Sample size n = 50
Sample mean \(\bar { x } \) = 75
Sample S.D. s = 10
Standard error (S.E) = \(\frac { s }{ \sqrt { n } } =\frac { 10 }{ \sqrt { 50 } } =\frac { 10 }{ 7.07 } \)
= 1.414
As the significance level is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for the population mean is \(\bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 75 - (1.96)(1.414) ≤ μ ≤ 75 + (1.96)(1.414)
⇒ 75-2.771 ≤ μ ≤ 75 + 2.771
⇒ 72.23 ≤ μ ≤ 77.77
Hence, the 95% confidence interval of the population mean is (72.23, 77.77)

Consider -2x² + 4x - 2 = -2(x²-2x+1)
= -2(x-1)²
∴ \({ e }^{ -2x^{ 2 }+4x-2 }=e^{ -2(x-1)^{ 2 } }\)
\({ e }^{ -\frac { 1 }{ 2 } \frac { (x-1)^{ 2 } }{ \frac { 1 }{ 4 } } }=e^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ 2 } } \right) ^{ 2 } }\)
i.e \(K{ e }^{ -2x^{ 2 }+4x-2 }=\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) }\)
⇒ \(Ke^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ 2 } } \right) ^{ 2 } }=\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) }\)
⇒ σ = \(\frac{1}{2}\), μ = 1 and K = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } \)
⇒ K = \(\frac { 1 }{ \frac { 1 }{ 2 } .\sqrt { 2\pi } } \Rightarrow K=\sqrt { \frac { 2 }{ \pi } } \).

When 2 coins are tossed, sample space S={HH, HT, TH, TT} ⇒ n(s) = 4
∴P(X = 5) = p(getting 2 heads) =\(\frac{1}{4}\)
P(X = 2) = p(getting 1 head) = \(\frac{2}{4}=\frac{1}{2}\)
P(X = 1) = p(getting no head) = \(\frac{1}{4}\)
Hence the probability distribution function is 

\(\therefore E(x)=\sum { { x }_{ i }{ p }_{ i }=1(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 2 } )+5(\frac { 1 }{ 4 } )\)
\(=\frac { 1 }{ 4 } +1+\frac { 5 }{ 4 } =\frac { 1+4+5 }{ 4 } \)
\(=\frac { 10 }{ 4 } =2.50\)
Hence the expected money to win is Rs. 2.50

Σa_i = 12 + 14 + 4 = 30
Σb_j = 9 + 10 + 11 = 30
Σa_i =  Σb_j
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I - allocation :

[∵ the highest penalty is 7, In C, least cost is 0 & min (11, 14) = 11] 
II - allocation :

[∵ the highest penalty is 4, In S1'least cost is 1& min (10, 12) = 10]
III - allocation :

[∵ In A, least cost is 2 & min (9, 3) = 3]
IV - allocation :

[∵ In A, least cost is 3 & min (6,4) = 4]
V - allocation :

[∵ min (2, 2) = 2]
Thus, the allocations are

∴ The transportation schedule is
S₁ → A, S₁ → B, S₂ → A, S₂ → C  S₃ → A
Hence, the total transportation cost is
= 2(5) + 10(1) + 3(2) + 11(0) + 4(3)
= 10 + 10 + 6 + 0 + 12 = Rs. 38

Here total supply = 300 + 400 + 500 = 1200
Total demand = 250 + 350 + 400 + 200 =1200
∴ Total supply = total demand
The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem
I-allocation:

[∵ Min (250, 300) = 250]
II-allocation:

[∵ Min (50,350) = 50]
III-allocation:

[∵ Min (300,400) = 300]
IV-allocation:

[∵ Min (400,100) = 100]
V-allocation:

[∵ Min (300, 500) = 300]
VI-allocation:

[∵ Min (200, 200) = 200]
Thus, the allocations are

∴ The transportation schedule is
A → P, A → Q, B → Q, B → R, C → R, C → S
Hence, the total transportation cost is
= 250 (3) + 50 (1) + 300 (6) + 100 (5) + 300 (3) + 200 (2)
= 750 + 50 + 1800 + 500 + 900 + 400
= Rs. 4400

Given 
f' (x) = 3x² - \(\frac { 2 }{ { x }^{ 3 } } \)
We know f(x) = ∫ f'(x) dx
= \(\int { \left( { 3x }^{ 2 }-\frac { 2 }{ { x }^{ 3 } } \right) } \)
\(f\left( x \right) =3\left( \frac { { x }^{ 3 } }{ 3 } \right) -2\left( \frac { { -x }^{ -2 } }{ -2 } \right) +c\)
\(f\left( x \right) ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 2 } } +c....(1)\)
Also, f(1) = 0
\(0={ 1 }^{ 3 }+\frac { 1 }{ { 1 }^{ 2 } } +c\)
= 1 + 1 + c
c = -2
\(\therefore f\left( x \right) ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 2 } } -2\)

Given (x²-yx²)dy + (y²+xy²)dx = 0
⇒ x²(1-y)dy+y²(1+x)dx = 0
⇒ x²(1-y)dy = -y²(1+x)dx
Separating the variables we get,
\(\frac { (1-y) }{ y^{ 2 } } dy=-\frac { (1+x) }{ x^{ 2 } } \)dx
⇒ \(\frac { 1 }{ { y }^{ 2 } } dy-\frac { 1 }{ y } dy=-\frac { 1 }{ x^{ 2 } } dx-\frac { 1 }{ x } dx\)
Integrating, \(\int { { y }^{ -2 } } dy-\int { \frac { 1 }{ y } } dy=-\int { \frac { 1 }{ x^{ 2 } } } dx-\int { \frac { 1 }{ x } } \)
\(-\frac { 1 }{ y } -logy=\frac { 1 }{ x } \) -log x + C
⇒ log x - log y = \(\frac { 1 }{ x } +\frac { 1 }{ y } \)+C
⇒ log \(log\left( \frac { x }{ y } \right) =\frac { x+y }{ xy } \)+C
⇒ \(\frac { x }{ y } =e^{ \frac { x+y }{ xy } +C }\)
⇒ \(\frac { x }{ y } =K.e^{ \frac { x+y }{ xy } }\)  [where e^C = K]

Since x = 0.2 lies at the beginning of the table, use Newton's foward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Here h = 1, x_o = 0, x = 0.2
⇒ x₀ + nh = 0.2 ⇒ 0 + n(1) = 0.2 ⇒ n = 0.2
The forward difference table is

 ∴ y = 176 +\(\frac { 0.2 }{ 1! } (9)+\frac { (0.2)(0.2-1) }{ 2! } (0)+\frac { (0.2)(0.2-1)(0.2-2) }{ 3! } (-1)+\frac { (0.2)(0.2-1)(0.2-2)(0.3-3) }{ 4! } (4)\)

= 176 + 1.8 - 0.048 - 0.1344 = 177.6176
ஃ Hence when x = 0.2, y = 177.6176.

The limits for one arch of the curve y = sin ax When y = 0 ⇒Sin ax = 0
⇒ sin ax = sin 0, sin \(\pi\)
⇒ ax = 0 or ax = \(\pi\)
⇒ x = 0, x = \(\frac{\pi}{a}\)
∴ The limits are from x = 0 to x = \(\frac{\pi}{a}\)
∴ Area =\(\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 0 }^{ a }{ sin\quad ax\quad dx } \)
\(={ \left[ -\frac { cos\quad ax }{ a } \right] }_{ 0 }^{ \frac { \pi }{ a } }\)
\(=-\frac { 1 }{ a } \left[ cos\quad a\times \frac { \pi }{ a } -cos(a)(0) \right] \)
\(=-\frac { 1 }{ a } \left[ cos\quad \pi -cos0 \right] \)
\(=-\frac { 1 }{ a } (-1-1)[\because cos0=1\ cos\pi =-1]\)
\(A=\frac { 2 }{ a } \) sq.units.

GivenAX B
\(\left( \begin{matrix} 2 & 4 \\ 4 & 3 \end{matrix} \right) ,\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left( \begin{matrix} 8 \\ 11 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} 2n+4 \\ 4n+3 \end{matrix} \right) =\left( \begin{matrix} 8 \\ 11 \end{matrix} \right) \)
Equating the corresponding entries on both sides, we get
2n +4 = 8
2n = 8-4
2n=4
\(n=\cfrac { 4 }{ 2 } \)
2 = 2

Given non-homogeneous equations are
x + 2y = 3,y - z = 2,x + Y + z = 1

Obviously,\(\rho (A)=2\) and \(\rho (A,B)\)
Hence \(\rho (A)=2\quad \rho\) (A, B) = 2
\(\therefore\) The system is consistent and has infinite number of solutions.

Let A =  \(\left( \begin{matrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ -2 & 4 & -4 \end{matrix} \right) \)
The order of A is 3 x 3
\(\therefore \rho (A)\le 3\) [Since minimum of (3,3) is 3]
Let us transform the matrix to an echelon form

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (A)=2\)

Cost of living index number = \({ \frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}}} \times100\)
C.L.I = \(\frac {174}{146.5} \times 100\)  = 118.77

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.

Let I = \(\int _{ 1 }^{ e }{ \frac { dx }{ x{ \left( 1+\log { x } \right) }^{ 3 } } } \)
Let t = 1+log x
\(\Rightarrow dt=\frac { 1 }{ x } dx\)
When x = 1, t = 1 + log 1 = 1 + 0 = 1
When x = e, t =1 + log e = 1 + 1 = 2
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { dt }{ { t }^{ 3 } } } =\int _{ 1 }^{ 2 }{ { t }^{ -3 } } dt\)
\(={ { \left[ \frac { { t }^{ -3+1 } }{ -3+1 } \right] }_{ 1 } }^{ 2 }\)
\(={ { \left[ \frac { { t }^{ -2 } }{ -2 } \right] }_{ 1 } }^{ 2 }={ { \left[ -\frac { 1 }{ { 2t }^{ 2 } } \right] }_{ 1 } }^{ 2 }\)
\(=-\frac { 1 }{ 2 } \left[ \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ { 1 }^{ 2 } } \right] \)
\(=-\frac { 1 }{ 2 } \left[ \frac { 1 }{ 4 } -1 \right] =-\frac { 1 }{ 2 } \left( \frac { -3 }{ 4 } \right) =\frac { 3 }{ 8 } \)

Given A =\(\left( \begin{matrix} -2 & 1 & 3 \\ 0 & 1 & 1 \\ 1 & 3 & 4 \end{matrix}\begin{matrix} 4 \\ 2 \\ 7 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ -2 & 1 & 3 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 1 }\rightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 7 & 11 \end{matrix}\begin{matrix} 7 \\ 2 \\ 18 \end{matrix} \right) { R }_{ 2 }\rightarrow { { R_{ 2 }+2{ R }_{ 1 } } }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 3 }\rightarrow { R_{ 3 }-{ 7R }_{ 2 } }\)
The last equivalent matrix is in echelon form and there are 3 non - zero rows.
\(\therefore \rho (A)=3\)

Let the amount invested in the two accounts be Rs. x and Rs. y respectively
By the given data, x + y = 8600   ..(1)
\(4\cfrac { 3 }{ 4 } \times \cfrac { x }{ 100 } +6\cfrac { 1 }{ 2 } \times \cfrac { y }{ 100 } =431.25\) \(\left[ \therefore interest=\cfrac { PNR }{ 100 } \right] \)
\(\Rightarrow \cfrac { 19x }{ 400 } +\cfrac { 13y }{ 3200 } =431.25\)
\(\Rightarrow \cfrac { 19x+26y }{ 400 } =431.25\)
19x + 26y = 172500   ...(2)
\(\Delta =\left| \begin{matrix} 1 & 1 \\ 19 & 26 \end{matrix} \right| =1(26)-1(19)\)
= 26-19 =7
\({ \Delta }x=\left| \begin{matrix} 8600 & 1 \\ 172500 & 26 \end{matrix} \right| =8600(26)-1(172500)\)
= 223600 - 172500 = 51100
\(\Delta y=\left| \begin{matrix} 1 & 8600 \\ 19 & 172500 \end{matrix} \right| =1(172500)-19(8600)\)
= 172500 - 163400 = 9100
\(x=\cfrac { \Delta x }{ \Delta } -\cfrac { 51100 }{ 7 } =7300\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 9100 }{ 7 } =1300\)
\(\therefore\) Investment in the interest of \(4\frac { 3 }{ 4 } \) % account is Rs. 7300 and investment in the rate of \(6\frac { 1 }{ 2 } \) account is Rs.1300. 

The matrix equation corresponding to the given system is
\(\begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 8 \end{matrix} \right) \)
A X = B

Number of non-zero rows is 2.
\(\rho (A)=\rho ([A,B])=2=\) Number of unknowns.
The given system is consistent and has unique solution.
Now, the given system is transformed into
\(\left( \begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ -2 \end{matrix} \right) \)
x + y = 5
y = 2
\(\therefore (1)\Rightarrow x+2=5\)
x = 3
Solution is x = 3, y = 2

Given P_d = 18− 2x − x² ; P_s = 2x − 3
We know that at equilibrium prices p_d = p_s
18− 2x − x² = 2x – 3
x² + 4x −21 = 0
(x − 3) (x + 7) = 0
x = –7 or 3
The value of x cannot be negative, x = 3
When x₀ = 3
ஃ p₀ = 18 − 2(3)  − (3)² = 3
CS = \(\int _{ 0 }^{ { x }_{ o } }{ f(x) } \) dx - x₀p₀
= \(\int _{ 0 }^{ 3 }{ (18-2x-{ x }^{ 2 }) } \)dx - 3 x 3
= \({ \left[ 18x-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }\)- 9
= 18(3) - (3)² - \(\left( \frac { { 3 }^{ 3 } }{ 3 } \right) \) - 9
CS = 27 units
PS = x₀P₀ - \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \)
= (3 \(\times\) 3) - \(\int _{ 0 }^{ 3 }{ (2x-3) } \)
= 9 - \(({ { { x }^{ 2 }-3x) } }_{ 0 }^{ 3 }\)
= 9 units
Hence at equilibrium price,
(i) the consumer’s surplus is 27 units
(ii) the producer’s surplus is 9 units.

Given probability of success p = 0.09
∴ q = 1 - p = 1 - 0.09 = 0.91
n = 1
Also P(atleast one success) = \(\frac { 1 }{ 3 } \) or more
∴ P(X≥1) = \(\frac { 1 }{ 3 } \)
⇒ 1-P(X < 1) = \(\frac { 1 }{ 3 } \)
⇒ P(X<1) =\(1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
⇒ P(X = 0) =\(\frac { 2 }{ 3 } \)
⇒ ⁿC_x p^xq^n-x = \(\frac { 2 }{ 3 } \)
Putting x = 0,
ⁿC₀ (0.09)⁰ (0.91)^n-0 = \(\frac { 2 }{ 3 } \)
⇒ (0.91)ⁿ = \(\frac { 2 }{ 3 } \) =0.6666
when (0.91) is Jultiplied 5 times we are getting 0.6240
∴ n = 5 or more
Here number of trails are 5 or more.