Lets say that a group of students all take 5 tests, with the average and standard deviations as follows:

  • test 1: mean 43/50, SD 8
  • test 2: mean 23/30, SD 4
  • test 3: mean 56/70, SD 12
  • test 4: mean 33/40, SD 8
  • test 5: mean 44/60, SD 10

Is there a way to find the standard deviation over all of the assignments such that a student can find out, cumulatively, in which percentile he is in?

  • $\begingroup$ Is a test out of $30$ to be counted less than a test out of $60$? And a technical question, for SD does one divide by $n$ or by $n-1$? (Here $n$ is class size, assumed constant. Both $n$ and $n-1$ are in common use.) $\endgroup$ – André Nicolas Apr 19 '14 at 0:06
  • $\begingroup$ Not to add too much complexity, but each test should have an equal weight on the final score. Also, I think dividing by n would be fine. $\endgroup$ – student Apr 19 '14 at 0:13

Since in a comment you say the tests should have equal weight, probably we should first scale the means and standard deviations, so they are all out of some common value. In your example, you could choose $100$ for each, or the least common multiple of the denominators.

We will assume that sample standard deviation was computed by dividing by the class size $n$, assumed constant. If it is done by dividing by $n-1$, the adjustment is straightforward.

After scaling has been done, let $\mu_1, \mu_2,\dots, \mu_k$ be the test means, and $s_1,s_2,\dots, s_k$ be the sample standard deviations. Multiplying the $\mu_j$ by $n$ gives us the class total for each test.

For any test $j$, consider the $n$ students, and let $m_{j,1},\dots, m_{j,n}$ be their marks (unknown to you). Then $$s_j^2=\left(\frac{1}{n}\sum_{t=1}^n m_{j,t}^2\right) -\mu_j^2.$$ You know $s_j$ and $\mu_j^2$, so we can rewrite the above as $$\sum_{t=1}^n m_{j,t}^2=n(s_j^2 +\mu_j^2).\tag{1}$$ The point is that you now know the sum of the squares of the (adjusted) marks in test $j$. By adding up over all tests from $1$ to $k$, you get the sum of the squares of the adjusted marks over all tests. This is $A$, where $$A=n\sum_1^k (s_j^2+\mu_j^2).\tag{2}$$ You also know the sum of the adjusted marks over all tests. This is $B$, where $$B=\sum_1^k n\mu_k.\tag{3}$$ To compare your student with the class, compute your student's adjusted mark. The class mean is $B/n$, and the class standard deviation is equal to $$\sqrt{\frac{A}{n} -\left(\frac{B}{n}\right)^2}.\tag{4}$$

If you want to compute out of $100$ instead of whatever the standardized values of the tests add up to, just divide the mean and standard deviaton obtained above by the appropriate factor.

  • $\begingroup$ The Cclass standard deviation in the last part should be the square root of the square of ((A/n)-(B/n)), right? $\endgroup$ – student Apr 21 '14 at 4:03
  • $\begingroup$ I think (4) is right as stated. But I am quite tired, can check the details tomorrow. It would be helpful to know what you think it should be. The variance should look like $\frac{A}{n}-\frac{B^2}{n^2}$. $\endgroup$ – André Nicolas Apr 21 '14 at 4:28

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