$I f a v a r i a b l e t a k e s t h e v a l u e 0, 1, 2, \dots . . . n w i t h f r e q u e n c i e s p r o p o r t i o n a l t o t h e b i n o m i a l c o e f f i c i e n t s$ ${}^{n}C_{0}, {}^{n}C_{1} {}^{n}C_{2}, \dots, {}^{n}C_{n}, then mean of distribution is$

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$\frac{n (n + 1)}{2}$

$\frac{n}{2}$

$\frac{n (n - 1)}{2}$

$\frac{2}{n}$

answer is A.

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Detailed Solution

$N = \sum_{i = 1}^{n} f_{i} = k (^{n} C_{0} +^{n} C_{1} + {......}^{n} C_{n}) = k \cdot 2^{n}$

$\begin{array}{l} \sum_{i = 1}^{n} f_{i} x_{i} = \sum_{i = 1}^{n} i . (k {}^{n}c_{i}) = k \cdot n {(2)}^{n - 1} \\ M e a n = \bar{x} = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{N} = \frac{n \cdot 2^{n - 1}}{2^{n}} = \frac{n}{2} \end{array}$

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