the times and the mean, now square each difference.
add up those squares of the differences.
Divide that sum by 7
Take the square root of that and you have your SD


you will still have to change your single decimal to
h:m:s
e.g. 2.578 hours
= 2 hours + .578 hours
= 2 hours + .578(60) minutes
= 2 hours + 34.68 minutes
= 2 hours + 34 minutes + .68(60) seconds
= 2 : 34 : 41

b)
repeat the entire for the margins

This is a very tedious process, work patiently.

here is a Khan Academy page for this procedure:

https://www.khanacademy.org/math/probability/data-distributions-a1/summarizing-spread-distributions/a/calculating-standard-deviation-step-by-step