Question

Sol. [Key Concept: Mean $= \frac{\sum\ r(r{- 1)\ }^{n}C_{r}}{2^{n}} = \frac{n(n - 1)}{4} = 60$ fixes $n$; then the median is the middle cumulative value.]

$$\overline{x} = \frac{0.\ ^{n}C_{0} + 0.\ ^{n}C_{1} + 2.\ ^{n}C_{2} + \ldots + n(n - 1).nC_{n}}{\ ^{n}C_{0} + \ ^{n}C_{1} + \ ^{n}C_{2} + \ldots + {nC}_{n}}$$

$$ {= \frac{\sum_{r = 0}^{n}\mspace{2mu} r(r - 1)^{n}C_{r}}{2^{n}}}$$

$$ {= \frac{{\sum_{}^{}}_{r = 1}^{n}\mspace{2mu} n(r - 1)^{n - 1}C_{r - 1}}{2^{n}}}$$

$${= \frac{{\sum_{}^{}}_{r = 2}^{n}\mspace{2mu} n(n - 1)^{n - 2}C_{r - 2}}{2^{n}} }{= \frac{n(n - 1)2^{n - 2}}{2^{n}} = \frac{n(n - 1)}{4}}$$

$$ {\Rightarrow \frac{n(n - 1)}{4} = 60}$$

$$\Rightarrow n^{2} - n - 240 = 0$$

$$ \Rightarrow n = 16, - 15$$ (Reject)

$$N = \sum f_{i} = \ ^{16}C_{0} + \ ^{16}C_{1} + \ldots.. + \ ^{16}C_{16} = 2^{16}$$

$$ N = 2^{16}$$ (Even)

Median $= \frac{\left( \frac{N}{2} \right)^{\text{th}\text{~}} + \left( \frac{N}{2} + 1 \right)^{\text{th}\text{~}}}{2}$ term

$$\frac{N}{2} = 2^{15} = 32768$$

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$$240\ \ ^{16}C_{16}$$

$$ \begin{matrix} x_{i} & f_{i} & \text{~}\text{CF}\text{~} \end{matrix}$$

$$ \begin{matrix} 0 & \ ^{16}C_{0} & 1 \end{matrix}$$

$$ \begin{matrix} 0 & \ ^{16}C_{1} & 16 \end{matrix}$$

$$ \begin{matrix} 2 & \ ^{16}C_{2} & 137 \end{matrix}$$

$$ {6\ \ ^{16}C_{3}\ 697}$$

$$ {12\ \ ^{16}C_{4}\ 2517}$$

$$ {20\ \ ^{16}C_{5}\ 6885}$$

$$ {30\ \ ^{16}C_{6}\ 14893}$$

$$ {42\ \ ^{16}C_{7}\ 26333}$$

$$ 56\ \ ^{16}C_{8}\ 39203 \rightarrow \frac{\text{ }N}{2}\&\left( \frac{\text{ }N}{2} + 1 \right)$$ lies

$${72\ \ ^{16}C_{9}\ \frac{n(n - 1)2^{n - 2}}{2^{n}} }{\vdots \ \vdots }$$⋮ ⋮

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⇒ median $= \frac{56 + 56}{2} = 56$