13.Statistics
hard

 ધારોકે માહિતી

$X$ $1$ $3$ $5$ $7$ $9$
આવૃતિ $(f)$ $4$ $24$ $28$ $\alpha$ $8$

 નો મધ્યક $5$ છે.જો માહિતીના મધ્યક સાપેક્ષ સરેરાશ વિચલન અને વિચરણ અનુક્રમે $m$ અને $\sigma^2$ હોય, તો $\frac{3 \alpha}{m+\sigma^2}=........$

A

$7$

B

$6$

C

$8$

D

$5$

(JEE MAIN-2023)

Solution

$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$

$\text { M.. }(\bar{x})=\frac{\sum f _{ i }\left| x _{ i }-\overline{ x }\right|}{\sum f _{ i }} \text { where } \sum f _{ i }=64+16=80$

$\text { M.D. }(\bar{x})=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}$

$=\frac{8}{5}$

$\text { Variance }=\frac{\sum f _{ i }\left( x _{ i }-\overline{ x }\right)^2}{\sum f _{ i }}$

$=\frac{4 \times 16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{352}{80}$

$\therefore \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{128}{80}+\frac{352}{80}}=8$

Standard 11
Mathematics

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