Let mean and variance of 8 numbers x , y , 10 , 12,6,12,4,8 , be 9 and 9.25 respectively. If x >

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13.Statistics

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Let the mean and variance of $8$ numbers $x , y , 10$, $12,6,12,4,8$, be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to $...........$.

A

$24$

B

$25$

C

$23$

D

$22$

(JEE MAIN-2023)

Solution

$\frac{x+y+52}{8}=9 \Rightarrow x+y=20$

For variance

$x-9, y-9,3,3,1,-5,-1,-3$

$\bar{x}=0$

$\therefore \frac{(x-9)^2+(y-9)^2+54}{8}-0^2=9.25$

$(x-9)^2+(11-x)^2=20$

$x=7 \text { or } 13 \therefore y=13,7$

$3 x-2 y=3 \times 13-2 \times 7=25$

Standard 11

Mathematics

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