The set X and Y have five elementseach . Given that ΣX=25,ΣY=55,ΣX^2 =165 and ΣY^2 =765 and a linear Function y= px + q tranforms the set X into the set Y,where p and q are positive constants. a) Find the mean and Variance of X and Y hence,or otherwise , b)find the values of p and q.


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Question Number 42299 by Rio Michael last updated on 22/Aug/18

$T h e s e t X a n d Y h a v e f i v e e l e m e n t s e a c h .$ $G i v e n t h a t Σ X = 25, Σ Y = 55, Σ X^{2} = 165 a n d Σ Y^{2} = 765$ $a n d a l i n e a r F u n c t i o n y = p x + q t r a n f o r m s t h e s e t X i n t o t h e s e t$ $Y, w h e r e p a n d q a r e p o s i t i v e c o n s t a n t s .$ $a) F i n d t h e m e a n a n d V a r i a n c e o f X a n d Y$ $h e n c e, o r o t h e r w i s e,$ $b) f i n d t h e v a l u e s o f p a n d q .$
	Answered by MJS last updated on 22/Aug/18

	$V = \frac{Σ x^{2}}{n} - {(\frac{Σ x}{n})}^{2}$ $mean = \bar{x} = \frac{Σ x}{n}$ $1 . n = 5 Σ X = 25 Σ X^{2} = 165$ $V = \frac{165}{5} - {(\frac{25}{5})}^{2} = 33 - 25 = 8$ $\bar{X} = \frac{25}{5} = 5$ $2 . n = 5 Σ Y = 55 Σ Y^{2} = 765$ $V = \frac{765}{5} - {(\frac{55}{5})}^{2} = 153 - 121 = 32$ $\bar{Y} = \frac{55}{5} = 11$ $Σ Y = p Σ X + q \Rightarrow 55 = 25 p + q \Rightarrow q = 55 - 25 p$ $\bar{Y} = p \bar{X} + q \Rightarrow 11 = 5 p + q \Rightarrow 11 = 5 p + 55 - 25 p \Rightarrow$ $\Rightarrow 20 p = 44 \Rightarrow p = \frac{11}{5} \Rightarrow q = 55 - 25 \times \frac{11}{5} = 0$ $y = \frac{11}{5} x$
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