Cost ofliving index  = \(\frac{\sum PV}{\sum V}\) = \(\frac {18994.1}{175}\) = 108.54

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

Sample size n = 50
Sample mean \(\bar { x } \) = 75
Sample S.D. s = 10
Standard error (S.E) = \(\frac { s }{ \sqrt { n } } =\frac { 10 }{ \sqrt { 50 } } =\frac { 10 }{ 7.07 } \)
= 1.414
As the significance level is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for the population mean is \(\bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 75 - (1.96)(1.414) ≤ μ ≤ 75 + (1.96)(1.414)
⇒ 75-2.771 ≤ μ ≤ 75 + 2.771
⇒ 72.23 ≤ μ ≤ 77.77
Hence, the 95% confidence interval of the population mean is (72.23, 77.77)

p = 0.4, n = 3
q =1 - P = 1 - 0.4 = 0.6
P(X = x) = nC_x p^xq^n-x
P(X≥1) = P(X = 1) + P(X = 2) + P(X = 3)
=3C₁ (0.4)¹ (0.6)² + 3C₂ (0.4)² (0.6) + 3C₃ (0.4)³ (0.6)⁰
=3 x \(\left( \frac { 4 }{ 10 } \right) \left( \frac { 36 }{ 100 } \right) +3\left( \frac { 16 }{ 100 } \right) \left( \frac { 6 }{ 100 } \right) +\left( \frac { 64 }{ 1000 } \right) \)
= \(\frac { 1 }{ 1000 } \)(432+288+64) =\(\frac { 784 }{ 1000 } \)
P(X≥1) = 0.784

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

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

Given 
f' (x) = 3x² - \(\frac { 2 }{ { x }^{ 3 } } \)
We know f(x) = ∫ f'(x) dx
= \(\int { \left( { 3x }^{ 2 }-\frac { 2 }{ { x }^{ 3 } } \right) } \)
\(f\left( x \right) =3\left( \frac { { x }^{ 3 } }{ 3 } \right) -2\left( \frac { { -x }^{ -2 } }{ -2 } \right) +c\)
\(f\left( x \right) ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 2 } } +c....(1)\)
Also, f(1) = 0
\(0={ 1 }^{ 3 }+\frac { 1 }{ { 1 }^{ 2 } } +c\)
= 1 + 1 + c
c = -2
\(\therefore f\left( x \right) ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 2 } } -2\)

Given slope = y + 2x
⇒ \(\frac { dy }{ dx } \) = y+2x
⇒ \(\frac { dy }{ dx } \)- y = 2x
This is of the form \(\frac { dy }{ dx } \)+ Py = Q where P = -1, Q = 2x
\(\\ \int { P } dx=\int { -1 } dx\) = -x
∴ I.F. = \(e^{ \int { P } dx }\) = e^-x
∴ The solution is y.\(e^{ \int { P } dx }=\int { Q } .e^{ \int { P } dx }\)dx+C
⇒ y.e^-x = \(\int { 2x } \).e^-xdx+C
Let u = x; dv = e^-x
u² = 1; v = -e^-x
v₁ = e^-x
⇒ ye^-x = 2[-xe^-x-1(e^-x)]+C
[Bernoulli's formula]
⇒ ye^-x = -2x e^-x -2e^-x+ C...(1)
Since the Curve passes through (0, 0), we get
⇒ 0 = 0-2e⁰ + C ⇒ C = 2
(1) becomes,
∴ ye^-x = -2xe^-x - 2e^-x + 2
ye^-x = -2xe^-x - 2e^-x + 2e^x.e^-x
= e^-x(2 e^x-2x-2)
ye^-x = 2e^-x(e^x-x-1)

Given \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)= 1
⇒ \(\frac { { b }^{ 2 }x^{ 2 }+{ a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ y }^{ 2 } } \)= 1
⇒ b²x²+a²y² = a²b²
Differentiating again w.r.t 'x' we get,
2b²x + 2a²y\(\frac { dy }{ dx } \)=0 ⇒ b²x+a²yy₁ = 0 ....(2)
Differentiating w.r.t 'x' we get,
b² + a²[yy₂ + y₁y₁] = 0
⇒ b² + a²[yy₂ + y₁²] = 0 ...(3)
Eliminating a² and b² from (1) and (3) we get
\(\left| \begin{matrix} x & y{ y }_{ 1 } \\ 1 & { y }_{ 1 }^{ 2 }+y{ y }_{ 2 } \end{matrix} \right| \) = 0
⇒ x(y₁²+yy₂) - yy₁ = 0
⇒ x\(\left( \left( \frac { dy }{ dx } \right) ^{ 2 }+y.\frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) -y\left( \frac { dy }{ dx } \right) \)= 0 which is the required differential equation.

Let us calculate the number of men whose wages is less than Rs. 35 by using Newton's forward interpolation formula
x_o + nh = x ⇒ 30 + n (10) = 35
⇒ 10n = 35 - 30 = 5
⇒ n = \(\frac{5}{10}\) = 0.5
The difference table is

\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
\(y(35)=9+\frac { 0.5 }{ 1! } (30)+\frac { (0.5)(0.5-1) }{ 2 } (5)+\frac { (0.5)(0.5-1)(0.5-2) }{ 6 } (2)\)
= 9 + 15 - 0.6 + 0.1 = 24 (approximately)
∴ Number of men getting wages between Rs. 30 and Rs. 35 is y(35) - y(30) = 24 - 9 = 15.

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.

Given non-homogeneous equations are
x + 2y = 3,y - z = 2,x + Y + z = 1

Obviously,\(\rho (A)=2\) and \(\rho (A,B)\)
Hence \(\rho (A)=2\quad \rho\) (A, B) = 2
\(\therefore\) The system is consistent and has infinite number of solutions.