The arithmetic mean of the following frequency distribution: Variable (X) : 0 1 2 3 …. nFrequency (f) : nC0 nC1 nC2 nC3 …. nCnis

Let X¯ denote the required mean. Then,X¯=∑r=0n r⋅nCr∑r=0n nCr=∑r=1n r⋅nrn−1Cr−1∑r=0n rnCr=n∑r=1n n−1Cr−12n⇒ X¯=n×2n−12n=n2 ∵∑r=1n n−1Cr−1=2n−1

The arithmetic mean of the following frequency distribution: Variable (X) : 0 1 2 3 …. nFrequency (f) : nC0 nC1 nC2 nC3 …. nCnis

Let X¯ denote the required mean. Then,X¯=∑r=0n r⋅nCr∑r=0n nCr=∑r=1n r⋅nrn−1Cr−1∑r=0n rnCr=n∑r=1n n−1Cr−12n⇒ X¯=n×2n−12n=n2 ∵∑r=1n n−1Cr−1=2n−1

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The arithmetic mean of the following frequency distribution:
Variable (X) $\begin{array}{lllllll} : 0 1 2 3 \dots . n \end{array}$
Frequency (f) $\begin{array}{lllllll} :^{n} C_{0}^{n} C_{1}^{n} C_{2}^{n} C_{3} \dots .^{n} C_{n} \end{array}$
is

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$\frac{2^{n}}{n}$

$\frac{2^{n}}{n + 1}$

$\frac{n}{2}$

$\frac{2}{n}$

answer is C.

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Detailed Solution

Let $\bar{X}$ denote the required mean. Then,

$\begin{array}{l} \bar{X} = \frac{\sum_{r = 0}^{n} r \cdot^{n} C_{r}}{\sum_{r = 0}^{n}^{n} C_{r}} = \frac{\sum_{r = 1}^{n} r \cdot {\frac{n}{r}}^{n - 1} C_{r - 1}}{\sum_{r = 0}^{n} r^{n} C_{r}} = \frac{n \sum_{r = 1}^{n}^{n - 1} C_{r - 1}}{2^{n}} \\ \Rightarrow \bar{X} = \frac{n \times 2^{n - 1}}{2^{n}} = \frac{n}{2} [∵ \sum_{r = 1}^{n}^{n - 1} C_{r - 1} = 2^{n - 1}] \end{array}$