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

  1. Less than $30$
  2. More than $30$
  3. Equal to $30$
  4. 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.


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  • $\begingroup$ If p=1/2 the binomial is symmetric and hence the mean is equal to the median. Since P(X<15)<1/2 the median and hence the mean also are greater than 15. Now the mean for the binomial is np=n/2 in this case.Now n/2 is greater than 15. so n>30. Choice 2 is the answer. You don't have enough information to determine n exactly. $\endgroup$ – Michael Chernick Feb 7 at 3:20
  • $\begingroup$ But isn't median=mean only when np is an integer, but here we don't know anything about n. But since probability=0.5. So, if n is even, n/2 is the unique median, and if n is odd then median lies between (n-1)/2 and (n+1)/2. So we can't really say that median=mean=n/2? $\endgroup$ – user233797 Feb 7 at 6:31

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.


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 $0$.

  • $\begingroup$ If $p=0$, then $P(X=0) = 1$. How are you getting that $P(X<15) < 0.5$ for any $n \geq 15$? In fact, $P(X<15) = 1$, not $0$ as you claim, for all $n >0$. $\endgroup$ – Dilip Sarwate Feb 6 at 17:09
  • $\begingroup$ @DilipSarwate an unfortunate typo, thanks. The 2nd sent. luckily assumes the inverse :) $\endgroup$ – gunes Feb 6 at 17:17
  • $\begingroup$ How P(X=0)=1? Isn't 0^0 not defined? $\endgroup$ – user233797 Feb 7 at 5:59
  • $\begingroup$ @user233797, the question is not for me but, yes, it is undefined, however, if $p=0$, number of successful events out of $n$ events will be equal to $0$ for certain, so probability of $P(X=0)$ will be $1$. (If it were p = 0, but it is p = 1 in the answer, after the typo correction) $\endgroup$ – gunes Feb 7 at 6:02

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