Question

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

Answer 1

Hint: In order to solve this question, first of all we will find the value of \[{\mu _1}\] which is given by direct formula as \[{\mu _1} = \dfrac{{\sum\limits_{}^{} {n{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }}\] and \[{\mu _1}\] is called as mean of the given distribution. Then we will find \[{\mu _2}\] which is the mean of the squares of the variables.Here we will use the direct formula of \[{\mu _2}\] as \[{\mu _2} = \dfrac{1}{{{2^n}}}\sum\limits_0^n {r\left( {r - 1} \right)} {}^n{C_r} + \dfrac{n}{2}\] .And finally we will use the formula of variance as $\,Variance\,{\sigma ^2}\, = {\mu _2} - {\left( {{\mu _1}} \right)^2}$ . 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 $0,1,2,3.........,n$ with frequencies proportional to the binomial coefficients $C\left( {n,0} \right)\,,\,C\left( {n,1} \right)\,,\,C\left( {n,2} \right)\,,........,C\left( {n,n} \right)$ respectively. And we have to find the variance. So, first of all let’s \[{\mu _1}\] which is the mean of the given distribution. As we know that,
${\mu _1} = \dfrac{{\sum\limits_{}^{} {r{}^n{C_r}} }}{{\sum {{}^n{C_r}} }}$
$ \Rightarrow \dfrac{{\sum\limits_{}^{} {r\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }}$
\[ \Rightarrow \dfrac{{\sum\limits_{}^{} {n{}^{n - 1}{C_{r - 1}}} }}{{\sum {{}^n{C_r}} }}\]

Now we know that ${\left( {1 + x} \right)^n} = {}^n{C_0} + x{}^n{C_1} + {x^2}{}^n{C_2} + ............ + {x^n}{}^n{C_n}$
Now on putting the value of $x = 1$. We get,
${2^n} = {}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ........ + {}^n{C_n}$
Therefore, on solving we have
$ \Rightarrow \dfrac{{n{2^{n - 1}}}}{{{2^n}}}$
As we know that ${2^{n - 1}} = \dfrac{{{2^n}}}{2}$
$ \Rightarrow \dfrac{{n{2^n}}}{{2 \times {2^n}}}$
On dividing, we get
$ \Rightarrow \dfrac{n}{2}$

Now we know that
\[{\mu _2} = \dfrac{1}{{{2^n}}}\sum\limits_0^n {\left\{ {\left( {r - 1} \right) + r} \right\}{}^n{C_r}} \]
\[ \Rightarrow \dfrac{1}{{{2^n}}}\sum\limits_0^n {r\left( {r - 1} \right)} {}^n{C_r} + \dfrac{n}{2}\]
As we know that ${}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}}$,
$ \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
$ \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 ${2^{n - 2}} = \dfrac{{{2^n}}}{4}$
$ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{{{2^n}}}.\dfrac{{{2^n}}}{4} + \dfrac{n}{2}$
Then on dividing, we get
$ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{4} + \dfrac{n}{2}$
Now, using formula
$Variance\,{\sigma ^2}\, = {\mu _2} - {\left( {{\mu _1}} \right)^2}$
$ \Rightarrow {\sigma ^2}\, =\dfrac{{n\left( {n - 1} \right)}}{4} + \dfrac{n}{2} - {\left( {\dfrac{n}{2}} \right)^2}$
On multiplying, we get
$ \Rightarrow {\sigma ^2}\, =\dfrac{{{n^2}}}{4} - \dfrac{n}{4} + \dfrac{n}{2} - {\left( {\dfrac{n}{2}} \right)^2}$
$ \Rightarrow {\sigma ^2}\, =\dfrac{{{n^2}}}{4} - \dfrac{{{n^2}}}{4} + \dfrac{n}{2} - \dfrac{n}{4}$
$ \therefore {\sigma ^2}\, =\dfrac{n}{4}$

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

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.