Sample mean \(\bar { x } =\frac { \Sigma x }{ n } \)
= \(\frac { 6.35+6.37+6.36+6.32+6.37 }{ 5 } \)
= \(\frac { 31.75 }{ 5 } \) = 6.35 mm

Sample variance = \(\frac { 1 }{ n-1 } \Sigma (x-\bar { x } )^{ 2 }=\frac { 0.0023 }{ 4 } \)
= 0.00055
n = 0.0055 mm²

Given P(X = 2) = 9 P(X = 4) + 90 P(X = 6)
Since X follows Poisson distribution with
P(X,λ) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } +\frac { { 90e }^{ -\lambda }.{ \lambda }^{ 6 } }{ 6! } \)

⇒ \(\frac { 1 }{ 2 } =\frac { 3{ \lambda }^{ 2 }+\lambda ^{ 4 } }{ 8 } \)
⇒ λ⁴+3λ² = 4
⇒ λ⁴+ 3λ²-4 = 0
Put λ² = t
⇒ t² + 3t-4 = 0  ⇒ (t-1)(t + 4) = 0
⇒ t = 1 or t = -4
∴ t = 1 [t = -4 is not possible]
∴ λ² = 1 ⇒ λ = 1
∴ Mean = 1
For Poisson distribution, mean = variance = 1

\(E(X)=\sum { xp(x)=-1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+2\left( \frac { 1 }{ 3 } \right) } \)
\(=\frac { -1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 2 }{ 3 } =\frac { -2+1+4 }{ 6 } \)
\(=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(E({ X }^{ 3 })={ \sum { x } }^{ 2 }p(x)\)
\(=1(\frac { 1 }{ 3 } )+0(\frac { 1 }{ 6 } )+1(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 3 } )\)
\(=\frac { 1 }{ 3 } +\frac { 1 }{ 6 } +\frac { 4 }{ 3 } =\frac { 2+1+8 }{ 6 } =\frac { 11 }{ 6 } \)
\(\therefore E{ (2X+3) }^{ 2 }=E(4{ X }^{ 2 }+12X+9)\)
\(=4\left( \frac { 11 }{ 6 } \right) +12\left( \frac { 1 }{ 2 } \right) +9\)
\(=\frac { 22 }{ 3 } +6+9=\frac { 22 }{ 3 } +15\)
\(=\frac { 22+45 }{ 3 } =\frac { 67 }{ 3 } \)
\(\therefore E(2X+3{ ) }^{ 2 }=\frac { 67 }{ 3 } \)

(i) Max (50, 10, 5) = 50
∴ A is the best alternative under maximin principle
(ii) Min (100, 80,40) = 40
∴ C is the best alternative under minimax principle

\(\int { \frac { cos2x-cos2\alpha }{ cosx-cos\alpha } } dx\)
= \(\frac { ({ 2cos }^{ 2 }x-1)-({ 2cos }^{ 2 }\alpha -1) }{ cos\quad x-cos\alpha } dx\)
= \(\frac { { 2cos }^{ 2 }x-cos^{ 2 }\alpha }{ cos\quad x-cos\alpha } \)

= \(2\int { (cos\quad x+cos\alpha )dx } \)
= \(2[sinx+cos\alpha .x]+c\)
= \(2sinx+2xcos\alpha +c\)

Given y = (A + Bx)e^3x ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= (A+Bx)e^3x(3)+e^3x(B)
⇒ \(\frac { dy }{ dx } \) = 3y + Be^3x  [Using (1)]
⇒ Be^3x = \(\frac { dy }{ dx } \)-3y
Differentiating again w.r.t 'x' we get,
\(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) \)+ Be3x(3)
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left[ \frac { dy }{ dx } -3y \right] \) [Using (2)]
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left( \frac { dy }{ dx } \right) \)-9y
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =6\left( \frac { dy }{ dx } \right) \)-9y which is the required differential equation.

From the graph, it is clear that when x = 27, the value of y is 24.8

Area \(=\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 3 }^{ 9 }{ \frac { 1 }{ x } dx } \)
\(={ [log\quad x] }_{ 3 }^{ 9 }\)
= log9-log3
\(=log\left( \frac { 9 }{ 3 } \right) \)
A = log 3 sq.units.

The order of A is 3 x 4
\(\therefore \rho (A)\le min\left( 3,4 \right) \)
\(\rho (A)\le 3\)
Consider the third order minor
\(\left| \begin{matrix} 1 & 2 & -4 \\ 2 & -1 & 3 \\ 8 & 1 & 9 \end{matrix} \right| =1\left| \begin{matrix} -1 & 3 \\ 1 & 9 \end{matrix} \right| -2\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right| -4\left| \begin{matrix} 2 & -1 \\ 8 & 1 \end{matrix} \right| \)
= 1(- 9 - 3) - 2(18 - 24) - 4(2 + 8)
= 1 (-12) - 2 (- 6) - 4 (10)
= - 12 + 12 - 40
= - 40::\(\neq \) 0.
There is a minor of order 3, which is not zero
\(\therefore \rho (A)=3\)