The standard deviation of the data: x 1 a a^2 .... a^n f ^nC_0 ^nC_1 ^nC_2 .... ^nC_2 is,

The standard deviation of the data: x 1 a a^2 .... a^n f ^nC_0 ^n…

MCQ

The standard deviation of the data:

x	1	a	$a^2$	....	$a^n$
f	$^nC_0$	$^nC_1$	$^nC_2$	....	$^nC_2$

is,

A
$\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
B
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
C
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
✓
None of these

✓

Answer

Correct option: D.

None of these

None of these

Solution:

$x_i$	$f_i$	$f_ix_i$	$x_i^2$	$f_ix_i^2$
1	$^nC_0$	$^nC_0$	1	1
a	$^nC_1$	$a^nC_1$	$a^2$	$a^2nC_1$
a	$^nC_2$	$a^2nC_2$	$a^4$	$a^{4 n}C_2$
a	$^nC_3$	$a^{3 n}C_3$	$a^6$	$a^{6 n}C_3$
: : :	: : :	: : :	: : :	: : :
$a^n$	$^nC_n$	$a^{n n}C_n$	$a^{2n}$	$a^{2n n}C_n$
	$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}=2^\text{n}$	$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}=(1+\text{a})^\text{n}$		$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$

Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$
$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$
$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$
$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$
$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$
$\sigma=\sqrt{\text{Variance}(\text{X})}$
$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$