# Consider the data on x taking

Question:

Consider the data on $x$ taking the values $0,2,4,8, \ldots, 2^{\text {n }}$ 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:

Mean $=\frac{\Sigma x_{i} f_{i}}{\Sigma f_{i}}$

$=\frac{0 \cdot{ }^{n} C_{0}+2 \cdot{ }^{n} C_{1}+2^{2} \cdot{ }^{n} C_{2}+\ldots+2^{n} \cdot{ }^{n} C_{n}}{{ }^{n} C_{0}+{ }^{n} C_{1}+\ldots .+{ }^{n} C_{n}}$

To find sum of numerator consider

$(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$.....(i)

Put $x=2 \Rightarrow 3^{n}-1=2 \cdot{ }^{n} C_{1}+2^{2} \cdot{ }^{n} C_{2}+\ldots . .+2{ }^{n} \cdot{ }^{n} C_{n}$

To find sum of denominator, put $x=1$ in (i), we get

$2^{n}={ }^{n} C_{0}+{ }^{n} C_{1}+\ldots . .+{ }^{n} C_{n}$

$\therefore \frac{3^{n}-1}{2^{n}}=\frac{728}{2^{n}} \Rightarrow 3^{n}=729 \Rightarrow n=6$