There is a concept in Bayesian network literature, called I-equivalence. Two Bayesian network structures are called I-equivalence if they encode the same set of conditional independencies. For example, the following three structures are I-equivalence, since they all encode $A$ is independent of $C$ given $B$:
$$A\rightarrow B \rightarrow C$$
$$A\leftarrow B \leftarrow C$$
$$A\leftarrow B\rightarrow C$$
But the following structure does not belong to the I-equivalence class of the above three structures:
$$A\rightarrow B \leftarrow C$$
This is because of the v-structure or the head-to-head node $B$. In the above structure, $A$ is NOT independent of $C$ given $B$. There is a very useful theorem for checking the I-equivalency of two structures:
Two Bayesian network structures are I-equivalence if and only if they have the same set of immoralities and the same skeleton. Immoralities are head-to-head nodes without any edge between the parents. For example, $A\rightarrow B \leftarrow C$ is an immorality but it is not an immorality if there is an edge between $A$ and $C$. The skeleton of a Bayesian network structure is simply its undirected version.
Obviously, the I-equivalence relation is an equivalence relation which partition the space of structures into equivalence classes. In the above examples, $A\rightarrow B \leftarrow C$ belongs to another class than the class of other three structures. In your example, $Age \rightarrow Edu$ and $Age \leftarrow Edu$ belong to the same equivalence class.
No Bayesian network structure learning algorithm can choose a structure from the equivalence class based on data alone. In other words, the structures in an equivalence class cannot be distinguished based on data alone. Therefore, a Bayesian network structure learning cannot favor $Age \rightarrow Edu$ over $Age \leftarrow Edu$ or vice versa based on the data alone.
Note that this does not mean that the structure learning algorithms cannot find any direction in the graph. For example, if according to the data it finds out that the skeleton of the graph is $A-B-C$ and also $A$ is not independent of $C$ given $B$, it will conclude that there should be a v-structure, that is the correct structure is $A\rightarrow B \leftarrow C$.
Although it is not possible to choose a structure in an equivalence class based on data alone, as Diego mentioned we can exploit other knowledge than data to find the direction of undirected edges. For example, in our recent work [1], we tried to use the experts' knowledge to find more accurate Bayesian network structures.
Hope this short summary answers your question. For more information, I encourage you to read chapter 3 of the excellent book by Koller and Friedman [2].
[1] Amirkhani, Hossein, et al. "Exploiting Experts' Knowledge for Structure Learning of Bayesian Networks." IEEE Transactions on Pattern Analysis and Machine Intelligence (2016).
[2] Koller, Daphne, and Nir Friedman. Probabilistic graphical models: principles and techniques. MIT press, 2009.
Best Answer
I use the following rule: There is a link from A to B if A is "one of the causes of" B. And generally, there is a link from A to B OR from B to A if there is a some correlation (not necessarily linear) between the values of A and B. The true rule is more complex, regarding conditional independence (see for example the tutorial on BN by Andrew Moore at http://www.autonlab.org/tutorials/bayesnet.html)
Thus, a reasonable solution would be:
a) all variables point to the bicycle variable, since for all you can make the argument that for example raining or not will definitely have an impact on whether or not a person will bike to work.
b) there are some relationship between the meteorological variables, although exactly which it is not totally clear to me. For example, if this refers to people living in temperate zones, then there is a link from temperature to rain, not a causal relation, but knowing the temperature changes your belief whether it is raining, since it will not rain if the temperature is below freezing! It is not clear to me how wind is related to the other variables: is it more or less likely that it will be windy if it is raining or not? If it is, then there is a link from rain to wind.
c) If you want to be complete, there are some links between the socioeconomic variables: the number of cars probably depends on the age, since older people are more likely richer than young people (and also much older people may have less cars because they drive less?). Thus a link from age to cars. Also there may be a link from gender to obese: are man more likely to be obese than woman? (I dont know). Also a possible link from age to obese (possibly the distribution of obese-ness is not constant across age).
(sorry I do not have at the moment any software that would allow me to draw the net and post it as part of the answer).