In most undergrad classes, E(X) is taught to be the "expected gain" by playing a game of bets.
For instance: If we consider a Bernoulli distribution with $p=0.3$ (Winning) and if you win, you get \$10 and you lose, you pay \$3, the E(X) = 10 x 0.3 - 3 x 0.7 = 3 - 2.1= \$0.9.
Is there a way this intuition carries forth to a continuous distribution?
I understand that E(X) can be considered are centre of mass for a distribution but I don't seem to understand what it means intuitively to say that a mean of a continuous distribution is 5.