How to calculate perceptiles in a discute r.v? > The problem said: > > Suppose the random variable $X$ has pmf > > $Pr(n)=\frac{1}{3} (\frac{2}{3})^n , n=0,1,2,···$ > > Find the median and the $70$th percentile. My work, **MEDIAN:** $0.5=(1/3) \cdot (2/3)^n$ then $n=0.608$ and for **$70$th perceptile.** $0.7=(1/3) \cdot (2/3)^n$ then $n=1.11$ * * * The solution said: median=$1$ and $70$th percentile=$2$. **My question** , is my procedure fine? or I missing something? Thanks.

The random variable $X$ is the number of **failures** until the first success, where the probability of success on any trial is $1/3$.

The random variable $X$ is $\ge n$ if there are $n$ failures in a row. Thus $\Pr(X\ge n)=(2/3)^n$.

As is the case for many discrete distributions (and even some continuous ones) the median is not completely determined. But $\Pr(X\ge 2)=(2/3)^2=4/9\lt 1/2$, so $\Pr(X\lt 2)=5/9$, and $\Pr(X\lt 1)=1/3$. By one of the conventions for median in this kind of situation, one would average and say the median is $(1+2)/2$.

For the $70$-th percentile, we want to find the $n$ such that $(2/3)^n\approx 0.30$. Note that $n=3$ brings us under $0.30$, while $n=2$ leaves us over $0.30$. One of the conventional answers for median is $(2+3)/2$.