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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Statistics

Median of $ ^{2n}C_0, ^{2n}C_1, ^{2n}C_2, ^{2n}C_3...... ^{2n}C_n$ ( where n is even ) is

$\begin {array} {1 1} (A)\;^{2n}C_{\large\frac{n}{2}} & \quad (B)\;\large\frac{^{2n}C_{n-1}}{2} \\ (C)\;\large\frac{^{2n}C_{n+1}}{2} & \quad (D)\;None \: of \: these \end {array}$

 

1 Answer

Given observation are
$ ^{2n}C_0, ^{2n}C_1, ^{2n}C_2......^{2n}C_n$
Here total number of observation = $ n+1$ which is odd.
$ \therefore $ Median = $ \bigg( \large\frac{(n+1)+1}{2} \bigg)th $ observations.
$ = \bigg( \large\frac{n}{2}+1 \bigg)th$ observations
$ = ^{2n}C_n$
Ans : (D)
answered Jan 31, 2014 by thanvigandhi_1
 

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