Is not always the accumulated probability until the expected value equal to 1/2?

Question

My problem is that doing calculations with exponential distribution, the result I get is not 1/2, and I think it should be that according to the intuitive concept of the expected value. The density function for an exponential distribution is:

f(y) = (1/β) * e^(-y/β)
0 < y <∞

The average is β (that is known in advance). Calculating the accumulative probability to the average:

F(β) = ∫{0 < y < β} (1/β)*e^(-y/β) dy

F(β) = - ∫{0 < y < β} e^(-y/β) * (-1/β) dy

F(β) = - [e^(-y/β)] {y from 0 to β}

F(β) = - [e^(-β/β) - e^(0/β)]

F(β) = 1 - e^(-1)

F(β) = 1 - 1/e

F(β) = (e-1)/ e

I think the error is conceptual-

Answer 1

You are asking whether the median of a random variable $X$, that is, the place $m$ such that $\Pr(X\le m)=\frac{1}{2}$, is the same as the mean $E(X)$. For a symmetric distribution, like the normal, the median is equal to the mean (if the mean exists). But for "most" distributions, including the exponential, the median is not equal to the mean.

The "accumulated probability" until $E(X)$ is $\frac{1}{2}$ precisely if the median is equal to the mean.