Suppose values taken by a variables X are such that a≤xi≤b where xi denotes the value of X in the ith case for i=1, 2,......,n. Then,

A
$\frac{a^{2}}{4} \leq Var (X)$
B
$(b - a)^{2} \geq Var (X)$
C
$a \leq Var (X) \leq b$
D
$a^{2} \leq Var (X) \leq b^{2}$

Solution:

We have,

$a \leq x_{i} \leq b; i = 1, 2, \dots, n$ …(i)

$\begin{array}{l} \Rightarrow \sum_{i = 1}^{n} a \leq \sum_{i = 1}^{n} x_{i} \leq \sum_{i = 1}^{n} b \\ \Rightarrow n a \leq \sum_{i = 1}^{n} x_{i} \leq n b \end{array}$

$\Rightarrow a \leq \frac{1}{n} \sum_{i = 1}^{n} x_{i} \leq b$

$\Rightarrow a \leq \bar{X} \leq b \Rightarrow - b \leq - \bar{X} \leq - a$ …(ii)

From (i) and (ii), we get

$a - b \leq x_{i} - \bar{X} \leq b - a; i = 1, 2, \dots, n$

$\begin{array}{l} \Rightarrow |x_{i} - \bar{X}| \leq b - a; i = 1, 2, \dots, n \\ \Rightarrow {(x_{i} - \bar{X})}^{2} \leq (b - a)^{2}; i = 1, 2, \dots, n \end{array}$

$\begin{array}{l} \Rightarrow \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2} \leq n (b - a)^{2} \\ \Rightarrow \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2} \leq (b - a)^{2} \Rightarrow Var (X) \leq (b - a)^{2} \end{array}$

Suppose values taken by a variables $X$ are such that $a \leq x_{i} \leq b$ where $x_{i}$ denotes the value of $X$ in the $i t h$ case for $i = 1, 2, . . . . . ., n .$ Then,

Solution:

Related content