Fisher's Ideal Index = \(\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}}\times100\)
\(= \sqrt\frac{1380\times1500}{510\times571}\times100\)
= 266.61
Time reversaltest:
\(P_{01}\times P_{10}=\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}\times{\Sigma p_0q_1}\times{\Sigma p_0q_0}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}\times{\Sigma p_1q_1}\times{\Sigma p_1q_0}}\)

= \(\sqrt{1}=1\)
Hence, time reversal test is satisfied.
Factor reversaltest:
\(P_{01}\times Q_{01}=\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}\times{\Sigma q_1p_0}\times{\Sigma q_1p_1}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}\times{\Sigma q_0p_0}\times{\Sigma q_0p_1}}\)

\(P_{01}\times Q_{01}=\frac{{\Sigma p_1q_1}}{{\Sigma p_0q_0}}\)
Hence, Fisher's ideal index satisfies factor reversal test also.

Given p = \(\frac { 55 }{ 100 } \)
∴ q = \(\frac { 45 }{ 100 } \) and n = 100
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { \frac { 55 }{ 100 } \times \frac { 45 }{ 100 } }{ 100 } } \)
= 0.0497
(a) As the level of significance α = 0.05 \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for proportion is given by \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (1.96) (0.0497) ≤ p ≤ 0.55 + (1.96) (0.0497)
⇒ 0.453 ≤ p ≤ 0.647
∴ 95% confidence interval for proportion is (0.45, 0.65)
(b) As the level of significance is α = 0.01, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for proportion is given by  \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (2.58) (0.0497) ≤ p ≤ 0.55 + (2.58) (0.0497)
⇒ 0.422 ≤ p ≤ 0.678
Hence, 99% confidence interval for proportion is (0.42, 0.68).

Given n = 4
p = probability of getting a head when a coin is tossed
= \(\frac { 1 }{ 2 } \)
∴ q=1-p =\(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
X is a ran dom verticical having
p(X = x) = nC_x p^xq^n-x, x = 0,1,2,3,4
a) atleast 2 heads
P(X ≥ 2) = 1-P(X<2)
= 1- [P(X = 0) + P(X = 1)]
=1-\(\left[ 4C_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }+4C_{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 3 } \right] \)
=1-\(\left[ 1\left( \frac { 1 }{ 16 } \right) +4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) \right] \)
=1-\(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X ≥ 2) = \(\frac { 11 }{ 16 } \)
b) atmost 2 heads
P(X≤2) = 1-P(X>2)
= 1 - [P(X = 3) + P (X = 4)]
= 1-\(\left[ { 4C }_{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }+{ 4C }_{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 0 } \right] \)
= 1-\(\left[ 4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) +1\left( \frac { 1 }{ 16 } \right) \right] \)
= \(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X≤2) =\(\frac { 11 }{ 16 } \).

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

Here column IV has no zero. Go to step 2.
Step 2:
Select the smallest 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 row with exactly one zero. Mark the zero by \(\Box \) and draw a vertical line. After examining all the rows examine the column with one zero. mark the zero by \(\Box\) and draw a horizontal line.

Here only 4 assignments have been made.
The numbers not lying on the line are

and min. of these numbers is 1.
Now subtract 1 from all these numbers and add 1 to the numbers on the intersecting line (ie. 6, 7, 2). Other numbers remains the same.
∴ The new cost matrix is

Now, repeat Step 3.
Thus, all the 5 assignments have been made.
The optimal assignment schedule and total cost is

\(E(X)=0\times \frac { 1 }{ 5 } +1(\frac { 2 }{ 5 } )+2\left( \frac { 1 }{ 5 } \right) +3\times \frac { 1 }{ 5 } \)
\(=\frac { 2 }{ 5 } +\frac { 2 }{ 5 } +\frac { 3 }{ 25 } =\frac { 7 }{ 5 } \)
\(\\ \therefore 2E(X)=\frac { 14 }{ 5 } ...(1)\)
\(E({ Y }^{ 2 })=0\times \frac { 1 }{ 5 } +{ 1 }^{ 2 }(\frac { 3 }{ 10 } )+{ 2 }^{ 2 }(\frac { 2 }{ 5 } )+{ 3 }^{ 2 }(\frac { 1 }{ 10 } )\)
\(=\frac { 3 }{ 10 } +\frac { 8 }{ 5 } +\frac { 9 }{ 10 } =\frac { 3+16+9 }{ 10 } \)
\(=\frac { 28 }{ 10 } =\frac { 14 }{ 5 } ..(2)\)
From (1) and (2), E(Y²) = 2 E(X).

Let I =  ഽ x³ sin (x⁴) dx
Put t = x⁴
⇒ dt = 4x³ dfx
\(\Rightarrow \frac { dt }{ 4 } ={ x }^{ 3 }dx\)
\(\therefore I=\int { sin } t.\frac { dt }{ 4 } =\frac { 1 }{ 4 } sint\quad dt\)
= \(-\frac { 1 }{ 4 } cos\quad t+c\)
= \(-\frac { 1 }{ 4 } cos\left( { x }^{ 4 } \right) +c\)      \(\left[ \because t={ x }^{ 4 } \right] \)

Given \(\frac { dc }{ dq } =\frac { { c }^{ 2 }+2cq }{ { q }^{ 2 } } \)
This is a homogeneous equation in e and q of order 2
∴ Put c = vq and \(\frac { dc }{ dq } =v+q\frac { dv }{ dq } \)
∴ v + q\(\frac { dv }{ dq } \) = v²a² + 2vq₀q =\(\frac { { q }^{ 2 }({ v }^{ 2 }+2v) }{ { q }^{ 2 } } \)v²+2vv
q\(\frac { dv }{ dq } \) = v²+2v-v = v²+v
Separating the variables we get,
\(\frac { dv }{ v+v } =\frac { dq }{ q } \)
Integrating \(\int { \frac { dv }{ v(v+1) } } =\int { \frac { dq }{ q } } \)
[ \(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
⇒ 1 = A(v+1)+B
put v = -1
1 = -B ⇒ B = -1
put v = 0
⇒ 1 = A ]
\(\int { \left( \frac { 1 }{ v } -\frac { 1 }{ v+1 } \right) dv } =\int { \frac { dq }{ q } } \)
⇒ log v -log (v + 1) = log q + log k
⇒ log\(\frac { v }{ v+1 } \) = logq.k
⇒ \(\frac { v }{ v+1 } \) = q.k
Replacing v by \(\frac { c }{ q } \), we get
\(\frac { c/q }{ c/q+1 } \) = q.k
⇒ \(\frac { c }{ c+q } \) = kq
⇒ c = kq(c+q)  .....(1)
Given when c = 1 and q = 1
⇒ 1 = k(1) (1+1) ⇒ 1 = 2 k ⇒ k = \(\frac { 1 }{ 2 } \)
∴ (1) ⇒ c = \(\frac{q}{2}\)(c+q)
∴ ⇒ 2c = q(c+q)

Let us calculate the number of students whose height is less than 90 ern using Newton's forward interpolation formula.
x_o+ nh = x ⇒ 60 + n(20) = 90 ⇒ 20n = 30
⇒ n = \(\frac32\) = 1.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is as follows:

∴y(90) = 250 + (1.5)(120)\(\frac { (1.5)(1.5-1) }{ 2! } (20)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (-10)+\frac { (1.5)(1.5-1)(1.5-2)(1.56-3) }{ 4! } (20)\)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59 = 424 (app)
ஃ Number of students whose height is between 80 cm and 90 cm is y(90) - y(80)
= 424 - 370
= 54.

Given C'(x) = 2 + 3e3x
\(\Rightarrow \int { C'(x) } =\int { (2+{ 3e }^{ 3x }) } dx\)
\(\Rightarrow C(x)=2x+\frac { { 3e }^{ 3x } }{ 3 } +k\)
\(\Rightarrow C(x)=2x+{ e }^{ 3x }+k\)
Since the fixed cost is Rs. 500, when x = 0, C = 500
\(\Rightarrow 500=0+{ e }^{ 0 }+k\quad [\because { e }^{ 0 }=1]\)
⇒ 500-1 = k ⇒ = 499
∴ C(x) = 2x + e^3x+499
Average cost dunction \(AC=\frac { C }{ x } \)
\(AC=\frac { 2x+{ e }^{ 3x }+499 }{ x } \)
\(=2+\frac { { e }^{ 3x } }{ x } +\frac { 499 }{ x } \)

The given non-homogeneous equations can be written as
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

Case (i):
When  \(k\neq 1\) and \(k\neq 2\)
\(\rho (A)=\rho (A,B)=3=\) Number of unknowns
\(\therefore \) The system has unique solution 
Case (ii):
When k = 1
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) \)
\(\rho (A)=\rho\) (A, B) = 1
\(\therefore \) The system is consistent and has infinitely many solutions.
Case (iii):
When k = - 2
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ -3 \end{matrix} \right) \)
\(\rho (A)=2\rho (A,B)\)= 3
\(\Rightarrow \rho (A)\neq 2\rho (A,B)\)
\(\therefore\) The system is inconsistent and has no solution.