Fisher's ideal index = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}} \times 100\)
=   \(\sqrt{\frac{505}{425} \times \frac {430}{470}} \times 100\) 
\(P^{F}_{01}\)   = 115.75   

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.

Let X be a random vertical. which is normally distributed
Given that μ = 4.35 and σ = 0.59
When X = 5.20, Z + \(\frac { X-\mu }{ \sigma } =\frac { 5.2-4.35 }{ 0.59 } \)
= \(\frac { 0.85 }{ 0.59 } \) = 1.44

∴ P(X > 5.20) = P(Z > 1.44)
= P(1.44 = P(0 = 0.5 - 0.4251
P(X > 5.20) = 0.749

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)

Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 :
Select a minimum element in each row and subtract this from all the elements in its row.

Here IV column has no zero. Go to step 2.
Step 2:
Select the minimum element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3:
Examine the rows with only one zero. Mark that. zero by ׈ and draw a vertical line.

Thus, all the assignments have been made.
The optimal assignment schedule and total cost is
 

\(I=\int { { e }^{ x } } \left( \frac { 1+sinxcosx }{ { cos }^{ 2 }x } \right) dx\)
= \(\int { { e }^{ x } } \left( \frac { 1 }{ { cos }^{ 2 }x } +\frac { sinxcosx }{ cosxcosx } \right) dx\)
I = ഽ e^x (sec² x + tan x) dx      ----(1)
Let f(x) = tan x
f^'(x) = sec² x dx
We know ഽe^x (f(x) + f'(x)) dx = e^x.f(x) +c
∴ I = ഽ e^x (f (x) + f' (x)) dx
= e^x. f(x) +c
= e^x tan x + c

The auxiliary equation is m² + 1 = 0
⇒  m² = -1
⇒ m = ±\(\sqrt { -1 } \) = ±i
Here α = 0, β = 1
∴ CF is e^0x [A cosx + B sinx]
∴ The general solution is
y = A cos x + B sin x ...(1)
Given when x = 0, y = 2
∴ 2 = A cos 0 + B sin 0
⇒ 2 = A+0 ⇒ A = 2
[∵ cos0 = 1 and sin0 = 0]
Also, when x = \(\frac { \pi }{ 2 } \), y = -2
∴ -2 = A\(cos\frac { \pi }{ 2 } +Bsin\frac { \pi }{ 2 } \)
⇒ -2 = A(0)(+B(1) ⇒ B = -2
[∵ \(cos\frac { \pi }{ 2 } \) = 0 and \(sin\frac { \pi }{ 2 } \)= 1]
Substituting the values of A & B in (1) we get,
y = 2 cos x - 2 sin x
⇒ y = 2 (cosx - sin x)

By data, we have
x_o = 40, x₁ = 50, x₂ = 60, x₃ = 70
y_o = 31, y₁ = 73, y₂ = 124, y₃ = 159.
Using Lagrange's formula, we get
\(y={ y }_{ 0 }\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } +{ y }_{ 1 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } +{ y }_{ 3 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \)
∴ y(42) = 31\(\frac { (-8)(-18)(-28) }{ (-10)(-20)(-30) } +73\frac { (2)(-18)(-28) }{ (10)(-10)(-20) } +124\frac { (2)(-8)(-28) }{ (20)(10)(-10) } +59\frac { (2)(-8)(-28) }{ (30)(20)(10) } \)
= 20. 832 + 36. 792 - 27. 776 + 7.632
y = 37. 48

Given, marginal cost, C' (x) \(\frac{x}{3000}+2.50\)
\(\int { C'(x) } =\int { \left( \frac { x }{ 300 } +2.50 \right) dx } \)
\(C(x)=\frac { { x }^{ 2 } }{ 6000 } +2.50x+k\)
When x = 0, c = 0 ⇒ k = 0
∴ c(x) = \(\frac{x^2}{6000}+2.50x\)
When x = 3000
Cost of production
\(=\frac { { (3000) }^{ 2 } }{ 6000 } +2.50(3000)\)

= 1500 + 7500
= Rs. 9000

Given non-homogeneous equations are
2x + 3y = 5  6X + 5y = 11
\(\Delta =\left| \begin{matrix} 2 & 3 \\ 6 & 5 \end{matrix} \right| =10-18=-8\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 5 & 3 \\ 11 & 5 \end{matrix} \right| =25-33=-8\)
\(\Delta y=\left| \begin{matrix} 2 & 5 \\ 6 & 11 \end{matrix} \right| =22-30=-8\)
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(\therefore \) Solution set is {1, 1}