Sample size n = 1000
Sample proportion of T.V. viewers
p=\(\frac { 320 }{ 1000 } \)=0.32
∴ q = 1 - P = 1 - 0.32 =0.68
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { (.32)(.68) }{ 1000 } } \)
S.E = 0.0147

Given mean 20 ⇒ np = 20
S.D = 4 ⇒ \(\sqrt { npq } \) = 4
∴ \(\frac { npq }{ np } =\frac { 16 }{ 20 } \Rightarrow q=\frac { 4 }{ 5 } \)
P = 1-q =\(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substitutingp and q in npq = 16, we get
\(n\times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } \) =16
n = \(\frac { 16\times 5\times 5 }{ 4 } \)  = 100
∴ Number of events = 100

P(X = 0) + P(X = 1) + P(X = 2)
= 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence the given distribution of probabilities is not a probability distribution.

Here total availability = 150 + 100 + 250 = 500
total requirement = 50 + 150 + 300 = 500
∴ Total availability = total requirement
∴ The given problem is a balanced transportation problem
Hence, there exists a feasible solution to the given problem
I - allocation:

[∵ least cost is 4 & min (50,150) = 50]
II - allocation:

[∵ least cost is 6 & min (150, 250) = 150]
III - allocation:

[∵ least cost is 8 & min (300, 100) = 100]
IV - allocation:

[∵ least cost is 9 & min (200,100) = 100]
V - allocation:

[∵ min (100, 100) = 100]
Thus, the allocations are

∴ The transportation schedule is
O₁ → D₁, O₁ → D₃, O₂ → D₃, O₃ → D₂, O₃ → D₃
Hence, the total transportation cost is
= 50(4) + 100(8) + 100(11) + 150(6) + 100(9)
= 200 + 800 + 1100 + 900 + 900
= Rs. 3900

∴  \(\int { \frac { 2{ cos }^{ 2 }x-cos2x }{ { sin }^{ 2 }x } } \)
= \(\int { \frac { 2{ cos }^{ 2 }x-\left( 2{ cos }^{ 2 }x-1 \right) }{ { sin }^{ 2 }x } } \)

= \(\int { \frac { 1 }{ { sin }^{ 2 }x } } dx\)
= ∫ cosec² x dx
= -cot x + c

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

Since only two values of yare given, the polynomial which fits the data is of degree 1.
Hence 2nd differences are zeros
∴ Δ²(y₀) = 0
⇒ (E-1)²y_o=0
⇒(E² - 2E + 1) y_o = 0
⇒y₂ - 2y₁ +y_o = 0
⇒34 - 2y₁ + 51 = 0
​​​​​​​⇒85 - 2y₁ =0
​​​​​​​⇒2y₁ + 85 ​​​​​​​⇒ y₁ = \(\frac{85}{2}\)
​​​​​​​⇒y₁ = 42.5 

The Y-axis is the ordinate at x = 0.
The area bounded by the ordinates at x = 0, x = 2 and the given curve is
\(A=\int _{ a }^{ b }{ y } dx\)
\(=\int _{ 0 }^{ 2 }{ ({ 4x }^{ 2 }-8x+6)dx } \)
\(={ \left[ \frac { { 4x }^{ 3 } }{ 3 } -\frac { 8{ x }^{ 2 } }{ 2 } +6x \right] }_{ 0 }^{ 2 }\)
\(=4\left( \frac { 8 }{ 3 } \right) -4(4)+6(2)\)
\(\frac { 32 }{ 3 } -6+12=\frac { 32 }{ 3 } -4\)
\(=\frac { 32-12 }{ 3 } =\frac { 20 }{ 3 } \)
∴ Area \(=\frac { 20 }{ 3 } \)sq.units

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions