A variable X takes values 0,0,2,6,12,20, , n(n-1) with frequencies ^n C_0, ^n C_1, ^n C_2,…

Question

A variable $X$ takes values $0,0,2,6,12,20, \ldots, n(n-1)$ with frequencies ${ }^n C_0,{ }^n C_1,{ }^n C_2,{ }^n C_3,{ }^n C_4,{ }^n C_5, \ldots,{ }^n C_n$, respectively. If the mean of this data is 60 , then its median is :

Accepted Answer

The given variable $X$ takes values

$0,0,2,6,12,20,\ldots ,n(n-1)$

with corresponding frequencies

${}^{n}{C}_{0},{}^{n}{C}_{1},{}^{n}{C}_{2},{}^{n}{C}_{3},{}^{n}{C}_{4},{}^{n}{C}_{5},\ldots ,{}^{n}{C}_{n}$

We notice that the values of $X$ follow the pattern

$X=r(r-1),\, r=0,1,2,\ldots ,n$

and the corresponding frequency is

${}^{n}{C}_{r}$

So this is a discrete frequency distribution with:

variate value ${x}_{r}=r(r-1)$

frequency ${f}_{r}={}^{n}{C}_{r}$

Use the formula for mean: Mean is given by

$\vec{X}=\frac{\sum {f}_{r}{x}_{r}}{\sum {f}_{r}}$

So,

$\vec{X}=\frac{\sum _{n}^{r=0}{}^{n}{C}_{r}\, r(r-1)}{\sum _{n}^{r=0}{}^{n}{C}_{r}}$

Given mean $=60$, so

$\frac{\sum _{n}^{r=0}{}^{n}{C}_{r}\, r(r-1)}{\sum _{n}^{r=0}{}^{n}{C}_{r}}=60$

Now use standard binomial identities:

$\sum _{n}^{r=0}{}^{n}{C}_{r}={2}^{n}$

and

$\sum _{n}^{r=0}r(r-1)\, {}^{n}{C}_{r}=n(n-1){2}^{n-2}$

Hence,

$\vec{X}=\frac{n(n-1){2}^{n-2}}{{2}^{n}}=\frac{n(n-1)}{4}$

Given this is $60$:

$\frac{n(n-1)}{4}=60$

$n(n-1)=240$

${n}^{2}-n-240=0$

$(n-16)(n+15)=0$

So,

$n=16$

Write the values and frequencies for $n=16$: Now total frequency is

$N=\sum _{16}^{r=0}{}^{16}{C}_{r}={2}^{16}=65536$

To find the median, we need the value of $X$ whose cumulative frequency just exceeds $\frac{N}{2}$.

$\frac{N}{2}=32768$

Now frequencies are:

${}^{16}{C}_{0},{}^{16}{C}_{1},{}^{16}{C}_{2},\ldots ,{}^{16}{C}_{16}$

Using symmetry of binomial coefficients:

${}^{16}{C}_{0}+{}^{16}{C}_{1}+\cdots +{}^{16}{C}_{7}=\frac{1}{2}({2}^{16}-{}^{16}{C}_{8})$

Since

${}^{16}{C}_{8}=12870$

we get

${}^{16}{C}_{0}+{}^{16}{C}_{1}+\cdots +{}^{16}{C}_{7}=\frac{65536-12870}{2}=26333$

Now add ${}^{16}{C}_{8}$:

$26333+12870=39203$

So the cumulative frequency crosses $32768$ at $r=8$.

Hence the median corresponds to

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

Final Answer
$56$

So, the correct option is A.

Solutions · 1