Summer 2016 - Q.12 - Grad Plus

# Summer 2016 – Q.12

12. a) Find the coefficient of correlation and two lines of regression using following data: [7M]

 x 1 2 3 4 5 y 2 5 3 8 7

Here n = 5

$latex \begin{array}{l}\overline x=\frac1n,\;\Sigma\;x=\frac15\times15=3\;and\\\\\overline y=\frac1n,\;\Sigma\;y=\frac15\times25=5\end{array}$

 x y x = x – 3 y = y – 5 x2 y2 xy 1 2 -2 -3 4 9 6 2 5 -1 0 1 0 0 3 3 0 -2 0 4 0 4 8 1 3 1 9 3 5 7 2 2 4 4 4 $latex \Sigma\;x=15$ $latex \Sigma\;y=25$ Total 10 26 13

Correlation coefficient

$latex \begin{array}{l}r=\frac{\Sigma\;xy}{\sqrt{\Sigma\;x^2}\sqrt{\Sigma\;y^2}}\\\\\;\;=\frac{13}{\sqrt{10\times26}}\\\\r=0.8062\end{array}$

The regression of coefficeint of y on x is

$latex \begin{array}{l}a_1=\frac{\Sigma\;xy}{\sqrt{\Sigma\;x^2}}\\\\\;\;\;\;=\frac{13}{10}\\\\\;\;\;\;=1.3\end{array}$

The regression of coefficient of x on y is

$latex \begin{array}{l}b_1=\frac{\Sigma\;xy}{\sqrt{\Sigma\;y^2}}\\\\\;\;\;\;=\frac{13}{26}\\\\\;\;\;\;=0.5\end{array}$

equation of regression of line of y on x is y = a1x

$latex \begin{array}{l}y-5=1.3\left(x-3\right)\\\\y=1.3x+1.1\end{array}$

equation of regression of line of x on y

$latex \begin{array}{l}x=b_1y\\\\\left(x-3\right)=0.5\left(y-5\right)\\\\x=0.5y+0.5\end{array}$

b) Solve the difference equation $latex y_{n+2}-3y_{n+1}+2y_n=2_n+1+2^n$. [6M]

The given equation in E-form is

$latex \left[E^2-3E+2\right]y_n=2_n+1+2^n$

The auxiliary equation is

$latex \begin{array}{l}E^2-3E+2=0\\\\\left(E-2\right)\left(E-1\right)=0\\\\E=2,\;1\end{array}$

since the roots of A.E are real and different

$latex \begin{array}{l}\therefore\;C.F=c_11^n+c_22^n\\\\\;\;\;\;\;\;\;\;\;\;\;\;=c_1+c_22^n\end{array}$

Also,

$latex P.I=\frac1{E^2-3E+2}\left(2n-1\right)+\frac1{\left(E^2-3E+2\right)}2^n—–\left(i\right)$

Now,

$latex \frac1{E^2-3E+2}\left(2n+1\right)=\frac1{\left(\triangle+1\right)^2-3\left(\triangle+1\right)+2}\left(2n+1\right)$

$latex \begin{array}{l}=\frac1{\triangle^2-\triangle}\left(2n+1\right)\\\\=\frac{-1}\triangle\left(1-\triangle\right)^{-1}\left(2n+1\right)\\\\=\frac{-1}\triangle\left(1+\triangle\right)\;\left(2n+1\right)\\\\=\frac{-1}\triangle\left[\left(2n+1\right)+\triangle\;\left(2n+1\right)\right]\\\\=\frac{-1}\triangle\left(2n+3\right)\\\\=\frac{-1}\triangle\left[2n^1+3n^0\right]\\\\=-\left[\frac{2n^2}2+\frac{3n^1}1\right]\\\\=-\left\{n\left(n-1\right)+3n\right\}\\\\=-\left(n^2+2n\right)\end{array}$

and

$latex \frac1{\left(E^2-3E+2\right)}2^n=\frac1{\left(E-2\right){\displaystyle\left(E-1\right)}}2^n$

$latex \begin{array}{l}=\frac1{\left(E-2\right)}\left[\frac1{\left(2-1\right)}2^n\right]\\\\=\frac{\displaystyle1}{\displaystyle\left(E-2\right)}2^n\\\\=n2^{n-1}\end{array}$

from equation (i) we get

$latex P.I=-n^2-2n+n\cdot2^{n-1}$

Hence the complete solution is

$latex \begin{array}{l}y_n=C.F+P.I\\\\y_n=c_1+c_22^n-n^2-2n+n\cdot2^{n-1}\end{array}$

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