Question

$\large\rm{QuésTion:-}$

The equation given of the two regression lines are 2x + 3yb-6 = 0 and 5x + 7y - 12 = 0.

Find :-

A) correction cofficient
B) σ_{x}/σ_{y}

Answer 1

$\blue{\underline{ \large \rm{QuésTíon :- }}}$

The equation given of the two regression lines are 2x + 3yb-6 = 0 and 5x + 7y - 12 = 0.

Find :-

A) correction cofficient

B) $\rm\frac{σ_{x}}{σ_{y}}$

$\blue{\underline{ \large \rm{SoluTíon :- }}}$

ᴡᴇ ᴀssᴜᴍᴇ ᴛʜᴀᴛ 2x + 3ʏ -6 = 0 ᴛᴏ ʙᴇ ᴛʜᴇ ʟɪɴᴇs ᴏғ ʀᴇɢʀᴇssɪᴏɴ ᴏғ ʏ ᴏɴ x ,

$\sf \: 2x + 3y - 6 = 0$

$\sf \implies \: x = - \frac{3}{2} y + 3$

$\boxed{\sf \implies \: bxy = - \frac{3}{2}}$

5x + 7ʏ - 12 = 0 ᴛᴏ ʙᴇ ᴛʜᴇ ʟɪɴᴇ ʟɪɴᴇ ᴏғ ʀᴇɢʀᴇssɪᴏɴ ᴏғ x ᴏɴ ʏ ,

$\sf \: 5x + 7y - 12 = 0$

$\sf \implies \: y \: = - \frac{5}{7} x + \frac{12}{7}$

$\boxed{\sf \implies \: byx \: = - \frac{5}{7} }$

Now,

$\sf \: r \: = \sqrt{bxy \times byx} = \sqrt{ \frac{15}{14} }$

$\sf \: byx = \frac{rσ_y}{σ_y ^{2} } = \frac{ - 5}{7} \: \: ,bxy \: = \frac{rσ_x}{σ_y} = \frac{ - 3}{2}$

$\sf {\implies\frac{σ_x ^{2} }{σ_y ^{2} }} = \frac{ \frac{3}{2} }{ \frac{5}{7} }$

$\sf {\implies\frac{σ_x ^{2} }{σ_y ^{2} }} = \frac{21}{10}$

$\green{ \boxed{\red{ \sf {\implies\frac{σ_x }{σ_y }} = \sqrt{\frac{21}{10}}}} }$

Answer 2

$\huge\blue{\boxed{\red{\mathfrak{\overbrace{\underbrace{\fcolorbox{red}{aqua}{\underline{\red{✯answer✯}}}}}}}}}$

Thanks for the A2A!

⚡First, it asks to find the intersection of  ${\underline{ \mathtt{\red{2x-3y-5=0} \green{}\mathtt\blue{} \purple{} \mathtt \orange{}\pink{}}}}\:$  and  ${\underline{ \mathtt{\red{} \green{}\mathtt\blue{7x-5y-2=0 } \purple{} \mathtt \orange{}\pink{}}}}\:$

⚡To do this, we equate them.

⚡${\underline{ \mathtt{\red{2x-3y-5=7x-5y-2} \green{}\mathtt\blue{} \purple{} \mathtt \orange{}\pink{}}}}\:$  get them equal

⚡${\underline{ \mathtt{\red{} \green{}\mathtt\blue{14x-21y-35=14x-10y-4} \purple{} \mathtt \orange{}\pink{}}}}\:$  get x’s ready to cancel

⚡${\underline{ \mathtt{\red{-11y-31=0 } \green{}\mathtt\blue{} \purple{} \mathtt \orange{}\pink{}}}}\:$

⚡${\underline{ \mathtt{\red{} \green{}\mathtt\blue{y=-\frac{-31}{11} } \purple{} \mathtt \orange{}\pink{}}}}\:$

⚡Then we substitute that in to the original equations.

⚡${\underline{ \mathtt{\red{2x-3(-31/11)-5=0} \green{}\mathtt\blue{} \purple{} \mathtt \orange{}\pink{}}}}\:$

⚡We solve that to get${\underline{ \mathtt{\red{} \green{}\mathtt\blue{ x=\frac{-19}{11} } \purple{} \mathtt \orange{}\pink{}}}}\:$

$Correlation$  coefficient (r)=−0.866

✨Refer the attachment for further sum

Hope it helps✌