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  • Consider the data on x taking the values with frequencies <img alt="^{n}C_{0},^{n}C_{1},^{n}C_{2}, .......^{n}C_{n}" sr

Consider the data on x taking the values 0,2,4,8, ......., 2^{n} with frequencies ^{n}C_{0},^{n}C_{1},^{n}C_{2}, .......^{n}C_{n} respectively. If the mean of this data is \frac{728}{2^{n}}, then n is equal to _________.
Option: 1 6
Option: 2 7
Option: 3 8
Option: 4 9

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\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{x} & 0 & 2 & 4 & 8 &... & 2^{\mathrm{n}} \\ \hline \mathrm{f} & { }^{\mathrm{n}} \mathrm{C}_{0} & { }^{\mathrm{n}} \mathrm{C}_{1} & { }^{\mathrm{n}} \mathrm{C}_{2} & { }^{\mathrm{n}} \mathrm{C}_{3} &... & { }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \\ \hline \end{array}

\\\text { Mean }=\frac{\sum x_{i} f_{i}}{\sum f_{i}}=\frac{\sum_{r=1}^{n} 2^{r}\;{ }^{n} C_{r}}{\sum_{r=0}^{n}{ }^{n} C_{r}} \\ \\\\\text { 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

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himanshu.meshram

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