Grand average = \(\frac {75.2+70.4+83.2+76.8}{4}\)
= \(\frac {305.6}{4}\) = 76.4
Seasonal index S.I = \(\frac {Quarterly average}{Grand average} \times 100\)
Hence, S.I for I quarter = \(\frac {75.2}{76.4} \times 100\) = 98.4
S.I for II quarter =  \(\frac {70.4}{76.4} \times 100\) = 92.14
S.I for III quarter =  \(\frac {83.2}{76.4} \times 100\) = 108.9
S.I for IV quarter =  \(\frac {76.8}{76.4} \times 100\) = 100.5

Given sample size n = 50
Sample mean \(\bar { x }\)  = 825
Population mean μ = 900
Population S.D. σ = 110
Null hypotheses: H₀: μ = 900
Alternative hypotheses: H₁: μ ≠ 900
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
=\(\frac { 825-900 }{ \frac { 110 }{ \sqrt { 50 } } } \) = -4.82
∴ |z| = -4.82
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here |z| > \(Z_{ \frac { \alpha }{ 2 } }\) as 4.82 > 1.96
Inference: As |z| > \(Z_{ \frac { \alpha }{ 2 } }\), H₀ is rejected. Hence, we can conclude that mean life time of the population of electric bulbs cannot be taken as 900 hours.

Given n = 20
P(getting 1 or 5) = \(\frac { 2 }{ 6 } \Rightarrow p=\frac { 1 }{ 3 } \)
∴ q = 1 - p = \(1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
Using Binomial distribution,
Mean = np = 120 \(\times\) \(\frac { 1 }{ 3 } \) = 40
Variance = npq = 40 \(\times\) \(\frac { 2 }{ 3 } =\frac { 80 }{ 3 } \)

When 2 coins are tossed, sample space S={HH, HT, TH, TT} ⇒ n(s) = 4
∴P(X = 5) = p(getting 2 heads) =\(\frac{1}{4}\)
P(X = 2) = p(getting 1 head) = \(\frac{2}{4}=\frac{1}{2}\)
P(X = 1) = p(getting no head) = \(\frac{1}{4}\)
Hence the probability distribution function is 

\(\therefore E(x)=\sum { { x }_{ i }{ p }_{ i }=1(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 2 } )+5(\frac { 1 }{ 4 } )\)
\(=\frac { 1 }{ 4 } +1+\frac { 5 }{ 4 } =\frac { 1+4+5 }{ 4 } \)
\(=\frac { 10 }{ 4 } =2.50\)
Hence the expected money to win is Rs. 2.50

Here total supply = 300 + 400 + 500 = 1200
Total demand = 250 + 350 + 400 + 200 =1200
∴ Total supply = total demand
The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem
I-allocation:

[∵ Min (250, 300) = 250]
II-allocation:

[∵ Min (50,350) = 50]
III-allocation:

[∵ Min (300,400) = 300]
IV-allocation:

[∵ Min (400,100) = 100]
V-allocation:

[∵ Min (300, 500) = 300]
VI-allocation:

[∵ Min (200, 200) = 200]
Thus, the allocations are

∴ The transportation schedule is
A → P, A → Q, B → Q, B → R, C → R, C → S
Hence, the total transportation cost is
= 250 (3) + 50 (1) + 300 (6) + 100 (5) + 300 (3) + 200 (2)
= 750 + 50 + 1800 + 500 + 900 + 400
= Rs. 4400

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

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

Given (x²-yx²)dy + (y²+xy²)dx = 0
⇒ x²(1-y)dy+y²(1+x)dx = 0
⇒ x²(1-y)dy = -y²(1+x)dx
Separating the variables we get,
\(\frac { (1-y) }{ y^{ 2 } } dy=-\frac { (1+x) }{ x^{ 2 } } \)dx
⇒ \(\frac { 1 }{ { y }^{ 2 } } dy-\frac { 1 }{ y } dy=-\frac { 1 }{ x^{ 2 } } dx-\frac { 1 }{ x } dx\)
Integrating, \(\int { { y }^{ -2 } } dy-\int { \frac { 1 }{ y } } dy=-\int { \frac { 1 }{ x^{ 2 } } } dx-\int { \frac { 1 }{ x } } \)
\(-\frac { 1 }{ y } -logy=\frac { 1 }{ x } \) -log x + C
⇒ log x - log y = \(\frac { 1 }{ x } +\frac { 1 }{ y } \)+C
⇒ log \(log\left( \frac { x }{ y } \right) =\frac { x+y }{ xy } \)+C
⇒ \(\frac { x }{ y } =e^{ \frac { x+y }{ xy } +C }\)
⇒ \(\frac { x }{ y } =K.e^{ \frac { x+y }{ xy } }\)  [where e^C = K]

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.

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.

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.