How to calculate the time of finishing all tasks with probability of 90% I have n tasks to do. I don't know how much time exactly it will take to finish given task so let's assume that each task on average k_n with standard deviation std_n to take (we can assume that the distribution if the time spent on given task is normal).
Now I need to perform all the tasks. How can I calculate what time is needed to finish all tasks with the probability of 90%?
 A: If we assume that the time for task $i$, $T_i\sim N(\mu_i,\sigma^2_i)$ and we assume that task times are independent (and tasks are sequential -- see Daniel's answer and his comments under it), then the overall times $U=\sum_i T_i\sim N(\mu_U,\sigma^2_U)$ where $\mu_U= \sum_i \mu_i$, and $\sigma^2_U=\sum_i \sigma^2_i$.
As a result, $P(U<\mu_U+1.28\,\sigma_U)=0.9$
Even if you don't have normal distributions for the times, if there are a very large number of tasks such that (say by considering the non-identically distributed version of the Berry-Esseen theorem) we might expect the standardized sum to be close to normally distributed in such a way that $P(U<\mu_U+1.28\,\sigma_U)\approx 0.9$ (the theorem gives a bound on the deviation in this kind of probability).
A: First of all, a normal distribution is unlikely to be an appropriate model for this situation, exponential will likely be a better approximation, and you can always put a prior on the rate parameter if you wanted to explicitly model uncertainty in the times taken for each task.
Anyway, the problem is fairly easy to solve if we assume i.i.d samples, but it looks like you want different marginal distributions for each task which makes the problem much harder.
You need to calculate the distribution for the random variable
$$
Y = \text{max}\{X_1, ..., X_n\}
$$
where $X_i$ is the r.v for task $i$. Then by the definition of $Y$, we have
$$
p(Y \le y) = p(X_1 \le y, ..., X_n \le y)
$$
because $Y \le y$ if and only if each task $X_i \le y$. Now as we assume independence we have
$$
p(Y \le y) = p(X_1 \le y)...p(X_n \le y) = F_1(y)...F_n(y)
$$
where $F_i$ is the c.d.f for task $i$. Now you can see why assuming the tasks are sampled from the same distribution would make this easier; we would have $F_i = F \forall i$ and hence $p(Y \le y) = [F(y)]^n$ which can be solved easily for normal or exponential.
If you do want different parameters for each task, then I'd say Monte Carlo methods are your only option.
A: This sounds like a homework question? I'll point you into the direction, but not give an exact answer. 
The question asks to sample from a normally distributed variable, with known average k_n and standard deviation std_n. The size of your sample is n. The answer involves the application of the central limit theorem. This theorem states that a sample average of size n is distributed with average k_n and standard deviation std_n/sqrt(n). 
With this information you can:
- find the 90% confidence interval of the sample mean using a two sided interval (hint: 1.65)
- from the lower and higher value calculate the total time instead of the average time for the tasks
HTH
