The mean of the values $0, 1, 2, . . . . ., n$ having corresponding weight $^{n} C_{0},^{n} C_{1},^{n} C_{2}, . . . . .,^{n} C_{n}$ respectively, is?

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

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

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\frac{n + 1}{2}

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\frac{n}{2}

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Solution

The correct option is D $\frac{n}{2}$
The solution mean is $¯ x = \frac{0 \cdot 1 + 1 \cdot^{n} C_{1} + 2 \cdot^{n} C_{2} + 3 \cdot^{n} C_{3} + . . . . + n \cdot^{n} C_{n}}{1 +^{n} C_{1} +^{n} C_{2} + . . . . +^{n} C_{n}}$

$= \frac{n \sum r = 0 r \cdot^{n} C_{r}}{n \sum r = 0^{n} C_{r}} = \frac{n \sum r = 1 r \cdot \frac{n}{r} \cdot^{n - 1} C_{r - 1}}{n \sum r = 0^{n} C_{r}}$

$= \frac{n n \sum r = 1^{n - 1} C_{r - 1}}{n \sum r = 0^{n} C_{r}}$

$= \frac{n \cdot 2^{n - 1}}{2^{n}} = \frac{n}{2}$

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