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Question: If the variable takes the values \(0,1,2,3.........,n\) with frequencies proportional to the binomia...

If the variable takes the values 0,1,2,3.........,n0,1,2,3.........,n with frequencies proportional to the binomial coefficients C(n,0) , C(n,1) , C(n,2) ,........,C(n,n)C\left( {n,0} \right)\,,\,C\left( {n,1} \right)\,,\,C\left( {n,2} \right)\,,........,C\left( {n,n} \right) respectively, then the variance of the distribution is
A. nn
B. n2\dfrac{{\sqrt n }}{2}
C. n2\dfrac{n}{2}
D. n4\dfrac{n}{4}

Explanation

Solution

In order to solve this question, first of all we will find the value of μ1{\mu _1} which is given by direct formula as μ1=∑nn−1Cr−1∑nCr{\mu _1} = \dfrac{{\sum\limits_{}^{} {n{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }} and μ1{\mu _1} is called as mean of the given distribution. Then we will find μ2{\mu _2} which is the mean of the squares of the variables.Here we will use the direct formula of μ2{\mu _2} as μ2=12n∑0nr(r−1)nCr+n2{\mu _2} = \dfrac{1}{{{2^n}}}\sum\limits_0^n {r\left( {r - 1} \right)} {}^n{C_r} + \dfrac{n}{2} .And finally we will use the formula of variance as  Variance σ2 =μ2−(μ1)2\,Variance\,{\sigma ^2}\, = {\mu _2} - {\left( {{\mu _1}} \right)^2} . And hence we will simplify it and get the required result.

Complete step by step answer:
We have given that the variable takes the values 0,1,2,3.........,n0,1,2,3.........,n with frequencies proportional to the binomial coefficients C(n,0) , C(n,1) , C(n,2) ,........,C(n,n)C\left( {n,0} \right)\,,\,C\left( {n,1} \right)\,,\,C\left( {n,2} \right)\,,........,C\left( {n,n} \right) respectively. And we have to find the variance. So, first of all let’s μ1{\mu _1} which is the mean of the given distribution. As we know that,
μ1=∑rnCr∑nCr{\mu _1} = \dfrac{{\sum\limits_{}^{} {r{}^n{C_r}} }}{{\sum {{}^n{C_r}} }}
⇒∑rnrn−1Cr−1∑nCr\Rightarrow \dfrac{{\sum\limits_{}^{} {r\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }}
⇒∑nn−1Cr−1∑nCr\Rightarrow \dfrac{{\sum\limits_{}^{} {n{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }}

Now we know that (1+x)n=nC0+xnC1+x2nC2+............+xnnCn{\left( {1 + x} \right)^n} = {}^n{C_0} + x{}^n{C_1} + {x^2}{}^n{C_2} + ............ + {x^n}{}^n{C_n}
Now on putting the value of x=1x = 1. We get,
2n=nC0+nC1+nC2+........+nCn{2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ........ + {}^n{C_n}
Therefore, on solving we have
⇒n2n−12n\Rightarrow \dfrac{{n{2^{n - 1}}}}{{{2^n}}}
As we know that 2n−1=2n2{2^{n - 1}} = \dfrac{{{2^n}}}{2}
⇒n2n2×2n\Rightarrow \dfrac{{n{2^n}}}{{2 \times {2^n}}}
On dividing, we get
⇒n2\Rightarrow \dfrac{n}{2}

Now we know that
{\mu _2} = \dfrac{1}{{{2^n}}}\sum\limits_0^n {\left\\{ {\left( {r - 1} \right) + r} \right\\}{}^n{C_r}}
⇒12n∑0nr(r−1)nCr+n2\Rightarrow \dfrac{1}{{{2^n}}}\sum\limits_0^n {r\left( {r - 1} \right)} {}^n{C_r} + \dfrac{n}{2}
As we know that nCr=nrn−1Cr−1{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}},
⇒12n∑0nr(r−1)n(n−1)r(r−1)n−2Cr−2+n2\Rightarrow \dfrac{1}{{{2^n}}}\sum\limits_0^n {r\left( {r - 1} \right)\dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}} + \dfrac{n}{2}}
On cancelling in numerator and denominator, we get
⇒n(n−1)2n.2n−2+n2\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{{{2^n}}}{.2^{n - 2}} + \dfrac{n}{2}

As there is no term of r we can remove the sigma.
We can also write 2n−2=2n4{2^{n - 2}} = \dfrac{{{2^n}}}{4}
⇒n(n−1)2n.2n4+n2\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{{{2^n}}}.\dfrac{{{2^n}}}{4} + \dfrac{n}{2}
Then on dividing, we get
⇒n(n−1)4+n2\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{4} + \dfrac{n}{2}
Now, using formula
Variance σ2 =μ2−(μ1)2Variance\,{\sigma ^2}\, = {\mu _2} - {\left( {{\mu _1}} \right)^2}
⇒σ2 =n(n−1)4+n2−(n2)2\Rightarrow {\sigma ^2}\, =\dfrac{{n\left( {n - 1} \right)}}{4} + \dfrac{n}{2} - {\left( {\dfrac{n}{2}} \right)^2}
On multiplying, we get
⇒σ2 =n24−n4+n2−(n2)2\Rightarrow {\sigma ^2}\, =\dfrac{{{n^2}}}{4} - \dfrac{n}{4} + \dfrac{n}{2} - {\left( {\dfrac{n}{2}} \right)^2}
⇒σ2 =n24−n24+n2−n4\Rightarrow {\sigma ^2}\, =\dfrac{{{n^2}}}{4} - \dfrac{{{n^2}}}{4} + \dfrac{n}{2} - \dfrac{n}{4}
∴σ2 =n4\therefore {\sigma ^2}\, =\dfrac{n}{4}

Hence, the correct option is (D)\left( D \right).

Note: Variance is the expectation of the squared deviation of a random variable from its mean. Variance is a measure of dispersion, meaning it is a measure of how far a set of numbers is spread out from their average value. The standard deviation is derived from variance and tells you, on average, how far each value lies from the mean. It’s the square root of variance. However, the variance is more informative about variability than the standard deviation, and it’s used in making statistical inference.