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If a variable x takes values 0,1,2,..,n ...

If a variable x takes values 0,1,2,..,n with frequencies proportional to the binomial coefficients `.^(n)C_(0),.^(n)C_(1),.^(n)C_(2),..,.^(n)C_(n)`, then var (X) is

A

`(n^(2)-1)/(12)`

B

`(n)/(2)`

C

`(n)/(4)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C

We have
`overline(X)=(0^(n)C_(0)+1^(n)C_(1)+2^(n)C_(2)+..+n^(n)C_(n))/(.^(n)C_(0)+.^(n)C_(1)+.^(n)C_(2)+..+.^(n)C_(n))=(underset(r=0)overset(n)(sum r^(n)C_(r )))/(underset(r=0)overset(n)(sum .^(n)C_(r )))`
`=(1)/(2^(n))underset(r=1)overset(n)(sum r)(n)/(r ).^(n-1)C_(r-1) " " [because underset(r=0)overset(n)(sum).^(n)C_(r )=2^(n), .^(n)C_(r )=(n)/(r ).^(n-1)C_(r-1)]`
`=(n)/(2^(n))underset(r=1)overset(n)(sum).^(n-1)C_(r-1)=(n)/(2^(n))2^(n-1)=(n)/(2)[because underset(r=1)overset(n)(sum).^(n-1)C_(r-1)=2^(n-1)]`
and `(1)/(N)sum f_(i)x_(1)^(2)=(1)/(2^(n))sum r^(2) .^(n)C_(r )=(1)/(2^(n)) underset(r=0)overset(n)(sum)[r(r-1)+r].^(n)C_(r )`
`=(1)/(2^(n)){underset(r=0)overset(n)(sum)r-(r-1).^(n)C_(r )+underset(r=0)overset(n)(sum r) .^(n)C_(r )}`
`=(1)/(2^(n)){underset(r=2)overset(n)(sum)r(r-1)(n)/(r )(n-1)/(r-1) .^(n-2)C_(r-2)+underset(r=1)overset(n)(sum r) (n)/(r ).^(n-1)C_(r-1)}`
`=(1)/(2^(n)){n(n-1)2^(n-2)+n2^(n-1)}=(n(n-1))/(4)+(n)/(2)`
`therefore "Var"(X)=(1)/(N)sum f_(i)x_(i)^(2)-overline(X^(2))=(n(n-1))/(4)+(n)/(2)-(n^(2))/(4)=(n)/(4)`
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