A variable X takes values 0, 2, 4, 6 ….,2n with frequencie nC0,nC1,nC2,…,nCn respectively. If the mean of this data is 7282n, then n is equal to

Solution:

It is given that

$\begin{array}{l} \frac{0 \cdot^{n} C_{0} + 2 \cdot^{n} C_{1} + 4 \cdot^{n} C_{2} + \dots + 2^{n} \cdot^{n} C_{n}}{^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}} = \frac{728}{2^{n}} \\ \Rightarrow \frac{\sum_{r = 1}^{n}^{n} C_{r} \cdot 2^{r}}{\sum_{r = 1}^{n}^{n} C_{r}} = \frac{728}{2^{n}} \end{array}$

$\begin{array}{l} \Rightarrow \frac{\sum_{r = 0}^{n}^{n} C_{r} 2^{r} - 1}{2^{n}} = \frac{728}{2^{n}} \\ \Rightarrow \frac{(1 + 2)^{n} - 1}{2^{n}} = \frac{728}{2^{n}} \\ \Rightarrow 3^{n} - 1 = 729 \Rightarrow 3^{n} = 729 \Rightarrow 3^{n} = 3^{6} \Rightarrow n = 6 \end{array}$

$A$ variable $X$ takes values 0, 2, 4, 6 … $., 2^{n}$ with frequencie $^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}$ respectively. If the mean of this data is $\frac{728}{2^{n}},$ then $n$ is equal to

Solution:

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