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 ?
Best Answer
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 $m=10\cdot 0.5$
$$\mathsf P(X\leq 5) = \sum_{x=0}^5 \binom{10}{x}0.5^{10} = \frac{319}{512}\\\mathsf P(X\geq 5) = \sum_{x=5}^{10} \binom{10}{x}0.5^{10} = \frac{319}{512}$$