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).

Let X denotes the weight of the male students
Given μ = 151, σ = 15 and N = 500
(i) between 120 and 155 pounds
P(120 < X < 155)
When X = 120, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 120-151 }{ 15 } \)
= \(\frac { -31 }{ 15 } \) = -2.067
When x = 155, Z = \(\frac { 155-151 }{ 15 } \)
= \(\frac { 4 }{ 15 } \) = 0.2667

∴ P(120< X < 155) = P(-2.067 < X < 0.2667)
= P (-2.067 < Z < 0) + P (0 < Z < 0.2667)
= P (0 < Z < 2.067) + P (0 < Z < 0.2667)
(By symmetry)
= 0.4803 + 0.1026 = 0.5829
Probability for a student weigh between 120 and 155 pounds is 0.5829
∴ Out of 500 students, number of students weighing between 120 and 155 pounds
= 500 \(\times\) 0.5829 = 291 students
(ii) more than 185 pounds
P(X > 185)
When X = 185, Z = \(\frac { 185-151 }{ 15 } \)
=\(\frac { 34 }{ 15 } \) = 2.2667

∴ P(X >185) = P (Z > 2.2667)
= P (2.2667 < Z < ∞)
= P(0 < Z < ∞) - (0 < Z < 2.2667)
= 0.5 - 0.4881
= 0.0119
i.e Probability for a student weighing above 185 is 0.0119
∴ out of 500 male students, number of students weighing more than 185 pounds.
= 500 \(\times\) 0.0119 = 6 students.

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

Let I = [sin (log x) + cos (log x)] dx
Put log x = t ⇒ x = e^t
⇒ dx = e^t. dt
I = ഽ(sin t + cos t). e^t dt
= ഽ e^t (sin t + cos t) dt
Let f(t) = sin t ⇒ f' (t) = cos t
I = ഽe^t (f (t) + f' (t)) dx
= e^t . f(t) +c
[∵ ഽ e^t [ (f (t) +f' (t) ] dt = e^t . f (t) +c]
= e^{log x}. sin (log x) +c
= x sin (log x) +c   [∵ e^{log x} =x]

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

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 \(C'(x)=20+\frac { x }{ 20 } \)
\(\Rightarrow \int { C'(x)dx= } \int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Given when x = 0, C = 200
⇒ k₁ = 200
\(\therefore C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +200\quad ---(1)\)
Also, R'(x) = 30
\(R(x)=\int { 30dx } +{ k }_{ 2 }=30x+{ k }_{ 2 }\)
When x = 0 R = 0 ⇒ K₂ = 0
∴ R(x) = 30x   ----(2)
Profit = Total revenue - total cost
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -200\)
\(\therefore P=10x-\frac { { x }^{ 2 } }{ 40 } -200---(3)\)
\(\frac { dp }{ dx } =10-\frac { 2x }{ 40 } =10-\frac { x }{ 20 } \)
\(\frac { dp }{ dx } =0\)
\(\Rightarrow 10-\frac { x }{ 20 } =0\)
\(\Rightarrow 10=\frac { x }{ 20 } \Rightarrow x=200\)
\(\frac { d^{ 2 }p }{ { dx }^{ 2 } } =-\frac { 1 }{ 20 } <0\)
∴ Profit is maximum when x= 200
∴ maximum profit \(P=10(200)-\frac { { (200) }^{ 2 } }{ 40 } -200[From(3)]\)
P = 2000 -1000 - 200
2000 - 1200
P = Rs .800.
Hence, the maximum profit is Rs. 800.

fix) = ax² + bx + c
\(f(1)=0\Rightarrow a\left( 1 \right) ^{ 2 }+b(1)+c=0\Rightarrow a+b+c=0\)  ...(1)
\(f(2)=-2\Rightarrow a\left( { 2 }^{ 2 } \right) +b(2)+c=-2\Rightarrow 4a+2b+c=2\)..(2)
\(f(3)-6\Rightarrow a(3^{ 2 })+b(3)+c=-6\Rightarrow 9a+3b+c=-6\)
Now \(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{matrix} \right| \)
= 1(2 - 3) - 1(4 - 9) + 1(12 - 18)
= \(-1+5-6=-2\neq 0\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system has unique solution
\(\Delta a=\left| \begin{matrix} 0 & 1 & 1 \\ -2 & 2 & 1 \\ -6 & 3 & 1 \end{matrix} \right| \)
= 0-1(-2+6)+ 1(-6+ 12)
= -4 + 6 = 2
\(\Delta b=\left| \begin{matrix} 1 & 0 & 1 \\ 4 & -2 & 1 \\ 9 & -6 & 1 \end{matrix} \right| \)
= 1(-2+6)+0+1(-24+ 18)
 = 4 - 6 = -2
\(\Delta c=\left| \begin{matrix} 1 & 1 & 0 \\ 4 & 2 & -2 \\ 9 & 3 & -6 \end{matrix} \right| \)
= 1 (-12 + 6) - 1( - 24 + 18) + 0
= -6 + 6 = 0

f(n) = (-1)x² + 1(x) + 0 
f(x) = x²+ x.