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

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.

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

Answered by MJS last updated on 22/Aug/18

V=((Σx^2 )/n)−(((Σx)/n))^2 mean=x^− =((Σx)/n) 1. n=5 ΣX=25 ΣX^2 =165 V=((165)/5)−(((25)/5))^2 =33−25=8 X^− =((25)/5)=5 2. n=5 ΣY=55 ΣY^2 =765 V=((765)/5)−(((55)/5))^2 =153−121=32 Y^− =((55)/5)=11 ΣY=pΣX+q ⇒ 55=25p+q ⇒ q=55−25p Y^− =pX^− +q ⇒ 11=5p+q ⇒ 11=5p+55−25p ⇒ ⇒ 20p=44 ⇒ p=((11)/5) ⇒ q=55−25×((11)/5)=0 y=((11)/5)x

$${V}=\frac{\Sigma{x}^{\mathrm{2}} }{{n}}−\left(\frac{\Sigma{x}}{{n}}\right)^{\mathrm{2}} \\ $$$$\mathrm{mean}=\overset{−} {{x}}=\frac{\Sigma{x}}{{n}} \\ $$$$\mathrm{1}.\:{n}=\mathrm{5}\:\:\Sigma{X}=\mathrm{25}\:\:\Sigma{X}^{\mathrm{2}} =\mathrm{165} \\ $$$${V}=\frac{\mathrm{165}}{\mathrm{5}}−\left(\frac{\mathrm{25}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{33}−\mathrm{25}=\mathrm{8} \\ $$$$\overset{−} {{X}}=\frac{\mathrm{25}}{\mathrm{5}}=\mathrm{5} \\ $$$$\mathrm{2}.\:{n}=\mathrm{5}\:\:\Sigma{Y}=\mathrm{55}\:\:\Sigma{Y}^{\mathrm{2}} =\mathrm{765} \\ $$$${V}=\frac{\mathrm{765}}{\mathrm{5}}−\left(\frac{\mathrm{55}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{153}−\mathrm{121}=\mathrm{32} \\ $$$$\overset{−} {{Y}}=\frac{\mathrm{55}}{\mathrm{5}}=\mathrm{11} \\ $$$$ \\ $$$$\Sigma{Y}={p}\Sigma{X}+{q}\:\Rightarrow\:\mathrm{55}=\mathrm{25}{p}+{q}\:\Rightarrow\:{q}=\mathrm{55}−\mathrm{25}{p} \\ $$$$\overset{−} {{Y}}={p}\overset{−} {{X}}+{q}\:\Rightarrow\:\mathrm{11}=\mathrm{5}{p}+{q}\:\Rightarrow\:\mathrm{11}=\mathrm{5}{p}+\mathrm{55}−\mathrm{25}{p}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{20}{p}=\mathrm{44}\:\Rightarrow\:{p}=\frac{\mathrm{11}}{\mathrm{5}}\:\Rightarrow\:{q}=\mathrm{55}−\mathrm{25}×\frac{\mathrm{11}}{\mathrm{5}}=\mathrm{0} \\ $$$$ \\ $$$${y}=\frac{\mathrm{11}}{\mathrm{5}}{x} \\ $$

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