How does the proportion of X > a compare to the proportion of Y > a? I am completing CS109 Data Science class available online for free. I'm not quite sure what they are asking for.

In general, if group A has larger values than group B on average, does this
      mean the largest values are from group A? 
Assume you have two list of numbers, X and Y, with distribution
      approximately normal. X and Y have standard deviation equal to 1, but the 
      average of X is different from the average of Y. If the difference in the
      average of X and the average of Y is larger than 0, how does the proportion 
      of X > a compare to the proportion of Y > a?

What proportion are they talking about here:

of X > a compare to the proportion of Y > a?

What do I need to discover?
Thank you.
 A: The distributions are supposed to look similar, because we are told they are "approximately normal" and both have the same standard deviation.  The one with larger mean ($\mu_A$, the average of $X$) therefore has been shifted to the right, just like the blue density curve (representing a histogram) relative to the red one in this figure (whose mean is $\mu_B$, the average of $Y$).

This should make it obvious that there is more area under the blue (right-hand) curve to the right of any given value $a$ compared to the area under the red (left-hand) curve.  The difference in areas is that of the shaded region.  Since areas represent proportions (the total area under either of these curves is $100\%$), the shaded area represents how much greater the proportion is in group $A$ compared to group $B$.
A: Whenever you have to compare two different normal distributions, you have to convert both to Standard Normal Distributions. That way both distributions are in the same scale:
for X Distribution:
Convert a into standard normal (Z value) = (a-X̅)/1= a-X̅
Pr (X>a) = 1 -Pr (Z<( a-X̅))

Similarly for Y distribution:
Pr (Y>a) = 1 -Pr (Z<( a-Y̅))

Now your original question is difference in proportions:
Pr (X>a) - Pr (Y>a) (Or Vice Versa)
Which is equivalent to=  
= [1 -Pr (Z<( a-X̅))] – [1 -Pr (Z<( a-Y̅))] 
=  [Pr (Z<( a-X̅))] – [Pr (Z<( a-Y̅))]                        
=  Pr (Z<( a-X̅-a+Y̅)
=  Pr (Z<( Y̅-X̅))

Since Z is Symmetric: Pr (Z<( Y̅-X̅)) = Pr (Z<( X̅-Y̅))
This you can read directly from the Z table
