JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 21)

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 _____________.
Answer
25

Explanation

$$ \begin{array}{|c|c|c|} \hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\ \hline x & x-9 & (x-9)^2 \\ \hline y & y-9 & (y-9)^2 \\ \hline 10 & 1 & 1 \\ \hline 12 & 3 & 9 \\ \hline 6 & -3 & 9 \\ \hline 12 & 3 & 9 \\ \hline 4 & -5 & 25 \\ \hline 8 & -1 & 1 \\ \hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\ \hline \end{array} $$

Now, mean $(\bar{x})=9$

$$ \begin{aligned} & \Rightarrow \frac{x+y+52}{8}=9 \\\\ & \Rightarrow x+y=20 \end{aligned} $$

Also, variance $=9.25$

$$ \begin{aligned} & \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\ & \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\ & \Rightarrow x^2+y^2-18 \times 20=-142 \\\\ & \Rightarrow x^2+y^2=218 \\\\ & \Rightarrow x^2+(20-x)^2=218 \\\\ & \Rightarrow x^2+400+x^2-40 x=218 \\\\ & \Rightarrow 2 x^2-40 x+182=0 \\\\ & \Rightarrow x=\frac{40 \pm 12}{4} \\\\ & \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\ & \text { But } x>y \\\\ & \therefore x=13 \text { and } y=7 \\\\ & \text { So, } 3 x-2 y=39-14=25 \end{aligned} $$

Concept :

(a) Mean $=\frac{\Sigma x_i}{n}$

(b) Variance $=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}$

Comments (0)

Advertisement