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 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 \(\times\) \(\frac { 2 }{ 3 } =\frac { 80 }{ 3 } \)

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

We have P(X=r)∝α^r
⇒P(X = r) = λα^r,r = 0,1,2,....
Since sum of all the probabilities in a probability distribution is 1.
P(X = 0)+P(X = 1)+P(P(X = 2)+... = 1
⇒ λα0+λα1+λα2+...= 1
⇒λ(1+α+α2+.....) = 1
⇒\(\lambda(\frac{1}{1-\alpha})\) = 1
⇒ λ = 1 - α
∴ P(X = r) = (1-α)α^r, r = 0,1,2,...
Hence P(X=0) = (1-α)α⁰ = (1-α)(1) = 1-α.

(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 = \(\int _{ 1 }^{ 2 }{ \frac { log\quad x }{ { x }^{ 2 } } } dx\)

u = log x and dv = \(\frac { 1 }{ { x }^{ 2 } } dx\quad ={ x }^{ -2 }dx\)
\(du=\frac { 1 }{ x } ;v=\frac { { x }^{ -2+1 } }{ { -2+1 } } =\frac { { x }^{ -1 } }{ -1 } =\frac { -1 }{ x } \)
Using integration by parts we get,
ഽu dv = uv - ഽv du
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { log\quad x }{ { x }^{ 2 } } } dx\)
= \({ \left[ -\frac { 1 }{ x } logx-\int { -\frac { 1 }{ x } .\frac { 1 }{ x } } dx \right] }_{ 1 }^{ 2 }\)
= \({ \left[ -\frac { 1 }{ x } logx+\int { \frac { 1 }{ { x }^{ 2 } } } dx \right] }_{ 1 }^{ 2 }\)
= \({ \left[ -\frac { 1 }{ x } logx-\frac { 1 }{ x } \right] }_{ 1 }^{ 2 }\)
= \(-{ \left[ \frac { 1 }{ x } logx+\frac { 1 }{ x } \right] }_{ 1 }^{ 2 }\)
= \(-\left[ \left( \frac { 1 }{ 2 } log2+\frac { 1 }{ 2 } \right) -\left( 1log1+\frac { 1 }{ { 1 }^{ 1 } } \right) \right] \)
= \(-\left[ \frac { 1 }{ 2 } log2+\frac { 1 }{ 2 } -0-1 \right] \)
\(\left[ \because log1=0 \right] \)
= \(-\left[ \frac { 1 }{ 2 } log2-\frac { 1 }{ 2 } \right] \)
= \(\frac { 1 }{ 2 } -\frac { 1 }{ 2 } log2\)
= \(\frac { 1 }{ 2 } (1-log2)\)

\(\int { \frac { { 8 }^{ 1+x }+{ 4 }^{ 1-x } }{ { 2 }^{ x } } } dx\)
= \(\int { \frac { { \left( { 2 }^{ 3 } \right) }^{ 1+x }+{ \left( { 2 }^{ 2 } \right) }^{ 1-x } }{ { 2 }^{ x } } } dx\)
= \(\int { \frac { { 2 }^{ 3+3x }+{ 2 }^{ 2-2x } }{ { 2 }^{ x } } } dx\)
= \(\int { \frac { { 2 }^{ 3+3x } }{ { 2 }^{ x } } } dx+\int { \frac { { 2 }^{ 2-2x } }{ { 2 }^{ x } } } +dx\)
= ഽ2^3+3x-xdx + ഽ 2^2-2x-x dx
= ഽ 2^2x+3 dx+ ഽ 2^2-3x dx
= \(\frac { { 2 }^{ 2x+3 } }{ 2log2 } +\frac { { 2 }^{ 2-3x } }{ -3log2 } +c\)
= \(\frac { { 2 }^{ 2x+3-1 } }{ log2 } -\frac { { 2 }^{ 2-3x } }{ 3log2 } +c\)
= \(\frac { { 2 }^{ 2x+2 } }{ log2 } -\frac { { 2 }^{ 2-3x } }{ 3log2 } +c\)

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.

From the graph, it is found that the population for 1986 was 24 thousands.

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.

The non-homogeneous equation are
2x - y + z = 7, 3x + y - 5z = 13, x + y + z = 5

Clearly \(\rho (A)=3\) and \(\rho (A,B)\) = 3 = Number of unknowns 
\(\therefore\) The given system is consistent and has unique solution.

(i) Max (3000, 1000, 6000) = 6000
∴ Debentures is the best using maximin principle
(ii) Max (10000, 8000, 6000) = 6000
∴ Debentures is the best using minimax principle.

First allocation:

Second allocation:

Third allocation:

Fourth allocation:

Fifth allocation:

Final allocation:

The transportation cost is
\( = (30 \times 6)+(5 \times 5)+(28 \times 11)+ (7 \times 9)+(25 \times 7)+(25 \times 13) \)
= 180 + 25 + 308 + 63 + 175 + 325  
= Rs. 1076

Tendency movements are due to nature, which repeat themselves periodically in every seasons. These variations repeat themselves in less than one year time. It is measured in an interval of time.
Seasonal variations may be influenced by natural force, social customs and traditions.

Given average = λ = \(\frac { 3 }{ 20 } \) = 0.15
P(will not be more than one failure)
= P(X ≤ 1)
= P(X = 0) + P(X = 1)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \)
= e^-λ (1+λ) =e^-0.15 (1+0.15)
= (0.8607) (1.15)
= (0.98981 [e^-0.15 =0.8607]
∴ Probability that there will not be more than one failure = 0.98981

Probability of getting a degree p = 0.4
∴ q = 1– p
= 1 - 0.4
= 0.6
(i) P (one will be a graduate) = P(X = 1) = 5C₁ (0.4)(0.6)⁴
= 0.2592
(ii) P ( atleast one will be a graduate) = 1–P (none will be a graduate)
= 1-5C₀(P⁰)(Q)5-0
= 1-5C₀(0.4)⁰(0.6)5
= 1-0.0777
= 0.9222

If the random variable X is the top face of a tossed, fair, six sided die, then the probability mass function of X is
P_x (x) = \(\frac{1}{6}\), for x = 1, 2, 3, 4, 5 and 6
The average toss, that is, the expected value of X is
\(E(X)=\sum _{ x }^{ }{ { P }_{ x }(x) } \)
\(E(X)=\left( 1\times \frac { 1 }{ 6 } \right) +\left( 2\times \frac { 1 }{ 6 } \right) +\left( 3\times \frac { 1 }{ 6 } \right) +\left( 4\times \frac { 1 }{ 6 } \right) +\left( 5\times \frac { 1 }{ 6 } \right) +\left( 6\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } (1+2+3+4+5+6)\)
\(=\frac { 7 }{ 2 } \)
= 3.5
Therefore, the expected toss of a fair six sided die is 3.5.

For discrete random variable:
p(x_i)≥0∀i and \(\sum _{ i=1 }^{ n }{ p({ x }_{ i }) } =1\) and X takes only
infinite number of values
For Continuous random variable:
f(X)≥0∀x and \(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \) and X takes
infinite number of values in the interval.

(i) Probability mass function:
If X is a discrete random variable with distinet values x₁, x₂, x_n then the function denoted by p_x (x) and defined by
P_X(x) = p(x)
= {p(x = xi) = p_i = p(xi)if x = x_i, i = 1, 2,...n
0 if x ≠ x_i
Here p(x_i)≥0∀i & \(\sum _{ i=1 }^{ n }{ p({ x }_{ i })=1 } \)
(ii) Probability density function:
The probability that a random variable X takes a value in th interval [t₁, t₂] [open or closed] is given by the integral of a function called the probability density function
\(P({ t }_{ 1 }\le X\le { t }_{ 2 })=\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ { f }_{ x }(x)dx } \)
Here f(x) ≥0∀ x and \(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
(iii) Probability distribution function:
Fora discrete distribution, the distribution function of a random variable X is defined as 
FX(x) = P(X ≤ x)for all x ∈ R.

Given marginal cost MC = C'(x) = \(\frac { { x }^{ 2 } }{ 200 } +4\)
\(\Rightarrow \int { MC=\int { C'(x)=\int { \left( \frac { { x }^{ 2 } }{ 200 } +4 \right) dx } } } \)
To find the cost of producing 200 air conditioners
\(C=\int _{ 0 }^{ 200 }{ \left( \frac { { x }^{ 2 } }{ 200 } +4 \right) dx } \)
\(={ \left[ \frac { 1 }{ 200 } \times \frac { { x }^{ 3 } }{ 3 } +4x \right] }_{ 0 }^{ 200 }\)
\(={ \left[ \frac { { x }^{ 3 } }{ 600 } +4x \right] }_{ 0 }^{ 200 }\)
\(=\left[ \frac { { (200) }^{ 3 } }{ 600 } +4(200) \right] -0\)
\(=\frac { 8000000 }{ 600 } +800\)
= 13333.33 + 800
= Rs. 14,133.33
Hence, the cost of producing 200 air conditioners is Rs. 14133.33

Given \(MC=300{ x }^{ \frac { 2 }{ 5 } }\)
\(\Rightarrow \frac { dC }{ dx } =300{ x }^{ \frac { 2 }{ 5 } }\)
\(\int { dC } =300\int { { x }^{ \frac { 2 }{ 5 } }dx } \)
\(\Rightarrow C=300\frac { { x }^{ \frac { 2 }{ 5 } +1 } }{ \frac { 2 }{ 5 } +1 } +k\)
\(\Rightarrow C=300\frac { { x }^{ \frac { 7 }{ 5 } } }{ \frac { 7 }{ 5 } } +k\)
\(\Rightarrow C=300\times \frac { 5 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+k\) ...(1)
Given fixed cost is zero ⇒ k = 0
∴(1) becomes,
\(C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+0\Rightarrow C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }\)
Average cost function \((AC)=\frac { C }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } \frac { { x }^{ \frac { 7 }{ 5 } } }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } -1 }\)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 2 }{ 5 } }\)

Given probability mass function is

∴  The cumulative distribution function F_x(x) is
F_x(0) = 0 if x < 3
F_x(3) = P(X = 3) = 0.3, for 3 ≤x<5
F_x(5) = P(X = 3)+P(X = 5) = 0.3+0.2 = 0.5, for 5≤X<8
F_x(8) = P(X = 3)+P(X+5)+P(X = 8)
= 0.3+0.2+0.3
= 0.8,8≤x<10
F_x(10) = P(X = 3)+P(X = 5)+P(X = 8)+P(X = 10)
= 0.3+0.2+0.3+0.2
= 1 for x ≥10
\(F_{x}(x)= \begin{cases}0, & \text { if } x<3 \\ 0.3, & \text { if } 3 \leq x < 1 \\ 0.5, & \text { if } 5 \leq x < 8 \\ 0.8, & \text { if } 8 \leq x < 10 \\ 1, & \text { if } x ≥ 10 \\ \end{cases}\)

The matrix equation corresponding to the system is
\(\left( \begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 10 \end{matrix} \right) \)
A X = B

\(\rho (A)=\rho ([A,B])=1<\)number of unknowns
\(\therefore \)The given system is consistent and has infinitely many solutions.
Now, the given system is transformed into the matrix equation.
\(\left( \begin{matrix} 2 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 0 \end{matrix} \right) \)
\(\Rightarrow 2x+y=5\)
Let us take y = k,k \(\in \)R
\(\Rightarrow 2x+k=5\)
\(x=\frac { 1 }{ 2 } (5-k)\) 
\(x=\frac { 1 }{ 2 } (5-k),y=k\ for\ all\ k\in R\)
Thus by giving different values for k, we get different solution.
Hence the system has infinite number of solutions.

The highest derivative is third order and its power is one
∴ order : 3,
degree : 1