13.Statistics
medium

For the frequency distribution :

Variate $( x )$ $x _{1}$ $x _{1}$ $x _{3} \ldots \ldots x _{15}$
Frequency $(f)$ $f _{1}$ $f _{1}$ $f _{3} \ldots f _{15}$

where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and

$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be 

A

$2$

B

$1$

C

$4$

D

$6$

(JEE MAIN-2020)

Solution

$\because \sigma^{2} \leq \frac{1}{4}( M – m )^{2}$

Where $M$ and $m$ are upper and lower bounds

of values of any random variable.

$\therefore \quad \sigma^{2}<\frac{1}{4}(10-0)^{2}$

$\Rightarrow 0<\sigma<5$

$\therefore \sigma \neq 6$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.