Median for a Binomial distribution. Say, $X$ is the number of heads obtained in $n=10$ trials, so, $X$~$Bin(10,0.5)$.
I am a little confused in it's median.
Not in calculation, but it's meaning.
For $n=10,p=0.5$ the binomial distribution is symmetrical and Mean=Median=Mode=$5$.
Now, when I see $Mean$=$5$, I say, out of $10$ trials, the expected number of heads equals $5$.
When I see $Mode$=$5$, I say the chances of obtaining $5$ heads on tossing $10$ coins is the highest.
Now I don't really know what to make of $Median=5$.
I mean $X$ is the number of heads in $10$ coin tosses. How to interpret the median here ? By definition median tells us about the middlemost value in the data.
But what does it tell here? Can anyone help ?
 A: The median $m$ is defined as any value where $\mathsf P(X\leq m)\geq \tfrac 12$ and $\mathsf P(X\geq m)\geq \tfrac 12$.   It is basically the value which divides the probability distribution.
For $X\sim\mathcal {Bin}(10,0.5)$ we have $m=10\cdot 0.5$
$$\mathsf P(X\leq 5) = \sum_{x=0}^5 \binom{10}{x}0.5^{10} = \frac{319}{512}\\\mathsf P(X\geq 5) = \sum_{x=5}^{10} \binom{10}{x}0.5^{10} = \frac{319}{512}$$
A: It tells you that in roughly 50% of all cases you will have less than 5 or 5 out of 10 heads, and in the other ~50% of the cases you have 5 or more heads. 
With 10 tosses the probability of having exactly 5 heads is pretty big, so the exact probability of having 5 or more heads is significantly larger than 50%. More precisely, therefore, 5 is the largest number of heads that you'd expect to be exceeded (with equality) in the majority of cases.
A: $m$ is by definition a median of the distribution induced by random variable $X$ if it satisfies: $$P(X\geq m)\geq0.5\wedge P(X\leq m)\geq0.5$$
If $X\sim\mathsf{Bin}(10,0.5)$ then only $m=5$ has this property.
