Q. Consider the data on $x$ taking the values $0,2,$ $4,8, \ldots, 2^{ n }$ with frequencies ${ }^{n} C_{0},{ }^{n} C_{1},{ }^{n} C_{2}, \ldots,{ }^{n} C_{n}$ respectively. If the mean of this data is $\frac{728}{2^{n}}$, then $n$ is equal to _______.
Solution:
x
0
2
4
8
$2^{n}$
f
${ }^{ n } C _{0}$
${ }^{ n } C _{1}$
${ }^{ n } C _{2}$
${ }^{ n } C _{3}$
${ }^{ n } C _{n}$
Mean $=\frac{\sum x _{ i } f_{ i }}{\displaystyle\sum f_{ i }}=\frac{\displaystyle\sum_{ r =1}^{ n } 2^{ r }{ }^{ n } C _{ r }}{\sum_{ r =0}^{ n }{ }^{ n } C _{ r }}$
Mean $=\frac{(1+2)^{ n }-{ }^{ n } C _{0}}{2^{ n }}=\frac{728}{2^{ n }}$
$\Rightarrow \frac{3^{ n }-1}{2^{ n }}=\frac{728}{2^{ n }}$
$\Rightarrow 3^{ n }=729$
$ \Rightarrow n =6$
| x | 0 | 2 | 4 | 8 | $2^{n}$ | |
| f | ${ }^{ n } C _{0}$ | ${ }^{ n } C _{1}$ | ${ }^{ n } C _{2}$ | ${ }^{ n } C _{3}$ | ${ }^{ n } C _{n}$ |