A variable X takes values 0, 0, 2, 6, 12, 20, , n(n-1) with frequencies n0, n1, n2, n3, , …

Question

A variable $X$ takes values $0, 0, 2, 6, 12, 20, \ldots, n(n-1)$ with frequencies $\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \binom{n}{3}, \ldots, \binom{n}{n}$, respectively. If the mean of this data is 60, then its median is:

Accepted Answer

Here $x_r = r(r-1)$, $f_r = \binom{n}{r}$, for $r=0,1,\ldots,n$.

$$\bar{X} = \frac{\sum_{r=0}^{n}\binom{n}{r}r(r-1)}{\sum_{r=0}^{n}\binom{n}{r}} = 60$$

Using standard identities $\sum\binom{n}{r} = 2^n$ and $\sum r(r-1)\binom{n}{r} = n(n-1)2^{n-2}$:

$$\bar{X} = \frac{n(n-1)2^{n-2}}{2^n} = \frac{n(n-1)}{4} = 60 \implies n(n-1)=240 \implies n^2-n-240=0$$
$$(n-16)(n+15)=0 \implies n=16$$

Total frequency $N = \sum_{r=0}^{16}\binom{16}{r} = 2^{16} = 65536$, so $\dfrac{N}{2} = 32768$.

Using symmetry, $\binom{16}{0}+\cdots+\binom{16}{7} = \dfrac{1}{2}(2^{16}-\binom{16}{8}) = \dfrac{65536-12870}{2} = 26333$.

Cumulative frequency up to $r=8$: $26333+12870 = 39203 > 32768$, and up to $r=7$ it's $26333 < 32768$. So the median corresponds to $r=8$:

$X = 8(8-1) = 56$

Solutions · 1