Symmetrical nature of binomial distribution If $X$ is a $binomial(n,p)$ random variable, and $P(X<15)<0.5$ Find $n$.


*

*Less than $30$

*More than $30$

*Equal to $30$

*None of the above.


I know that binomial distribution is symmetric when probability=$0.5$ and $P(X<median)=P(X>median)<0.5$ But I am not able to interpret anything about $n$. Please help me with the same. 
 A: The median of a random variable $X$ is defined to be any number $m$ such that $P(X \leq m) \geq \frac 12$ and $P(X \geq m) \geq \frac 12$. For a binomial random variable, the median is known to be one of $\lfloor np \rfloor$ and $\lceil np \rceil$ (see the Wikipedia entry for binomial distributions for a citation for this result) and of course, if $np$ is an integer, then the floor and ceiling are the same and the median is $np$.  So, given that $P(X<15) = P(X \leq 14)$ is strictly smaller than $0.5$, we know for sure that the median must be at least $15$.  Can the median be exactly $15$? Sure, as @gunes's answer points out, $(n, p) = (15,1)$ will work as will $(n, p) = \left(N, \frac{15}{N}\right)$ for any $N \geq 15$ to give a median of $15$. 
So, why the distractors regarding $30$ in the choice of answers?  Well, they might be there to entrap people into thinking of the symmetric case $(n,p) = (30,\frac 12)$ which has a median of $15$. More likely, I suspect that the OP forgot to tell us that it is given that $p = \frac 12$ which makes the binomial pmf symmetric about $\frac n2$ (see the title of the question). So, a $\mathsf{Binomial}\,(n,\frac 12)$ random variable $X$ with the property that $P(X < 15) < \frac 12$ must have $n \geq 30$: a smaller $n$ will not do since, for example, $P(X < 15) = P(X \leq 14)$ equals $\frac 12$ for a $\mathsf{Binomial}\,(29,\frac 12)$ random variable $X$. Of course, $n \geq 30$ is not one of the choices listed in the possible answers which makes me suspect that the OP didn't bother to way that it was the minimum value of $n$ that was being asked about.
A: If there is no restriction on $p$, we can set it as $p=1$. So, any $n\geq15$ gives $P(X<15)<0.5$ because actually $P(X<15)=0$. If $p=1$, we only have $P(X=n)=1$. So, it is not restricted to $<30,>30$ or $=30$. We can also set $p$ arbitrarily small to penetrate each of the cases you listed, i.e. we don't need to set it to $0$.
