Okay, let try this one again. Sorry for the formatting issue:

Mean hrs= number of subjects/total hours driving (200/1580) = 7.9
Standard Deviation for hours driving: Needs to equal 1.05 I can get that
by
taking 12702(power1580,2)/200 which = 220
taking that sum (220) and making another calculation. sqrt(220/200-1)
Those combined net the standard deviation, but there has to be an easier
way.... Any help would be appreciated.


x f xf
x2f
Hrsdriving #Subjects Tot.hrsdriving
3.5 2 7
24.5
4.5 2 9
40.5
5.5 4 22
121
6.5 22 143
929.5
7.5 64 480
3600
8.5 90 765
6502.5
9.5 14 133
1263.5
10.5 2 21
220.5

56 200 1580
12702




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