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

Question

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

Accepted Answer

$\begin{gathered} {x}_{i} \quad  ({x}_{i}-\vec{x}) \quad  ({x}_{i}-\vec{x}{)}^{2} \ x \quad  x-9 \quad  (x-9{)}^{2} \ y \quad  y-9 \quad  (y-9{)}^{2} \ 10 \quad  1 \quad  1 \ 12 \quad  3 \quad  9 \ 6 \quad  -3 \quad  9 \ 12 \quad  3 \quad  9 \ 4 \quad  -5 \quad  25 \ 8 \quad  -1 \quad  1 \ x+y+92 \quad  \quad  (x-9{)}^{2}+(y-9{)}^{2}+54 \end{gathered}$

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

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

Also, variance $=9.25$

$\begin{gathered} \Rightarrow \frac{(x-9{)}^{2}+(y-9{)}^{2}+54}{8}=9.25 \ \Rightarrow {x}^{2}+{y}^{2}+81+81-2	imes 9(x+y)=20 \ \Rightarrow {x}^{2}+{y}^{2}-18	imes 20=-142 \ \Rightarrow {x}^{2}+{y}^{2}=218 \ \Rightarrow {x}^{2}+(20-x{)}^{2}=218 \ \Rightarrow {x}^{2}+400+{x}^{2}-40x=218 \ \Rightarrow 2{x}^{2}-40x+182=0 \ \Rightarrow x=\frac{40\pm 12}{4} \ \Rightarrow x=13x=7\Rightarrow y=7y=13 \ x>y \ 	herefore x=13y=7 \ 3x-2y=39-14=25 \end{gathered}$

Concept :

(a) Mean $=\frac{\Sigma {x}_{i}}{n}$

(b) Variance $=\frac{\Sigma {({x}_{i}-\vec{x})}^{2}}{n}$

Solutions · 1