# Symmetrical nature of binomial distribution [closed]

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 $$P(Xmedian)<0.5$$ But I am not able to interpret anything about $$n$$. Please help me with the same.

## closed as off-topic by kjetil b halvorsen, mkt, mdewey, Peter Flom♦Feb 7 at 11:11

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• 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. – Michael Chernick Feb 7 at 3:20
• 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? – 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 $$0$$.
• 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 $n >0$. – Dilip Sarwate Feb 6 at 17:09
• @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) – gunes Feb 7 at 6:02