Let the required equation of the straight line trend is
y = a + bX
Since Σx = 0. \(a = \frac{\Sigma y}{x} = \frac{16.9}{5}\)
\(b = \frac{\Sigma xy}{\Sigma x^2} = \frac{13.3}{10} = 1.33\)
Hence, the straight line trend is
y = 3.38 + 1.33 (x - 2002)
∴ The trend for 2007 is
y_t = 3.38 + 1.33 (2007 - 2002)
⇒  y_t = 3.38 + 1.33 (5)
⇒  y_t = 3.38 + 6.65
⇒  y_t = 10.03

Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)

Given n = 10, p = \(\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Let
X denote the number of defective bolts chosen
∴ X = 2
(i) Using binomial distribution
P(X = 2) = \(10{ C }_{ 2 }\left( \frac { 1 }{ 5 } \right) ^{ 2 }\left( \frac { 4 }{ 5 } \right) ^{ 8 }\)
= \(\frac { 10\times 9 }{ 2\times 1 } \left( \frac { { 4 }^{ 8 } }{ { 5 }^{ 10 } } \right) =45\left( \frac { 4^{ 8 } }{ { 5 }^{ 10 } } \right) \)
(ii) Using Poisson distribution
λ = np = 10 \(\times\) \(\frac { 1 }{ 5 } \) = 2
P(X = x) \(\times\)\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
x = 0,1,2,......n
∴ P(X = 2) = \(\frac { e^{ -2 }(2^{ 2 }) }{ 2 } =e^{ -2 }\left( \frac { 4 }{ 2 } \right) \)
= 2e^-2
= 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Here Σa_i = 22 + 15 + 8 = 45
Σbj = 7 + 12 + 17 + 9 = 45
Σa_{i =} Σbj
∴ The given problem is balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I-allocation:

[∵ the max penalty is 4. In II, least cost is 2 & min (12,22) = 12]

II-allocation:

[∵ the max penalty is 3. In B, least cost is 1 & min (17,15) = 15]
III-allocation:


[∵ the max penalty is 3. In III, least cost is 4 & min (2, 10) = 2]
IV-allocation:

[∵ the max penalty is 2. In A, least cost is 3 & min (9, 8) = 8]
V-allocation:

[∵ the max penalty is 1. In C, least cost is 4 & min (7, 8) = 7]
VI-allocation:

[∵ min (1,1) = 1]
Thus, the allocations are

∴ The transportation schedule is
A → II, A → III, A → IV, B → III, C → I and C → IV
Hence, the total transportation cost is
= 12(2) + 2(4) + 8(3) + 15(1) + 7(4) + 1(5)
= 24 + 8 + 24 + 15 + 28 + 5 = Rs.104

i) E(X) = Σx_ip_i
= 0(0.1)+1(0.25)+2(0.3)+3(0.2)+4.(0.15)
= 2.05
E(X²) = ∑x_i²p_i
= 0(0.1)+1(0.25)+4(0.3)+9(0.2)+16(0.15)
= 5.65
Now, V(X) = E(X²)-[E(X)]²
= 5.65 - (2.05)² = 1.4475
ii) \(V\left( \frac { X }{ 2 } \right) =\frac { 1 }{ 4 } v(X)\quad [\because V(aX)={ a }^{ 2 }V(X)]\)
\(=\frac { 1 }{ 4 } (1.4475)\)
\(V\left( \frac { X }{ 2 } \right) =0.361875\)

Let I = \(\int { \frac { 1 }{ { 3x }^{ 2 }+13x-10 } } dx\)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x-\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x+\frac { 169 }{ 36 } -\frac { 169 }{ 36 } -\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-\frac { 289 }{ 36 } } } \)

= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-{ \left( \frac { 17 }{ 16 } \right) }^{ 2 } } } \)
= \(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| } +c \right] \)
= \(\frac { 1 }{ 3 } \times \frac { 1 }{ 2\times \frac { 17 }{ 6 } } log\left| \frac { x+\frac { 13 }{ 6 } -\frac { 17 }{ 6 } }{ x+\frac { 13 }{ 6 } +\frac { 17 }{ 6 } } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { x-\frac { 2 }{ 5 } }{ x+5 } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { 3x-2 }{ 3(x+5) } \right| +c\)

(y-x)\(\frac { dy }{ dx } \) = a²
⇒ \(\frac { dy }{ dx } =\frac { { a }^{ 2 } }{ y-x } \Rightarrow \frac { dx }{ dy } =\frac { y-x }{ { a }^{ 2 } } \)
⇒ \(\frac { dx }{ dy } =\frac { y }{ { a }^{ 2 } } -\frac { x }{ a^{ 2 } } \)
⇒ \(\frac { dx }{ dy } +\frac { x }{ { a }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } \)
This is of the form \(\frac { dx }{ dy } \)+Px = Q
where P = \(\frac { 1 }{ { a }^{ 2 } } \) and Q = \(\frac { 1 }{ { a }^{ 2 } } \)y
∴ \(\int { P } dy=\int { \frac { 1 }{ { a }^{ 2 } } dy } =\frac { 1 }{ { a }^{ 2 } } \)
I.F = \(e^{ \int { pdy } }=e^{ y/{ a }^{ 2 } }\)
∴ The solution is \(\int { x. } e^{ \int { pdy } }=\int { Q.e^{ \int { pdy } } } dy\)       
⇒ x.e^y/a2 =\(\int { \frac { 1 }{ { a }^{ 2 } } y } \) e^y/a2dy+C....(1)
Put \(\frac { 1 }{ { a }^{ 2 } } \)y = t ⇒ dy = a²dt
[∵ u = t; d = e^t]
u¹= 1; v = e^t
v₁ = e^t
\(\int { u } dv\) = uv-u¹v₁]
∴ (1) ⇒ x.e^y/a2 = a²\(\int { te^{ t } } \)dt
= a²[te^t-e^t]+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }\) = a².e^t(t-1)+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }=a^{ 2 }.e^{ \frac { y }{ { a }^{ 2 } } }\left( \frac { y }{ { a }^{ 3 } } -1 \right) \)   [∵ t = \(\frac { y }{ { a }^{ 2 } } \)]

Since e^1.75 lies at the beginning of the table, we can use Newton's forward interpolation formula
∴ x_o + nh = x ⇒ 1.7 + n(0.1) = 1.75
⇒ n(0.1) = 1.75 - 1.7 = 0.05
⇒ n = \(\frac{0.05}{0.1}\) = 0.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is

∴ \(y\left( { e }^{ 1.75 } \right) =5.74+\frac { 0.5 }{ 1! } (0.576)+\frac { (0.5)(0.5-1) }{ 2! } (0.06)+\frac { (0.5)(0.5-1)(0.5-2) }{ 3! } (0.007)\)
y(e^1.75) = 5.474 + 0.288 - 0.0075 + 0.0004375
= 5.7549375

Given that η_d=2
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =2\)
\(\Rightarrow \frac { dx }{ x } =-2\frac { dp }{ p } \)
Integrating both sides,
\(\int { \frac { dx }{ x } =-2\int { \frac { dp }{ p } +logk } } \)
⇒ log x+2log p=log k
log x.p²=log k[∵ a log b=log b^a]
⇒ xp²=k
When x=4, p=
4(1)²=k
⇒ k=4
∴ xp²=4
⇒ xp²=\(\frac{4}{x}\)
⇒ \(p-\frac { \sqrt { 4 } }{ x } =\frac { 2 }{ \sqrt { x } } \)
Revenue \(R=px=\frac { 2 }{ \sqrt { x } } .x\)

=2√x
\(R=2\sqrt { x } and\quad p=\frac { 2 }{ \sqrt { x } } \)

The given non-homogeneous equations can be written as
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

Case (i):
When  \(k\neq 1\) and \(k\neq 2\)
\(\rho (A)=\rho (A,B)=3=\) Number of unknowns
\(\therefore \) The system has unique solution 
Case (ii):
When k = 1
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) \)
\(\rho (A)=\rho\) (A, B) = 1
\(\therefore \) The system is consistent and has infinitely many solutions.
Case (iii):
When k = - 2
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ -3 \end{matrix} \right) \)
\(\rho (A)=2\rho (A,B)\)= 3
\(\Rightarrow \rho (A)\neq 2\rho (A,B)\)
\(\therefore\) The system is inconsistent and has no solution.