Answer:
the standard deviation for the sample is 6.83%
Explanation:
∑ ( (x1-X)^2 ) / N-1
whre X is the mean of the sample and N is the number of samples.
Year Return
2010 21.00%
2009 12.50%
2008 25.00%
Average: (0.21 + 0.125 + 0.25 )/3 = 0.195
Difference between average and actual result squared:
(0.21-0.195)^2+(0.125 -0.195)^2+(0.25 -195)^2 =
(0.015)^2+(-0.07)^2+(0.055)^2 = 0.00815
1/(N-1) = 1/(3-1) = 1/2
standard deviation = √(0.00815/2) = 0.06837397166 = 6.83%
