I'm confused about finding the standard deviation of various weighted objects in a problem. For example, consider an academic setting, with four assessments as follows:

(30pts weight)
Quiz 1 - Avg: 75% Std: +/- 35%
Quiz 2 - Avg: 39.6% Std: +/- 25.4%
Quiz 3 - Avg: 69.4% Std: +/- 22.4%

(20pts weight)
Midterm - Avg: 63.62% +/- 13.09%

(50pts total)
Overall - Avg: 62% +/- 10.89%


Could one then calculate the std's for the course average, by treating the std's like the avg, and just weight them? That's how I got the std above for the overall grade.

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Do you want the (weighted) standard deviation of the individual assessment scores, or do you want the standard deviation of the overall student course averages? These are two different things. The first has 75, 39.6, 69.4, 63.62 as the raw data values, and the second has the n course averages for the n students as the raw data values. I'm guessing you want the latter? –  Mike Spivey Nov 16 '11 at 20:12
The latter, since I can compute the class avg, but I am not given the class avg, standard deviation. –  eWizardII Nov 16 '11 at 20:36

Since you are combining scores using formula $Y = \sum_k \omega_k X_k$, using the linearity of expectation, and Bienaymé formula, assuming the performances at each test are not correlated: $$\mathbb{E}(Y) = \sum_k \omega_k \mathrm{E}(X_k) \qquad \mathbb{V}(Y) = \sum_k \omega_k^2 \mathbb{V}(X_k)$$

I am getting combined average $(61.83 \pm 9.88) \%$.

As MikeSpivey and MichaelLugo justly point out, the assumption of zero correlation poorly reflects actual state of the matter. So assume that $\mathbb{Cor}(X_i,X_j) = \rho_{ij}$. Then $$\mathbb{V}(Y) = \sum_{i,j} \omega_i \omega_i \mathbb{Cov}(X_i, X_j) = \sum_k \omega_k^2 \mathbb{V}(X_k) + \sum_{i < j} \omega_i \omega_j \rho_{ij} \sqrt{\mathbb{V}(X_i)\mathbb{V}(X_j) }$$ Since the correlation is clearly positive, the total variance is bigger in the correlated case. I did some computations assuming equal correlation between test.

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I'm not sure that the assumption of uncorrelated assessment scores is valid. –  Mike Spivey Nov 16 '11 at 19:51
Thanks, I guess one thing I forgot to clarify though is I don't know what the actual overall standard deviation is I just did the following: $$\frac{\frac{35+22.4+25.4}{300}*30+\frac{13.09}{100}*20}{50}=21.8%$$ Actually made a mistake in the above question asked, but that's okay then? And then I can use the equation you gave and get an std? –  eWizardII Nov 16 '11 at 19:52
@MikeSpivey I agree, since a students performance ideally is related to the grades they are receiving over time. Like failing all the quiz - and then getting a failing midterm score. –  eWizardII Nov 16 '11 at 19:55
I'm sure that the assumption of uncorrelated assessment scores is not valid. –  Michael Lugo Nov 16 '11 at 19:58
In fact, to further emphasize the remarks from the two Michaels, this sort of example is precisely the canonical one used to demonstrate regression to the mean. –  cardinal Nov 16 '11 at 20:12

Since we can't assume that the assessment scores are uncorrelated, there's no way to calculate the standard deviation of the student course averages solely with the information in the question. (Added: However, we can obtain bounds. See below.)

Provided you have the right additional information, though, there are two ways you could do the calculation.

1. If you know the covariances or correlations of the assessment scores, you could use the formula Sasha mentions in his answer.
2. If you know the course average for each student, you could apply the usual standard deviation formula $$s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2},$$ where the $x_i$ are the $n$ individual student course averages and $\bar{x}$ is the overall course average ($\bar{x} = 62\%$, in your example). (The fact that the $x_i$ are weighted averages is irrelevant; you just want the standard deviation of a set of $n$ data values, and how those values are obtained does not affect the standard deviation calculation.)

I'm guessing that (2) is easier. If it were me, I would already have each student's course average in a spreadsheet. It would take about 5 seconds to tell the spreadsheet to find their standard deviation.

(Added, based on comment of OP below.)

Let's rewrite the general formula for the variance of a weighted sum of variables as $$\text{Var(Y)} = {\bf w}^T C {\bf w},$$ where ${\bf w} = (0.2(0.35), 0.2(0.254), 0.2(0.224), 0.4(0.1309))$ (i.e., the column vector of weighted standard deviations), and $C$ is the correlation matrix (i.e., $C_{ij} = \rho_{ij}$).

If the information in the question is all that you have available, you can use this formula to obtain bounds on the standard deviation of the overall course average. Since students who do well on one assessment tend to do well on others, and students who do poorly on one assessment tend to do poorly on others, we can assume positive correlations. So we get a lower bound on $s_Y$ by taking $\rho_{ij} = 0$ for all $i \neq j$ and an upper bound by taking $\rho_{ij} = 1$ for all $i,j$. (We have $\rho_{ii} = 1$ for all $i$ in any situation, since a variable has a correlation of $1$ with itself.)

These calculations are straightforward; I get $11.1 \% \leq s_Y \leq 21.8\%$. Note that the lower bound is the one with the assumption of independence, and the upper bound is the one obtained by taking a weighted average of the four assessments, as the OP notes in the comments on Sasha's answer. (I get a slightly different answer for the independence assumption than Sasha; I think he misread the data.)

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Thanks, though I wouldn't have been able to pursue the route of the excel spreadsheet since I am the student, and don't have access to everyone's scores. –  eWizardII Nov 17 '11 at 5:15
@eWizardII: If you don't have access to the individual scores (and you would need them to calculate the covariances, too, if you don't already know those), then the best you can do is make an estimate. You could also get an upper bound by using $\rho_{ij} = 1$ in the formula in Sasha's answer. (He doesn't have the formula quite right, though; look at the Wikipedia formula linked to in my answer instead, with the relation $\text{Cov}(X_i,X_j) = \rho_{ij} \sqrt{V(X_i) V(X_j)}$.) –  Mike Spivey Nov 17 '11 at 5:32