Question

║⊕QUESTION⊕║
Do not worry about your difficulties in Mathematics. I can assure you mine are still greater.
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CLASS 11
STATISTICS
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If a variable takes the values of 0, 1, 2, .... n with frequencies proportional to the binomial coefficients C(n,0), C(n,1), C(n,2),....C(n,n) respectively, then the variance of the distribution is
a) root n / 2
b) n/4
c) n/2
d) n

Answer 1

Answer: n/4

Step-by-step explanation:

$\text{The terms are :}\ 0.{{n}\choose{0}} ,1. {{n}\choose{1}},2. {{n}\choose{0}} ,...\\ \\ {{n}\choose{r}} \ \text{represents the binomial coefficients } ^nC_r \\ \\ \text{Mean} = \frac{{\sum}_{r=0}^{n}\ r. {{n}\choose{r}}}{{\sum}_{r=0}^{n}\ {{n}\choose{r}}} = \frac{n\cdot 2^{n - 1}}{2^n} = \frac{n}{2}$

$\mu' =\frac{{\sum}_{r=0}^{n}\ r^2. {{n}\choose{r}}}{{\sum}_{r=0}^{n}\ {{n}\choose{r}}} = \frac{n(n+1)(2^{n-2})}{2^n} =\frac{n(n+1)}{4} \\ \\ \text{Variance}\ (\sigma^2) = \mu' - (\mu)^2 = \frac{n}{4}$

Derivation of Summation of Binomial Coefficients used :

$\text{i)} \sum\limits_{k=0}^n {n\choose k} \\ \\ \\ \text{We know that:} \\ \\ (1+t)^n=\sum_{k=0}^n\binom{n}{k}t^{k} \\ \\ \implies \sum\limits_{k=0}^n {n\choose k} = (1+1)^n = 2^n \\ \\ \\ \text{ii)} \sum\limits_{k=0}^n k. {n\choose k} = \sum\limits_{k=0}^n k. \frac{n}{k} {n-1\choose k-1} = n \ \sum\limits_{k=0}^n {n-1\choose k-1} = n.(2)^{n-1}$

$\text{iii)} \sum\limits_{k=0}^n {n\choose k} \\ \\ \\ \sum\limits_{k=0}^n {n\choose k} .x^k = (1+x)^n \\ \\ \text{Differentiating with respect to x and multiplying by x we get : }\\ \\ \sum\limits_{k=0}^n k. {n\choose k} x^k = nx(1+x)^{n-1} \\ \\ \text{Differentiating again with respect to x and substituting x = 1 we get} \\ \\ \sum\limits_{k=0}^n k^2. {n\choose k} = n(n+1)2^{n-2}$

Comment if any doubts or Error.

Answer 2

Answer:

ANSWER IS OPTION B

I HOPE IT'S HELP YOU

SORRY IF IN MY ANSWER ANY MISTAKE