Expected Value of Binomial Distribution What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads  when a coin is tossed 5 times?
Here is what I think :  Let  $X$  is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$
$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$
But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.

Since $X\sim Binom(5,\frac{1}{2})$, the range of $X$ is $\{0,1,2,3,4,5\}$, which implies that $2X - 5$ is not necessarily always positive. Therefore in your proof, it is not true that $E(\left|2X - 5\right|) = E(2X - 5)$.
