CFA Practice Question

CFA Practice Question

The sum of the squares of 1,200 observations equals 9,830. The sum of the observations equals 1,510. The population standard deviation of the observations equals ______.
A. 5.31
B. 4.22
C. 2.57
Explanation: For N observations, it is easy to show that: population variance*N = (sum of squares) - N*(mean2)

The mean equals 1,510/1,200 = 1.258. Hence, population variance = (9,830 - 1,200*1.2582)/1,200= 6.608. The standard deviation then equals sqrt(6.608) = 2.57.

Note: You should be careful to note the difference between population variance and sample variance. The formula for sample variance is: sample variance*(N-1) = (sum of squares) - N*(mean2).

You can expect an exam question that asks for population variance, with the choices given containing both the population and the sample variances.

User Contributed Comments 17

User Comment
Rguerra Where did you get that equation from? pop var*N = (sum of squares) - N*(mean2)?

Shouldn´t: pop var*N = Sum(X - mean)squared? In this case: pop var*N = (sum of squares) - 2*X*mean) + mean(square). I don´t see how you get to the first equation.
PedroEdmundo What about 2*mean*SumX?
bhawa agreed with Rguerra and Pedro, can someone please explain from where the equation came from in the answer given??
SunilRaja This is how it shud be done.
variance = (Sum(X-Mean X)squared)/N

Now expand (A-B)squared = A squared + B squared - 2AB

So variance = {Sum(Xsquared)+Sum(Mean Xsquared) - 2 Sum(X*Mean X)}/N

Now Sum(Mean Xsquared) = Sum(Mean Xsquared * 1)
= Mean Xsquared * Sum(1) = Mean Xsquared * N
since Mean Xsquared is not affected by Sum and is constant and Sum(1) = 1+1+.....N times = N

Similarly Sum(X*Mean X) = Mean X * Sum(X)

Mean X = Sum(X)/N = 1510/1200 = 1.26

We are given Sum(X) and Sum(Xsquared)

(MeanX)squared = 1.26*1.26 = 1.5876
variance={9830 + 1.5876*1200 - 2*1.26*1510}/1200
= 7929.92/1200 = 6.608

standard deviation = square root(variance)
= 2.57
antony Thanks for the explanation. That helped.
surob Wow, hope this is not going to be a question in the real test.
Given that the sample is large relative to the population (i.e. (1200/1510) = 79.5%), I assumed that the sum of the squares of of mean variances for the sample is approximately equal to that of the population.

Therefore, population variance = sum of squares / n
population variance = 9830 / 1510 = 6.51
square root = 2.55

Though this is not the exact answer, it is different enough to A and B, and close enough to C, to confirm that C is correct.
grezavi Xocrevilo is right, we are not expected to the Algebra, the logic fits for large sample theory
apiccion Here's how I did it.

Var(x) = sum(x-Xbar)^2/n <= sum(X)^2/n = 9830/1200 = 8.19

Therefore, std(x) <= sqrt(8.19) = 2.86

Looking at the answers, there is only one number which is less than 2.86. Pick that number.
moritz Great solution, apiccion! Thx!
7Ricky Variance = mean of squares - square of means
= E[x^2]-E[x]^2
=(9830/1200) - (1510/1200)^2
sqr rt=2.57

Not sure if the equation is in CFA text but should be in any intro stat book
Ifi2703 Very useful eplanation by 7Ricky - will help for time management in exam knowing such tips...thanks!
bidisha Thanks ricky
Procbaby1 7Ricky that's really helpful!
birdperson thanks ricky.
Rsanches 7Rick, awesome explanation.
ashish100 7Ricky!

Where are you been in the last 197 hours? (time i spent on AN)

Thank you!
You need to log in first to add your comment.