Trial Completion time is not right skewed but it should be Main Question:
I am wondering why my completion time data for trials is not as skewed as the completion time for the tasks that makeup the trial? 
Details:
I am comparing two user interfaces on task completion time. The participants were given a set of tasks to complete with UI-A and the same set of tasks to complete with UI-B. For each interface participants completed the tasks twice. The order in which the tasks appeared was randomized and counterbalanced. I have noticed that if I look at the completion time for each task within the trials I get a right skewed plot (as expected with RT).

Histogram of completion times for the individual tasks withing the trial
However, if I summarize the data i.e., look at the time to complete all the tasks in a trial then I get something more normal looking. The point of looking at trial completion time is to determine overall which interface is more efficient.

Histogram of Trial Completion Time
I had expected a right skewed plot since the task times within a trial are right skewed. Now I am wondering why the data is not as skewed? Doing some research on response time I came across a paper titled Effective Analysis of Reaction Time Data. Based on this article it appears that the tasks within the trial can be seen as raw response time and the summarized trial data could be looked at as aggregating the data. However, this does not sit right with me since I only summarized the task completion times to get total trial times. 
Also, based on the aforementioned article one approach to normalizing response time data is to perform a log transformation on the completion times to reduce, but it is not clear to me if I should perform the log calculations on the individual task times or on the summarized trial times? Based on the trial completion time plot it may not even be necessary to do any transformations since T-Test is fairly robust against data that is not exactly normal
 A: Your second histogram is clearly right skew. It's just not as right skew as the components you started with.
This is exactly what you should expect. 
The distributions of sums of random variables will tend to be less skew than the variables that make up the sum.
For example, consider samples from two and four identical distributions shaped like the first histogram, and the distribution of their sums:

This is fairly typical of the kind of thing you see, though if one component of the sum is much more variable than the others it will tend to "dominate" the shape; nevertheless the sum will still tend to be less skew than the most skew component. When you add an independent random component to another, you get a convolution of their densities, which results a kind of "smearing out" or blurring - peaks become rounder, shapes become more symmetric.
If you add a lot of terms, you (slowly) tend to get closer to normally-distributed totals.
If we assume independence of the component times and look at the skewness measure based on third moments we can even quantify how much this tends to happen.
Specifically if we consider cumulants (which are related to moments) we can use the fact that the cumulants of independent random variables "add" (the $k$th cumulant of the sum is the sum of $k$th cumulants). 
[The first cumulant is the mean, the second cumulant is the variance, and the third cumulant is the third central moment; the fourth cumulant is not the fourth central moment, but we'll just use the first three for now]
If you have variables drawn independently from two distributions, the first with mean $\mu_1$, variance $\sigma^2_1$ and third central moment $\theta_1$, and the second with corresponding values $\mu_2$, $\sigma^2_2$ and $\theta_2$, then the mean, variance and third central moment of the sum will be$\mu_1+\mu_2$, $\sigma^2_1+\sigma^2_2$ and $\theta_1+\theta_2$ respectively.
The skewness coefficients of the components will be $\theta_1/\sigma_1^3$ and $\theta_2/\sigma_2^3$ and the skewness of the sum of components will be $\frac{\theta_1+\theta_2}{(\sigma_1^2+\sigma_2^2)^{3/2}}$.
Imagine for simplicity that the standard deviations of both distributions is about the same. Then the third-moment-skewness of the sum would be about 70% of the average of those two skewness values. If you added 100 of them (again assuming about equal standard deviations) the skewness would reduce to about 10% of the average skewness.
