13.Statistics
medium

बारंबारता बंटन

चर $( x )$ $x _{1}$ $x _{1}$ $x _{3} \ldots \ldots x _{15}$
बारंबारता $(f)$ $f _{1}$ $f _{1}$ $f _{3} \ldots f _{15}$

जहाँ $0 < x _{1} < x _{2} < x _{3} < \ldots < x _{15}=10$ तथा $\sum_{ i =1}^{15} f _{ i }>0$ है, का मानक विचलन, निम्न में से कौन-सा नहीं हो सकता ?

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

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