Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)

Let X denote the height of the student
Given P (X < 40) = 10% = \(\frac { 10 }{ 100 } \) =0.1
P(X> 90) = 10% = \(\frac { 10 }{ 100 } \) =0.1

∴ P(40 < X < 90) = P(-∞ < X < ∞) - [P(X < 40) + P(X < 90)]
= 1 - (0.1 + 0.1)
= 1 - 0.2 = 0.8
∴ out of 800 students, number of students scored between 40 and 90 = 800 x 0.8
= 640 students.

Here Σa_i = 22 + 15 + 8 = 45
Σbj = 7 + 12 + 17 + 9 = 45
Σa_{i =} Σbj
∴ The given problem is balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I-allocation:

[∵ the max penalty is 4. In II, least cost is 2 & min (12,22) = 12]

II-allocation:

[∵ the max penalty is 3. In B, least cost is 1 & min (17,15) = 15]
III-allocation:


[∵ the max penalty is 3. In III, least cost is 4 & min (2, 10) = 2]
IV-allocation:

[∵ the max penalty is 2. In A, least cost is 3 & min (9, 8) = 8]
V-allocation:

[∵ the max penalty is 1. In C, least cost is 4 & min (7, 8) = 7]
VI-allocation:

[∵ min (1,1) = 1]
Thus, the allocations are

∴ The transportation schedule is
A → II, A → III, A → IV, B → III, C → I and C → IV
Hence, the total transportation cost is
= 12(2) + 2(4) + 8(3) + 15(1) + 7(4) + 1(5)
= 24 + 8 + 24 + 15 + 28 + 5 = Rs.104

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X<0)=P(X=-2)
= 1/4

Let I = \(\int { \frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } } \)
Consider
\(\frac { { x }^{ 2 }+1 }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\frac { A }{ x-1 } +\frac { B }{ { \left( x-1 \right) }^{ 2 } } +\frac { C }{ x+3 } \quad ----(1)\)
\(\frac { { x }^{ 2 }+1 }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\quad \frac { A(x-1)(x+3)+B(x+3)+C{ (x-1) }^{ 2 } }{ { (x-1) }^{ 2 }(x+3) } \)
⇒ x² + 1 = A (x - 1)(x+3) + B (x + 3) + C (x-1)²
When X = 1
1 + 1 = B (1 + 3) ⇒ 2 = b (4)
⇒ \(B=\frac { 2 }{ 4 } \Rightarrow B=\frac { 1 }{ 2 } \)
When x = -3
(-3)² + 1 = C (-4)² ⇒ 10 = 16C
⇒ \(C=\frac { 10 }{ 16 } \)
\(C=\frac { 5 }{ 8 } \)
When x = 0,
1 = -3A +3B + 3C
\(1=-3A+3\left( \frac { 1 }{ 2 } \right) +\frac { 5 }{ 8 } \)
\(3A=-1\frac { 3 }{ 2 } +\frac { 5 }{ 8 } =\frac { -8+12+5 }{ 8 } \)
= \(\frac { 9 }{ 8 } \)
\(A=\frac { 9 }{ 8 } \)
From (1), \(\frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\frac { \frac { 3 }{ 8 } }{ x-1 } +\frac { \frac { 1 }{ 2 } }{ { (x-1) }^{ 2 } } +\frac { \frac { 5 }{ 8 } }{ x+3 } \)
\(I=\frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } \)
= \(\frac { 3 }{ 8 } \int { \frac { dx }{ x-1 } + } \frac { 1 }{ 2 } \int { \frac { dx }{ { (x-1) }^{ 2 } } } +\frac { 5 }{ 8 } \int { \frac { dx }{ x+3 } } \)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| +\frac { 1 }{ 2 } \int { { (x-1) }^{ -2 } } dx+\frac { 5 }{ 8 } log\left| x+3 \right| +c\)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| +\frac { 1 }{ 2 } \frac { { (x-1) }^{ -1 } }{ -1 } +\frac { 5 }{ 8 } log\left| x+3 \right| +c\)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| -\frac { 1 }{ 2(x-1) } +\frac { 5 }{ 8 } log\left| x+3 \right| +c\)

For market clearance, Q_d = Q_s
\(\Rightarrow 42-4p-4\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =-6+8p\)
\(\Rightarrow 48-12p-4\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =0\)
\(\Rightarrow \frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =4\frac { dp }{ dt } -12p=-48\)
The auxiliary equation is m² - 4m - 12 = 0
⇒ (m - 6) (m + 2) = 0
⇒ m = -2, 6
The roots are real and different
∴ C.F. is Ae^-2t + Be^6t
P.I. = \(\frac { 48 }{ (D-6)(D+2) } { e }^{ 0t }=\frac { -48 }{ (0-6)(0+2) } \)
= \(\frac { -48 }{ -12 } \)
∴ The general solution is
P = C.F. + P.I.
⇒ P = Ae^-2t + Be^6t + 4

Let the percentage of lead be x and temperature be y.
Given

Find y when x = 84
Since the temperature required is at the end of the table we apply Newton's backward interpolation formula.
∴ x_n + nh = 84 ⇒ 90 + n(10) = 84
⇒ 10n = 84 - 90 = -6
[∵ h = 10 & x_n = 90]
⇒ n = \(\frac{-6}{10}\) = -0.6
The difference table is

∴ The Newton's backward interpolation formula is
y_(y=xn+nh) = \(\frac { n }{ 1! } \Delta { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3) }{ 4! } { \triangledown }^{ 4 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3)(n+4) }{ 5! } { \triangledown }^{ 5 }{ y }_{ n }\)
∴ y(84) = 304 + (-0.6) (28) + \(\frac { (-0.6)(-0.6+1) }{ 2 } (2)+0+0+\frac { (-0.6)(-0.6+1)(-0.6+3)(-0.6+4) }{ 5! } (4)\) [∵ ∇³ & ∇⁴ are zero]
⇒ y(84) = 304 - 16.8+ (-0.6) (0.4) + \(\frac { (-0.6)(0.4)(1.4)(2.4)(3.4) }{ 120 } (4)\)
⇒ y(84) = 304 -16.8 - 0.24 - 0.09139
⇒ y(84) = 286.86
Hence, the melting point of the alloy containing 84 percent lead is 286.86°C.

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one.
To balance it, introduce dummy columns with all the entries zero.
∴ The revised assignment problem is

Step 1 : Select the smallest element in each row and subtract this from all the elements in its row.

Since each row and column has atleast one zero, assignments can be made.
Step 2 : Examine the rows with atleast one row.
Row A & B have exactly one row. Mark them by ׈ and mark X by other zeros in its column.
Row C & F also has only one zero.
Here only 4 vacant space can be assigned to 4 trucks.
∴ The optimal assignments schedule and total cost is

∴ The optimal assignment (minimum) cost = Rs. 12

Laspeyre's pnce index number
\(P^{L}_{01}\) = \(\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times 100\)
= \(\frac {239945.23}{480244.71} \times 100 \)
Paasches index number
\(P^{p}_{01}\) =  \(\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times 100\)
= \(\frac {325150.48}{645496.92} \times 100 \)
= 50.37
Fisher's index number
\(P^{F}_{01}\)  = \(\sqrt{ \frac {\sum p_{0}q_{0}}{\sum p_{0}q_{0}} \times ​\frac{​ {\sum p_{0}q_{0}} }{{\sum p_{0}q_{0}}}{}}\times 100\)
= \(\sqrt\frac {239945.23 \times 325150.48}{480244.71 \times 645496.92} \times 100\)
= \(\sqrt\frac {489.84 \times 570.22}{693 \times 803.43} \times 100\)
= 50.1

Sample size n =100, Sample mean \(\bar x\) = 390 pages, Population mean \(\mu\) = 400 pages
Population SD \(\sigma\) = 20 pages
The sample is a large sample and so we apply Z -test
Null Hypothesis:
There is no significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₀ : \(\mu\) = 400
Alternative Hypothesis:
There is significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₁:\(\mu\neq\) 400 (two tailed test)
The level of significance \(\alpha\) = 1% = 0.01
Applying the test statistic
\(Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) ; \)
\(Z=\frac{390-400}{\frac{20}{\sqrt{100}}}=\frac{-10}{2}=-5, \therefore|Z|=5\)
Thus the calculated value |Z| = 5 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=2.58\)
Comparing the calculated and table values, we found Z > \({ Z }_{ \frac { \sigma }{ 2 } }\) i.e., 5 > 2.58
Inference: Since the calculated value is greater than table value i.e., \(Z>{ Z }_{ \frac { \sigma }{ 2 } }\) at 1% level of significance, the null hypothesis is rejected and Therefore we concluded that \(\mu \neq400\) and the manufacturer’s claim is rejected at 1% level of significance.

Let the probability of student having reading habit
p = 40% = \(\frac { 40 }{ 100 } \) = 0.4
⇒ q = 1-p = 1-0.4 = 0.6
n = 9
(i) P (none of those who have selected having reading habit)
= P(X = 0)
= ⁹C₀(0.4)⁰ (0.6)⁹
= (1)(1)(0.6)⁹   [∵ ⁿC_x p^xq^n-x = p(x), n = 9, x = 0]
= (0.6)⁹   [ ∵ ⁹C₀ = 1 ]
= 0.01008
(ii) P (all those who have selected have newspaper reading habit)
= P(X = 9)
= ⁹C₉(0.4)⁹ (0.6)⁰   [∵ p(x) =ⁿC_x p^xq^n-x, n = 9, x = 9]
= (0.4)⁹   [ ∵ ⁹C₉ = 1 ] 
= 0.000261
(iii) Two thirds of 9 = \(\frac{2}{3}\) x 9 = 6
∴ P(atleast two third have newspaper reading habit)
= P (atleast 6 have newspaper reading habit)
= P(X≥6)
= P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)
= ⁹C₆ (0.4)⁶ (0.6)³ + ⁹C₇ (0.4)⁷ (0.6)² + ⁹C₈ (0.4)⁸ (0.6)¹ + ⁹C₉ (0.4)⁹ (0.6)⁰
= (0.4)⁶ [⁹C₆ (0.6)³ + ⁹C₇ (0.4) (0.6)² + ⁹C₈ (0.4)² (0.6) + (0.4)³ ]
= (0.4)⁶ [⁹C₃ (0.216) + ⁹C₂ (0.144) + ⁹C₁ (0.096)+0.64     [∵ ⁿC_r = ⁿC_n-r ]
= (0.4)⁶ \(\left[ \frac { 9\times 8\times 7 }{ 3\times 2\times 1 } (0.216)+\frac { 9\times 8 }{ 2\times 1 } (0.144)+9(0.096)+.064 \right] \)
= (0.4)⁶ [18.144 + 5.184 + 0.864 + 0.064]
= (0.4)⁶ [24.256] = (0.0041) (24.256)
∴ P(X≥6) = 0.0994

Given MC = \(\frac{dC}{dx}\alpha x\)
\(\Rightarrow \frac { dC }{ dx } ={ k }_{ 1 }x\)
\(\Rightarrow dC={ k }_{ 1 }xdx\)
\(\Rightarrow \int { dC={ k }_{ 1 }\int { x } dx } \)
\(\Rightarrow C={ k }_{ 1 }\frac { { x }^{ 2 } }{ 2 } +{ k }_{ 2 }...(1)\)
Given fixed cost is Rs. 5000
∴ When x = 0, C = 5000
⇒ 5000 = k₁(0) + k₂ = 5000
∴ (1)becomes C=k₁\(\frac{x^2}{2}+5000\) ...(2)
Also it is given that when x = 50, C = Rs. 5625
\(\therefore (2)5625={ k }_{ 1 }\frac { { x }^{ 2 } }{ 2 } +5000\)
\(\Rightarrow 5625-5000={ k }_{ 1 }\times \frac { { (50) }^{ 2 } }{ 2 } \)
\(\Rightarrow 625={ k }_{ 1 }\times \frac { (50)\times (50) }{ 2 } \)

\(C=\frac { 1 }{ 2 } \left( \frac { { x }^{ 2 } }{ 2 } \right) +5000\)
⇒C = \(\frac{x^2}{4}\) + 5000

The present value of payment for t year = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
Present value of 12 years = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
= 72000\({ \left[ \frac { { e }^{ -0.09t } }{ { -0.09 } } \right] }_{ 0 }^{ \\ 12 }\)
= \(\frac { 72000 }{ -0.09 } \left[ { e }^{ -0.09(12) }-{ e }^{ 0 } \right] \)
= \(-8,00,000[{ e }^{ -1.08 }-{ e }^{ 0 }]\)
= −8,00,000 [0.3396 −1]
= 5,28,320
Cost of the machine = 5,00,000 − 30,000
= 4,70,000
Hence it not advisable to rent the machine
It is better to buy the machine.

Let ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.

Given sample size n = 64
Sample mean \(\bar { x } \) = 51250
Null hypotheses: H₀: Population mean μ = 50000 Alternative hypotheses: H₁ : μ≠ 50,000
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 51250-50000 }{ \frac { 2000 }{ \sqrt { 64 } } } \)=5
∴ z = 5
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z > z_α as 5 < 1.96
Inference: As z > z_α, H₀ is rejected. Hence, we can conclude that the sample mean differs significantly from the population mean.

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

(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 \(I=\int { \frac { { sec }^{ 2 }x }{ 3+tanx } } dx\)
Put 3 + tan x = t
⇒ 0 + sec²x dx = dt
⇒ sec² x dx = dt
\(\therefore I={ \int { \frac { dt }{ t } } }\)
= log |t| + c
= log |3 + tan x| + c
[∵ t = 3 + tan x]

Given y = (A + Bx)e^3x ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= (A+Bx)e^3x(3)+e^3x(B)
⇒ \(\frac { dy }{ dx } \) = 3y + Be^3x  [Using (1)]
⇒ Be^3x = \(\frac { dy }{ dx } \)-3y
Differentiating again w.r.t 'x' we get,
\(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) \)+ Be3x(3)
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left[ \frac { dy }{ dx } -3y \right] \) [Using (2)]
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left( \frac { dy }{ dx } \right) \)-9y
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =6\left( \frac { dy }{ dx } \right) \)-9y which is the required differential equation.

Given y₃ = 2, y₄ = −6, y₅ = 8, y₆ = 9 and y₇ = 17
Δ⁴y₃ = (E−1)⁴y₃
= (E⁴ − 4E³ + 6E² − 4E+1)y₃
= E⁴y₃ − 4E³y₃ + 6E²y₃− 4Ey₃ + y₃
= y₇ − 4y₆ + 6y₅ −4y₄+ y₃
= 17 – 4(9) + 6(8) –4(–6) + 2
= 17 – 36 + 48 + 24 + 2 = 55

Using aggregate expenditure method,
Cost of living index number 
C.L.I = \(\sqrt \frac {\sum p_{1}q_{0}}{{\sum p_{0}q_{0}}}\times100\)
= \(\sqrt \frac {8600}{6510}\times100\)

= 132.10

\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ x \sin x } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ udv } \)
\(={ \left( uv \right) }_{ 0 }^{ \frac { \pi }{ 2 } }-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ vdu } \)
\(={ \left[ -x \cos x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos x } dx\)
\(=0+{ \left[ \sin x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=1\)

Probability of bus running late is denoted as p = 1-0.8 = 0.2
Probability of bus running according to the schedule is q = 0.8
Also given that n = 10
The binomial distribution is p(x) = 10C_x(0.2)^x(0.8)^10-x
(i) probability that exactly one is late P(x = 1) = 10C₁pq⁹
= 10C₁(0.2)(0.8)⁹
(ii) probability that at least one is late
= 1 – probability that none is late
= 1 – p(x = 0)
= 1– (0.8)¹⁰

Transition probability matrix
T = \(_{ B }^{ A }\left( \begin{matrix} \overset { A }{ 0.9 } & \overset { B }{ 0.1 } \\ 0.3 & 0.7 \end{matrix} \right) \)
At equilibrium, (A B) T = (AB) where A + B = 1
(A B) \(\left( \begin{matrix} 0.9 & 0.1 \\ 0.3 & 0.7 \end{matrix} \right) \) = (A B)
0.9A + 0.3B = A
0.9A + 0.3(1−A) = A
0.9A−0.3A + 0.3 = A
0.6A + 0.3 = A
0.4A = 0.3
A = \(\frac { 0.3 }{ 0.4 } =\frac { 3 }{ 4 } \)
B = 1-\(\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
Hence the market share of brand A is 75% and the market share of brand B is 25%

C.L.I = \(\frac{\Sigma p_1q_0}{\Sigma p_0q_0}\times100\)

C.L.I = \(\frac{4362.50}{3505}\times100\) = 124.46

Let p be the probability of happening of an event.
Given p = 0.002 = \(\frac { 2 }{ 1000 } \)
Also n = 1000
∴ Mean = np = \(1000\times \frac { 2 }{ 1000 } \) = 2
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P( event happens exactly twice)
= P(X = 2)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -2 }.{ (2 }^{ 2 }) }{ 2 } \)
= e^-2(2) = 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Since three values of f(x) are given, we assume that the polynomial is of degree two.
⇒ Δ³(f(x₀)) = 0
⇒ ∆³(y_o) = 0
⇒ (E - 1)³ y_o= 0
⇒ (E³ - 3E² + 3E - 1) y_o= 0
⇒ y₃- 3y₂+ 3y₁ - y_o= 0
⇒ 157 - 3 (126) + 3y₁ - 100 = 0
⇒ y₁ = 107
∴ The missing term is 107.

Given C'(x) = 85 + \(\frac{375}{x^2}\).
We know C(x) ഽC'(x) + k
The cost of producing 10 incremental units after 15 units have been produced
\(C'(x)=85+\frac { 375 }{ { x }^{ 2 } } \)
\(C(x)=\int { C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ \left( 85+\frac { 375 }{ { x }^{ 2 } } \right) } dx\)
\({ \left[ 85+\frac { 375 }{ { x } } \right] }_{ 15 }^{ 25 }\)
\(\left( 85(25)-\frac { 375 }{ 25 } \right) -\left( 85(15)-\frac { 375 }{ 15 } \right) \)
= (2125 - 15) - (1275 - 25)
= 2110 - 1250 = Rs. 860

Transition probability matrix

Given present market shares are 15% for A and 85% for B
\(\therefore\) Market shares after one year 
= \(\left( \cdot 15\cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((-15)(-65)+(-85)(-45) ·15x·35+·85x·55)
= (-0975 + 0.3825 .0525 + 4675)
= (0 .48 0.52)
\(\therefore\) Market shares after one year for A is 48% and for B is 52%

To assign the different jobs to the different machines (one job per machine) to minimize the overall cost is known as assignment problem.

The auxiliary equation is m² - 4m + 4 = 0
⇒ (m - 2)² = 0
⇒ m 2,2
The roots are real and equal
∴ Complementary function CF is (Ax + B)e^2x
∴ The general solution is y = (Ax + B)e^2x

The error of rejecting H_o when it is true is type I error.

\(F(x)=\int _{ -\infty }^{ x }{ f(t)dt } \)
\(=\int _{ 100 }^{ x }{ \frac { 100 }{ { t }^{ 2 } } dt,\quad x\ge 100 } \)
\(={ \left[ \frac { 100 }{ -t } \right] }_{ 100 }^{ x },\quad x\ge 100\)
\(F(x)=\left[ 1-\frac { 100 }{ x } \right] ,\ge 100\)

\(\int { \left( 3x+x \right) \left( 2-5x \right) } dx\)
\(=\int { \left( 6-15x+2x-{ 5x }^{ 2 } \right) } dx\)
\(=\int { \left( 6-13x-{ 5x }^{ 2 } \right) } dx\)
\(=6x-\frac { { 13x }^{ 2 } }{ 2 } -\frac { { 5x }^{ 3 } }{ 3 } +c\)