No of years = 7 (odd), by omitting the middle year we have
 

Given sample size n = 64
Sample mean \(\bar { x } \) = 51250
Null hypotheses: H₀: Population mean μ = 50000 Alternative hypotheses: H₁ : μ≠ 50,000
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 51250-50000 }{ \frac { 2000 }{ \sqrt { 64 } } } \)=5
∴ z = 5
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z > z_α as 5 < 1.96
Inference: As z > z_α, H₀ is rejected. Hence, we can conclude that the sample mean differs significantly from the population mean.

Let X denote the life of army shoes.
Given μ = 8, σ = 2 and N = 5000
When μ = 12, Z = \(\frac { X-\mu }{ \sigma } =\frac { 12-8 }{ 2 } \) = 2
∴ P(X≤12) = P(Z≤2)
= P(-∞ = P(-∞ =0.5+0.4772
P(X≤12) = 0.9772

∴ The probability for a shoe need to be replaces is 0.9772.
∴ Out of 5000 pairs of shoes, number of pairs need to be replaced = 5000 \(\times\) 0.9772
 = 4886.

Given n = 16, S.D = 2 ⇒ \(\sqrt { npq } \) =2
⇒ npq = 4
∴ 16(pq) = 4 ⇒ pq =\(\frac { 4 }{ 16 } =\frac { 1 }{ 4 } \)
⇒ q = \(\frac { 1 }{ 4p } \)...(1)
Since p + q = 1 ⇒ p+\(\frac { 1 }{ 4p } \)=1
⇒ \(\frac { 4{ p }^{ 2 }+1 }{ 4p } \) = 1
⇒ 4p²+1 = 4p ⇒ 4p² -4p+1 = 0
⇒ (2p-1)² = 0 ⇒ 2p-1 = 0
⇒ 2p = 1 ⇒ p = \(\frac { 1 }{ 2 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Mean = np = 16 x \(\frac { 1 }{ 2 } \) = 8

Given f(x) = 4k(x-1)³,1 ≤ x ≤ 3
Since f(x) is a p.d.f., \(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ 4k({ x-1) }^{ 3 }dx=1\Rightarrow { \left[ \frac { 4k(x-1{ ) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 3 }=1 } \)
\(\Rightarrow k({ 2 }^{ 4 }-{ 0 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
\(p(-2\le x\le 2)=\int _{ -2 }^{ 2 }{ f(x)dx } \)
\(=\int _{ 1 }^{ 2 }{ 4k({ x-1) }^{ 3 }dx } \)
\(=\frac { 4 }{ 16 } \int _{ 1 }^{ 2 }{ { (x-1) }^{ 3 }dx } \)
\([\therefore k=\frac { 1 }{ 16 } ]\)
\(=\frac { 1 }{ 4 } { \left[ \frac { { (x-1) }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 16 } ({ 1 }^{ 4 }-0)=\frac { 1 }{ 16 } \)

Σa_i = 12 + 14 + 4 = 30
Σb_j = 9 + 10 + 11 = 30
Σa_i =  Σb_j
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I - allocation :

[∵ the highest penalty is 7, In C, least cost is 0 & min (11, 14) = 11] 
II - allocation :

[∵ the highest penalty is 4, In S1'least cost is 1& min (10, 12) = 10]
III - allocation :

[∵ In A, least cost is 2 & min (9, 3) = 3]
IV - allocation :

[∵ In A, least cost is 3 & min (6,4) = 4]
V - allocation :

[∵ min (2, 2) = 2]
Thus, the allocations are

∴ The transportation schedule is
S₁ → A, S₁ → B, S₂ → A, S₂ → C  S₃ → A
Hence, the total transportation cost is
= 2(5) + 10(1) + 3(2) + 11(0) + 4(3)
= 10 + 10 + 6 + 0 + 12 = Rs. 38

(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

Let I =  ഽ sin³ x cos x dx
Put t = sin x 
⇒ dt = cos x dx
∵ I = t³.dt = \(\frac { { t }^{ 4 } }{ 4 } +c\)
= \(\frac { { sin }^{ 4 }x }{ 4 } +c\)

Given (x+y)²\(\frac { dy }{ dx } \) = 1
put x+y = z
⇒ 1+\(\frac { dy }{ dx } =\frac { dz }{ dx } \)
\(\frac { dy }{ dx } =\frac { dz }{ dx } \)-1
∴ (1) becomes,
z²\(\left( \frac { dz }{ dx } -1 \right) \)= 1
⇒ z²\(\frac { dz }{ dx } \)-z² = 1
⇒ z²\(\frac { dz }{ dx } \) = 1 + z²
Separating the variables we get,
\(\left( \frac { { z }^{ 2 } }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Adding and Subtracting 1 in the numerator, we get
\(\left( \frac { 1+{ z }^{ 2 }-1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( \frac { 1+{ z }^{ 2 } }{ 1+{ z }^{ 2 } } -\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( z-\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Integrating, \(\int { dz } -\int { \frac { dz }{ 1+{ z }^{ 2 } } } \)
[∵ \(\int { \frac { dz }{ 1+{ z }^{ 2 } } } \) = tan^-1x+y]
⇒ (z-tan^-1(2) = x+C
⇒ (x+y)-tan^-1(x+y) = x + C
⇒ y-tan^-1(x+y) = C

Since 1995 lies at the table of the table, use Newton's backward interpolation formula.
Also, xn + nh ⇒ x  2001 + n (10) = 1995
⇒ 10n = 1995 - 2001 ⇒ n = \(\frac{-6}{10}\) ⇒ n = -0.6
\(\Rightarrow { y }_{ n }+\frac { n }{ n! } \nabla { y }_{ n }+\frac { n(n+1) }{ 2! } { \nabla }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n-2) }{ 3! } { \nabla }^{ 3 }{ (y }_{ n })\)
The difference table is

∴ \(y=101+\frac { (0.6) }{ 1! } (8)+\frac { (-0.6)(-0.6+1) }{ 2! } (-4)+\frac { (-0.6)(-0.6+1)(-0.6+2) }{ 3! } (-1)+\frac { (-0.6)(-0.6+1)(-0.6+2)(-0.6+3) }{ 4! } (-3)\)
= 101 - (0.6)8 + \(\frac { (-0.6)(0.4) }{ 2 } (-4)+\frac { (-0.6)(0.4)(1.4) }{ 6 } (-1)+\frac { (0.6)(0.4)(1.4)(2.4)(-3) }{ 24 } \)
= 96.8368
Hence, population for the year 1995 is 96.837 thousands.

The limits for one arch of the curve y = sin ax When y = 0 ⇒Sin ax = 0
⇒ sin ax = sin 0, sin \(\pi\)
⇒ ax = 0 or ax = \(\pi\)
⇒ x = 0, x = \(\frac{\pi}{a}\)
∴ The limits are from x = 0 to x = \(\frac{\pi}{a}\)
∴ Area =\(\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 0 }^{ a }{ sin\quad ax\quad dx } \)
\(={ \left[ -\frac { cos\quad ax }{ a } \right] }_{ 0 }^{ \frac { \pi }{ a } }\)
\(=-\frac { 1 }{ a } \left[ cos\quad a\times \frac { \pi }{ a } -cos(a)(0) \right] \)
\(=-\frac { 1 }{ a } \left[ cos\quad \pi -cos0 \right] \)
\(=-\frac { 1 }{ a } (-1-1)[\because cos0=1\ cos\pi =-1]\)
\(A=\frac { 2 }{ a } \) sq.units.

\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
\(\Rightarrow \left( \begin{matrix} x0+0 \\ 0+0+z \\ 0+y+0 \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow x=2\quad z=-1\quad y=3\)
\(\therefore\) Solution set is {2,3, -1}