Cost ofliving index  = \(\frac{\sum PV}{\sum V}\) = \(\frac {18994.1}{175}\) = 108.54

Let y= ax + b be the best line of fit.
The normal equations are
a\(\sum\)x + nb = \(\sum\)y
a\(\sum\)x² + b\(\sum\)x = \(\sum\)xy
10a+ 5b = 15
⇒ 2a+b = 3 ...(1)
30a + 10b = 43
⇒  3a+b =  4.3 ....(2)
(1)-(2) ⇒ - a = -1.3
⇒  a = 1.3
Substituting a = 1.3 in (1) we get,
2(1.3) + b = 3
⇒ 2.6 + b = 3
⇒ b = 3 - 2.6 = 0.4
∴ The line of best fit isy = 1.3x + 0.4

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.

Sample size n = 500
Proportion of bad apples in the sample
= \(\frac { 45 }{ 500 } \) = 0.09
p = 0.09
∴ q = 1 - p = 1-0.09 = 0.91
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { (.09)(.91) }{ 500 } } \)
= 0.0128
As the significance level is α = 0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ The confidence limits for the population proportion of bad apples are given by
p-z_α (S.E.) ≤ p ≤ p + z_α (S.E.)
⇒ 0.09 - 2.58 (0.0128) ≤ p ≤ .09 + 2.58 (0.0128)
⇒ 0.09 - 0.033 ≤ p ≤ .09 + 0.033
⇒ 0.057 ≤ p ≤ 0.123
Thus, the bad apples in the consignment lie between (0.057, 0.123).

Average rate of particles emitted in
20 minutes = 5
∴ λ = 5
Let X be the number of particles emitted in 20 minutes
∴ P(X = x) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \), x = 0,1,2,....n
∴ P(X≥2) = 1-P(X<2)
= 1-[P(X = 0) + P(X = 1)]
= 1-\(\left[ \frac { e^{ -5 }.5^{ 0 } }{ 0! } +\frac { e^{ -5 }.5^{ 1 } }{ 1! } \right] \)
= 1-e^-5 (1+5)
= 1-e^-5(6)
= 1-(0.0067)(6)
= 1-0.0402
P(X ≥ 2) = 0.9598

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 = \(\frac { 1 }{ \frac { 1 }{ 2 } .\sqrt { 2\pi } } \Rightarrow K=\sqrt { \frac { 2 }{ \pi } } \).

Given n = 16, S.D = 2 ⇒ \(\sqrt { npq } \) =2
⇒ npq = 4
∴ 16(pq) = 4 ⇒ pq =\(\frac { 4 }{ 16 } =\frac { 1 }{ 4 } \)
⇒ q = \(\frac { 1 }{ 4p } \)...(1)
Since p + q = 1 ⇒ p+\(\frac { 1 }{ 4p } \)=1
⇒ \(\frac { 4{ p }^{ 2 }+1 }{ 4p } \) = 1
⇒ 4p²+1 = 4p ⇒ 4p² -4p+1 = 0
⇒ (2p-1)² = 0 ⇒ 2p-1 = 0
⇒ 2p = 1 ⇒ p = \(\frac { 1 }{ 2 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Mean = np = 16 x \(\frac { 1 }{ 2 } \) = 8

Let X denote the number of white balls drawn from the urn.
Since there are 4 white balls, X can take values 0,1,2,3,4.
P(X = 0)=p(gettmg no white balls)\(=\frac { { 6C }_{ 4 } }{ 10{ C }_{ 4 } } \)
\(\frac{1}{14}\)
P(X = 1) = P(getting one white ball and 3 red balls) \(\frac { { 4C }_{ 4 }\times { 6C }_{ 3 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 2) = P(getting two white balls and 2 red balls) \(\frac { { 4C }_{ 2 }\times { 6C }_{ 2 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 3) = P(getting 3 white balls and 1 red ball) \(\frac { { 4C }_{ 3 }\times { 6C }l }{ { 10 }C_{ 4 } } =\frac { 4 }{ 35 } \)
P(X = 4) = P(getting 4 white balls) =\(\frac { 4{ C }_{ 4 } }{ 10{ C }_{ 4 } } \)
=\(\frac{1}{210}\)
Thus the probability distribution of X is

Let X denote the number of aces in a pack of 52 playing cards.
Since there are 4 aces in a pack, and we are drawing 3 ace cards, X can take values 0, 1, 2, 3.
P(X = 0) = P(getting no ace card) \(\frac { 48{ C }_{ 3 } }{ 52{ C }_{ 3 } } \)
\(=\frac { 4324 }{ 5525 } \)
P(X=1) = P(getting one ace card and 2 other cards)
\(\\ =\frac { 4{ C }_{ 1 }\times 48{ C }_{ 2 } }{ 52{ C }_{ 3 } } =\frac { 1128 }{ 5525 } \)
P(X=2) = P(getting 2ace card and one other card)
\(=\frac { { 4C }_{ 2 }\times { 48C }_{ 1 } }{ 52{ C }_{ 3 } } =\frac { 72 }{ 5525 }\)
P(X=3) = P(getting 3 ace card) = \(\frac { { 4C }_{ 3 } }{ { 52C }_{ 3 } } \)
\(=\frac { 1 }{ 5525 } \)
Hence, the probability distribution function is

\(E(X)=\int _{ 0 }^{ 3 }{ x.f(x)dx=\int _{ 0 }^{ 3 }{ x\left( \frac { dx }{ 9 } \right) dx } } \)
\(=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ { x }^{ 2 }dx=\frac { 2 }{ 9 } .{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 } } \)
\(=\frac { 2 }{ 27 } ({ 3 }^{ 3 }-0)=\frac { 2 }{ 27 } (27)=2\)
\(\therefore E(3X+8)=3.E(X)+8\quad [\because E(8)=8]\)
= 3(2) + 8 = 6 + 8
E(3X + 8) = 14

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

\(E(X)=\sum { xp(x)=-2(\frac { 1 }{ 4 } ) } +2(\frac { 1 }{ 4 } )+5(\frac { 1 }{ 2 } )\)
\(=\frac { -2 }{ 4 } +\frac { 2 }{ 4 } +\frac { 5 }{ 2 } =\frac { 5 }{ 2 } \)
∴ 4E(X²)-V(2X)=4E(X²)-4.V(X)
=4E(X²)-4[E(X²)-E(X)²]
=4E(X²)-4E(X²)+4[E(X)]²
\(=4{ \left( \frac { 5 }{ 2 } \right) }^{ 2 }[\because E(X)=\frac { 5 }{ 2 } ]\)
\(=4\left( \frac { 25 }{ 4 } \right) =25\)

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

Let p(t) denotes the amount of money in the account at time t. Then the differential equation governing the growth of money is
\(\frac { dp }{ dt } =\frac { 12 }{ 100 } p\Rightarrow \frac { dp }{ dt } =0.12p\)
Separating the variables,
\(\frac { dp }{ dt } =0.12\)
Integrating, \(\int { \frac { dp }{ P } } =\int { 0.12 } \)dt+c
⇒ log p = 0.12t + C
⇒ P = e^0.12t+c ⇒ P = e^0.12t . e^c
⇒ P = e^0.12t.C1  .....(1)
When t = 0, p = 0 ⇒ p = e⁰ (c₁) ⇒ c₁ = p
When t = 5, and p = 25000
25000 = e^12(5).p       [∵ c₁ = p]
⇒ 25000 = e^0.6p
⇒ \(\frac { 2500 }{ { e }^{ 0.6 } } \) = p
⇒ 25000 (e^-0.6) = p
⇒ 25000 (0.5488) = p
⇒ Rs. 13720
Hence, to get an amount of Rs. 25000 at the end of 5 years, Rs. 13720 must be invested.

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 slope = y + 2x
⇒ \(\frac { dy }{ dx } \) = y+2x
⇒ \(\frac { dy }{ dx } \)- y = 2x
This is of the form \(\frac { dy }{ dx } \)+ Py = Q where P = -1, Q = 2x
\(\\ \int { P } dx=\int { -1 } dx\) = -x
∴ I.F. = \(e^{ \int { P } dx }\) = e^-x
∴ The solution is y.\(e^{ \int { P } dx }=\int { Q } .e^{ \int { P } dx }\)dx+C
⇒ y.e^-x = \(\int { 2x } \).e^-xdx+C
Let u = x; dv = e^-x
u² = 1; v = -e^-x
v₁ = e^-x
⇒ ye^-x = 2[-xe^-x-1(e^-x)]+C
[Bernoulli's formula]
⇒ ye^-x = -2x e^-x -2e^-x+ C...(1)
Since the Curve passes through (0, 0), we get
⇒ 0 = 0-2e⁰ + C ⇒ C = 2
(1) becomes,
∴ ye^-x = -2xe^-x - 2e^-x + 2
ye^-x = -2xe^-x - 2e^-x + 2e^x.e^-x
= e^-x(2 e^x-2x-2)
ye^-x = 2e^-x(e^x-x-1)

Given cos²x dy + y.e^tanx dx=0
⇒ cos²x dy = -y e^tanx dx   [∵ t = tanx, dt = sec²x dx, ∴ \(\int { { e }^{ t } } dt\) = e^tanx]
⇒ \(\frac { dy }{ y } =-\frac { { e }^{ tanx } }{ cos^{ 2 }x } \)dx
⇒ \(\frac { dy }{ y } \) = -sec²x.e^tanx dx
Integrating, \(\int { \frac { dy }{ y } } =-\int { sec^{ 2 }x } .e^{ tanx }dx\)
log y = -e^tanx + C
⇒ log y + e^tanx = C

Given \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)= 1
⇒ \(\frac { { b }^{ 2 }x^{ 2 }+{ a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ y }^{ 2 } } \)= 1
⇒ b²x²+a²y² = a²b²
Differentiating again w.r.t 'x' we get,
2b²x + 2a²y\(\frac { dy }{ dx } \)=0 ⇒ b²x+a²yy₁ = 0 ....(2)
Differentiating w.r.t 'x' we get,
b² + a²[yy₂ + y₁y₁] = 0
⇒ b² + a²[yy₂ + y₁²] = 0 ...(3)
Eliminating a² and b² from (1) and (3) we get
\(\left| \begin{matrix} x & y{ y }_{ 1 } \\ 1 & { y }_{ 1 }^{ 2 }+y{ y }_{ 2 } \end{matrix} \right| \) = 0
⇒ x(y₁²+yy₂) - yy₁ = 0
⇒ x\(\left( \left( \frac { dy }{ dx } \right) ^{ 2 }+y.\frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) -y\left( \frac { dy }{ dx } \right) \)= 0 which is the required differential equation.

Since 82 lies at the beginning of the table, we can use Newton's forward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Also x₀ + nh = 82 ⇒ 75 + n(5) = 82 ⇒ 5n = 82 - 75 = 7
⇒ n = \(\frac75\) = 1.4
The difference table is

\(y=2459+\frac { 1.4 }{ 1! } (-441)+\frac { (1.4)(1.4-1) }{ 2! } (-397)+\frac { (1.4)(1.4-1)(1.4-2) }{ 3! } (457)\)
= 2459 - 617.4 - 111.6 - 25.592
y = 1704. 408 when x = 82.

From the graph, it is clear that when x = 27, the value of y is 24.8

Here, the area lies below the x-axis
\(\therefore \ A=\int _{ -2 }^{ 0 }{ (-y)dx } =\int _{ 0 }^{ -2 }{ ydx } \)
\(\left[ \because \int _{ a }^{ b }{ f()dx } =-\int _{ b }^{ a }{ f(x)dx } \right] \)
\(={ \int _{ 0 }^{ -2 }{ (x-1)dx=\left[ \frac { { x }^{ 2 } }{ 2 } -x \right] } }_{ 0 }^{ -2 }\)
\(\left( \frac { { (-2) }^{ 2 } }{ 2 } -(2) \right) -0\)
\(=\frac { 4 }{ 2 } +2=2+2\)
A = 4 sq.units

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.

Transition probability matrix

Shares after one week
\(\left( \cdot 6\cdot 4 \right) \left( \begin{matrix} \cdot 7 & \cdot 3 \\ \cdot 6 & \cdot 8 \end{matrix} \right) \)
= (-6\(\times\)·7+-4x·2 ·6\(\times\)·3+·4\(\times\)·8)
= z:(-42+·08 ·18+·32)= (·50 ·50)
\(\Rightarrow\) A = 50% and B = 50%
Shares after two weeks \(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 7 & \cdot 3 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·7+·5\(\times\).2 ·5\(\times\)·3 +·5\(\times\)·8)
= (-35+·10 ·15 + 40) = (-45 ·55)
A = 45% and B = 55%

The non-homogeneous equations are
2x - 3y - 1 = 0, 5x + 2y - 12 = 0
\(\Delta =\left| \begin{matrix} 2 & -3 \\ 5 & 2 \end{matrix} \right| =4+15=19\neq 0\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 1 & -3 \\ 12 & 2 \end{matrix} \right| =2+36=38\)
\(\Delta y=\left| \begin{matrix} 2 & 1 \\ 5 & 12 \end{matrix} \right| =24-5=19\)
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 38 }{ 19 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 19 }{ 19 } =1\)
\(\therefore \) Solution set is {2, 1}

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.