Given two independent random variables (which describe two workers' durations to complete a task) $t_1, t_2$ which are normally distributed with ($\mu_1=20,\sigma_1=5$) and ($\mu_2=30,\sigma_2=10$). A task is given at random to worker 1 or 2 with equal probability of $p=0.5$.

A task is completed in $t=10$. Now, I would like to find the probability of this task being completed by worker 2.

It would be great if someone could guide me through this task. So far I came up with the following thought-process:

1) I assume workers durations to be independent, thus $p(t)=p(t_1)p(t_2)$ (But not sure what to do with the equal probability of 0.5)

2) My goal is to find $p(\text{worker}_2\vert t=10)$ but I am not sure how?

Ideas I had:

  • Use $p(t)=\mathcal{N}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2 )$

There are two type of workers, $W = 1$ and $W=2$. Given an observation of completion time $t$, we can use Bayes' rule to compute the probability that this observation was produced by a worker of type $i$: \begin{equation} p(W=i|t) = \frac{p(t|W=i)\,p(W=i)}{p(t)} , \end{equation} where \begin{equation} p(t) = \sum_{i=1}^2 p(t|W=i)\,p(W=i) . \end{equation}

In the setup we are told $p(W=1) = p(W=2) = 1/2$, $p(t|W=1) = \textsf{N}(t|20,5^2)$, and $p(t|W=2) = \textsf{N}(t|30,10^2)$. Therefore, \begin{equation} p(W=2|t=10) = \frac{\textsf{N}(10|30,10^2)\times \frac{1}{2}}{\textsf{N}(10|20,5^2)\times \frac{1}{2} + \textsf{N}(10|30,10^2)\times \frac{1}{2}} = \frac{1}{3} . \end{equation}

  • $\begingroup$ My teacher always tells us that the normal distribution is zero for a single value (e.g. like here t=10) - why does this not apply here? $\endgroup$ – Oliver Apr 26 '18 at 9:07
  • $\begingroup$ The probability of a single value is indeed zero. But the density at a single value is not zero. The answer involves the densities of a single value for two different distributions. $\endgroup$ – mef Apr 26 '18 at 12:24
  • $\begingroup$ It is confusing, then, to call these quantities "probabilities" in your answer and then claim in a comment that they are "densities." In fact, some of them are probabilities and some of them are densities! $\endgroup$ – whuber Feb 3 at 21:05

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