Consider the data on x taking the values 2^0, | vedclass

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(D) The mean is given by $	ext{Mean} = \frac{\sum_{i=0}^{n} x_i f_i}{\sum_{i=0}^{n} f_i}$.Here,$x_i = 2^i$ and $f_i = {}^nC_i$.So,$	ext{Mean} = \fra

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$6$

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(D) The mean is given by $	ext{Mean} = \frac{\sum_{i=0}^{n} x_i f_i}{\sum_{i=0}^{n} f_i}$.Here,$x_i = 2^i$ and $f_i = {}^nC_i$.So,$	ext{Mean} = \frac{\sum_{i=0}^{n} 2^i {}^nC_i}{\sum_{i=0}^{n} {}^nC_i}$.Using the binomial theorem,$\sum_{i=0}^{n} {}^nC_i = (1+1)^n = 2^n$.Also,$\sum_{i=0}^{n} 2^i {}^nC_i = (1+2)^n = 3^n$.Thus,$	ext{Mean} = \frac{3^n}{2^n}$.Given $	ext{Mean} = \frac{728}{2^n}$,we have $\frac{3^n}{2^n} = \frac{728}{2^n}$.This implies $3^n = 728$. However,checking the summation range,the values are $2^0, 2^1, \ldots, 2^n$. The sum $\sum_{i=0}^n 2^i {}^nC_i = 3^n$. If the first term $x=0$ is excluded or if the sequence starts from $2^1$,the sum is $3^n - {}^nC_0 = 3^n - 1$.Given $\frac{3^n - 1}{2^n} = \frac{728}{2^n}$,we get $3^n - 1 = 728$,so $3^n = 729$.Since $3^6 = 729$,$n = 6$.

Consider the data on $x$ taking the values $2^0, 2^1, 2^2, \ldots, 2^n$ with frequencies ${}^nC_0, {}^nC_1, {}^nC_2, \ldots, {}^nC_n$ respectively. If the mean of this data is $\frac{728}{2^n}$,then $n$ is equal to

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