13.Statistics
hard

$6$ અવલોકનો $a$, $b,$ $68,$ $44,$ $48,$ $60$ ના મધ્યક અને વિચરણ અનુક્કમે $55$ અને $194$ છે. જો $a > b,$ તો $a +$ $3 b=$..........................

A

$200$

B

$190$

C

$180$

D

$210$

(JEE MAIN-2024)

Solution

$\mathrm{a}, \mathrm{b}, 68,44,48,60$

Mean $=55$       $a>b$

Variance $=194$    $a+3 b$

$\frac{a+b+68+44+48+60}{6}=55$

$\Rightarrow 220+a+b=330$

$\therefore a+b=110 \ldots . .(1)$

Also,

$\sum \frac{\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{n}}=194 $

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2+(68-55)^2+(44-55)^2$

$+(48-55)^2+(60-55)^2=194 \times 6$

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2+169+121+49+25=1164$

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2=1164-364=800$

$\mathrm{a}^2+3025-110 \mathrm{a}+\mathrm{b}^2+3025-110 \mathrm{~b}=800$

$\Rightarrow \mathrm{a}^2+\mathrm{b}^2=800-6050+12100$

${a}^2+\mathrm{b}^2=6850 \ldots \ldots .(2)$

Solve $(1) \& (2);$

$a=75, b=35$

$\therefore$ $a+3 b=75+3(35)=75+105=180$

Standard 11
Mathematics

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