# Solve for standard deviation

The question is: “What is the Sharpe ratio for an investment in the Nopat Fund over the five years from 2005-2009? What is the Sharpoe ratio for an inestment in the Emfund over the six years 2004-2009? Which Sharpe ratio is more preffered?” I am trying to compute standard deviation with the “Nopat annual HPR” I start off by computing the mean which is 6.28%. This is correct in the book. However, when I compute the standard deviation I get a value of 6.06 when the answer is 6.77. To come to this, I perform the following computations: (-5.44% - 6.28%)^2 +(11.08%-6.28%)^2 +(6.66% - 6.28%)^2 +(10.48% - 6.28%)^2 +(8.61% - 6.28%)^2 / 5**==> **.01374 + .00230 + .00001 + .00176 + .00054 = .01835 / 5 = .00367. When I take the square root to get standard deviation I get 6.06%. The book says it is 6.77%.

Furthermore, when I try and compute the standard deviation for the emfund, I run into the same problem. I start by finding the mean of the fund return which is 4.4 (as confirmed by the book) but then when I try to compute the standard deviation, I get the wrong answer: (3% - 4.4%)^2 + (4% - 4.4%)^2 + (4.3%-4.4%)^2 + (5% - 4.4%)^2 + (4.1% - 4.4%)^2 + (6% - 4.4%)^2 / 6

==>

.00020 + .00002 + 0 + .00004 + .00001 + .00026 = .00053 / 6 = .00009

When I take the square root I get .00936. The book has the standard deviation as 1.01%

In a previous problem using the information from the attached link, the book asked for us to find the sample SD “using the emfunds returns as a sample” (meaning that in this case, the sample mean is no different than the regular mean). Their result was 1.01%. Sample SD is calculated by dividing the variance by N-1 and not N or in this case, 5 instead of 6. The books result for Sample standard deviation was 1.01% but they also list 1.01% as the standard deviation for

For both problems I come fairly close to getting the correct SD. I must be making a small but significant mistake but I cannot find it.

Thanks.

Your problem is one of population vs. sample.

sqrt[(-5.44% - 6.28%)^2 +(11.08%-6.28%)^2 +(6.66% - 6.28%)^2 +(10.48% - 6.28%)^2 +(8.61% - 6.28%)^2 / 5]

gives you 6.06% as Population Standard deviation.

What you should be doing is

SQRT[(-5.44% - 6.28%)^2 +(11.08%-6.28%)^2 +(6.66% - 6.28%)^2 +(10.48% - 6.28%)^2 +(8.61% - 6.28%)^2 / 4] which gives you 6.77%

(Remember (n-1) in the denominator)

Similar issue on the other one as well, I believe.

Hello CPK,

I understand the concept of sample means (and calculating using [n-1] as opposed to [n] as with population means) but why does this problem call for them?

Thanks!

you are looking at a sample of the data - not the entire period of time the fund was active. It is for the 5 years that you are looking at… so a sample.