Winter 2016 - Q.12 - Grad Plus

# Winter 2016 – Q.12

12. a) In a partially distributed laboratory analysis of a correlation data, the following results only are eligible: $latex \sigma_x^2=9$ Regression equations: 8x – 10y + 66 = 0,
40x – 18y = 214 what were. [7M]
i) The mean values of x and y.
ii) Coefficient of correlation between x and y.
iii) Standard Deviation of y.

i) Since both the line of reqression pass through the point $latex \left(\overline x,\;\overline y\right)$ Therefore we have

$latex \begin{array}{l}8\overline x-10\overline y+66=0—–\left(i\right)\\\\40\overline x-18\overline y-214=0—–\left(ii\right)\end{array}$

Multiplying equation (i) by 5

$latex \therefore40\overline x-50\overline y+330=0—–\left(iii\right)$

substracting equation (iii) from (ii) we get

$latex \begin{array}{l}32\overline y-544=0\\\\\therefore\overline y=\frac{544}{32}\\\\\;\;\;\;\;\;=17\end{array}$

∴ from (i)
$latex 8\overline x-170+66=0$

∴$latex \begin{array}{l}8\overline x=104\\\\\overline x=13\\\\\overline y=17\end{array}$

ii) variance of

x = σx2 = 9

$latex \therefore\sigma\;x=3$

from (i) and (2) The equations of line of regression can be written as

$latex \begin{array}{l}y=0.8x+6.6\\\\\;\;\;=a_0+a_1x\end{array}$ and

$latex \begin{array}{l}x=0.45y+5.35\\\\\;\;\;=b_0+b_1y\end{array}$

∴ The regression coefficient of y on x is

$latex \frac{r\sigma\;y}{\sigma\;x}=0.8—–(iv)$

The regression co-efficient of x on y is

$latex \frac{r\sigma\;x}{\sigma\;y}=0.45—–(v)$

multiplying (iv) and (v) we get

$latex r^2=0.8\times0.45=0.36$

∴ coefficient of correlation = r = 0.6

iii) from (iv)

$latex \begin{array}{l}\sigma\;y=\frac{0.8\sigma\;x}r\\\\\;\;\;\;\;\;=\frac{0.8\times3}{0.6}\\\\\;\;\;\;\;\;=4\end{array}$

$latex \therefore\sigma\;y=4$

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

we have $latex y_{n+2}-2y_n=2^n$

$latex \begin{array}{l}\therefore\left(E^2-2E+4\right)y_n=2^n\\\\\therefore A\cdot E\;is\;E^2-2E+4=0\\\\\therefore E=\frac{2\pm\sqrt{4-16}}2\\\\\;\;\;\;\;\;\;=\frac{2\pm\sqrt{-12}}2\\\\\;\;\;\;\;\;\;=\frac{2\pm2\sqrt3}2\end{array}$

$latex \begin{array}{l}\therefore\left(E^2-2E+4\right)y_n=2^n\\\\\therefore A\cdot E\;is\;E^2-2E+4=0\\\\\therefore E=\frac{2\pm\sqrt{4-16}}2\\\\\;\;\;\;\;\;\;=\frac{2\pm\sqrt{-12}}2\\\\\;\;\;\;\;\;\;=\frac{2\pm2\sqrt3}2\\\\E\;\;=1\pm i\sqrt3\\\\\therefore r=\sqrt{1+3}\\\\\;\;\;\;\;\;=2\end{array}$

and

$latex \begin{array}{l}\theta=\tan^{-1}\left(\frac{\sqrt3}1\right)\\\\\;\;\;=\frac\pi3\end{array}$

$latex \begin{array}{l}\therefore c.f=2^n\left(c_1\;\cos\;\frac{n\pi}3+c_2\sin\frac{n\pi}3\right)\\\\P.I=\frac1{E^2-2E+4}2^n\\\\\;\;\;\;\;\;=\frac1{4-2\cdot2+4}\\\\\;\;\;\;\;\;=\frac142^n\\\\P.I=\frac{2^n}4\end{array}$

Hence complete solution is

y= c·f + P.I

$latex \begin{array}{l}y_n=2^n\left[c_1\;\cos\;\frac{n\pi}3+c_2\sin\frac{n\pi}3\right]+\frac{2^n}4\\\end{array}$

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