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

Let I = \(\int { \frac { 1 }{ { 3x }^{ 2 }+13x-10 } } dx\)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x-\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x+\frac { 169 }{ 36 } -\frac { 169 }{ 36 } -\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-\frac { 289 }{ 36 } } } \)

= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-{ \left( \frac { 17 }{ 16 } \right) }^{ 2 } } } \)
= \(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| } +c \right] \)
= \(\frac { 1 }{ 3 } \times \frac { 1 }{ 2\times \frac { 17 }{ 6 } } log\left| \frac { x+\frac { 13 }{ 6 } -\frac { 17 }{ 6 } }{ x+\frac { 13 }{ 6 } +\frac { 17 }{ 6 } } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { x-\frac { 2 }{ 5 } }{ x+5 } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { 3x-2 }{ 3(x+5) } \right| +c\)

Given x²\(\frac { dy }{ dx } \)= y²+2xy
⇒ \(\frac { dy }{ dx } =\frac { { y }^{ 2 }+2xy }{ x^{ 2 } } \)
The numerator and denominator are homogeneous function of degree
∴ put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

= v²+2v
⇒ v + x\(\frac { dv }{ dx } \) = v²+ 2v
⇒ x\(\frac { dv }{ dx } \) = v²+ 2v-v = v²+ v
Separating the variables,
\(\frac { dv }{ { v }^{ 2 }+v } =\frac { dx }{ x } \Rightarrow \frac { dv }{ v(v+1) } =\frac { dx }{ x } \)
[\(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
1 = A(v+1)+Bv
put v = -1
1 = -B
put v = 0
1 = A
∴ \(\left( \frac { 1 }{ v } +\frac { 1 }{ v+v } \right) dv=\frac { dx }{ x } \)
Integrating
\(\int { \frac { dv }{ v } } -\int { \frac { dv }{ v+1 } } =\int { \frac { dx }{ x } } \)
⇒ log v - log (v + 1) - log x + log c
⇒ log\(\left( \frac { v }{ v+1 } \right) \)= log(xc)
⇒ \(\frac { v }{ v+1 } \)
Replacing v by \(\frac { y }{ x } \) we get,
\(\frac { \frac { y }{ x } }{ \frac { y }{ x } +1 } =xc\Rightarrow \frac { \frac { y }{ x } }{ \frac { x+y }{ x } } \) = xc
⇒ \(\frac { y }{ x+y } \) = xc
⇒ y = cx(x+y)  ....(1)
Given, when x = -1, y = 1
∴ 1 = c(1) (1+1) ⇒ = 2c ⇒ c = \(\frac { 1 }{ 2 } \)
∴ (1) becomes, y = \(\frac { x }{ 2 } \)(x+y)
⇒ 2y = x(x+y)

To find the population for the year 1905 (i.e) the value of y at x = 1905
Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\cfrac { n }{ n! } \Delta { y }_{ 0 }+\cfrac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\cfrac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 } + ...\)
To find y at x = 1905
\(\therefore\) x₀+nh = 1905 , x₀ = 1891, h = 10
1891+n(10) = 1905 \(\Rightarrow\) n = 1.4

y_(x=1905) = \(98,752+(1.4)(33533)+\frac { (1.4)(0.4) }{ 2 } (2258)+\frac { (1.4)(0.4)(-0.6) }{ 6 } (-10435)+\frac { (1.4)(0.6)(-0.6)(-1.6) }{ 24 } (41358)\)
=  98,752 + 46946.2 + 632.4 + 584.36 + 1389.63
= 1,48,304.43
= 1,48,304

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

Here column 2 has no zero. Go to Step 2.
Step 2 : Select the smallest element (10) and subtract it from all the elements in its column.

Step 3 : Examine the rows with only one zero mark that zero by Ԡ. Mark other zeros in its column by X.
Row 1 and Row 3 contains only one zero. Mark the other zeros by X
Column 2 contains exactly one zero. Mark it by Ԡ
Thus, all the 3 assignments have been made.
Hence, the optimal assignment schedule and total cost is

Total cost = Rs. 280
Thus, the optimal assignment (minimum) cost  =  Rs. 280

\(\overline {\overline{X}} = \frac {5.10+4.98+5.02+4.96+4.96+5.04+4.94+4.92+4.92+4.98}{10}\)
\(\overline {\overline{X}}\) = \(\frac {49.82}{10} = 4.982\)
\(\overline {\overline{R}} = \frac {0.3+.4+.2+.4+.1+.1+.8+.5+.3+.5}{10}\)
\(\overline {R}\) = \(\frac {3.6}{10}\) = 0.36
The control limits of mean chart are
UCL  = \(\overline {\overline{X}} + A_{2} \overline{R}\)
= 4.982 + .577(.36)
= 4.982+ .208 = 5.19
CL = 4.982
LCL =  \(\overline {\overline{X}} - A_{2} \overline{R}\)
= 4.982 - .208 = 4.774
The control limits of R-chart are
UCL = D₄\(\overline {R}\) = 2.114 (.36) = 0.761
CL = \(\overline {R}\)  = .36
LCL = D₃\(\overline {R}\)  = 0

Mean Chart

 Range chart

Conclusion: The above diagram shows all the control lines with the data points plotted.
Since one in Range chart, lie outside the control limits, we can say that the process is out of control.

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 442 }{ 10 } =44.2\)
\(\bar { R } =\frac { \sum { R } }{ n } =\frac { 58 }{ 10 } =5.8\)
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } \)
= 44.2 + 0.483(5.8) = 47.00
\(CL=\overset { = }{ X } =44.2\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =44.2-0.483(5.8)=41.39\)
The above diagram shows all the three control lines with the data points plotted, since four points falls out of the control limits, we can say that the process is out of control.

Given: n = 169, \(\bar x\) =1350 hours, \(\sigma\) =100 hours, since the level of significance is (100-90)% =10% thus \(\alpha\) is 0.1, hence the significant value at 10% is \({ Z }_{ \frac { \alpha }{ 2 } }\)=1.645
\(S.E.=\frac { \sigma }{ \sqrt { n } } =\frac { 100 }{ \sqrt { 169 } } =7.69\)
Hence 90% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \sigma }{ 2 } }<\mu <\bar { x } +{ Z }_{ \frac { \sigma }{ 2 } }SE\)
\(\begin{gathered} 1350-(1.645 \times 7.69) \leq \mu \leq 1350+(1.645 \times 7.69) \end{gathered}\)
\(1337.35 \leq \mu \leq 1362.65\)
Hence the mean life time of light bulbs is expected to lie between the interval (1337.35, 1362.65).

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 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)+P(X=0)
=\(\frac{1}{4}\)+\(\frac{1}{4}\)=\(\frac{1}{2}\)

\(Let\ I=\int { { e }^{ x } } \left[ \frac { x-1 }{ { \left( x+1 \right) }^{ 3 } } \right] \)
\(=\int { { e }^{ x } } \left[ \frac { x-1+1-1 }{ { \left( x+1 \right) }^{ 3 } } \right] dx\)
[Adding and subtracting 1 in the numerator]
\(=\int { { e }^{ x } } \left[ \frac { x+1-2 }{ { \left( x+1 \right) }^{ 3 } } \right] dx\)
\(=\int { { e }^{ x } } \left( \frac { x+1 }{ { \left( x+1 \right) }^{ 3 } } -\frac { 2 }{ { \left( x+1 \right) }^{ 3 } } \right) dx\)
\(=\int { { e }^{ x } } \left( \frac { 1 }{ { \left( x+1 \right) }^{ 3 } } -\frac { 2 }{ { { \left( x+1 \right) } }^{ 3 } } \right) dx\)
\(Let\ f\left( x \right) =\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } ={ \left( x+1 \right) }^{ -2 }\)
\(\Rightarrow f^{ ' }\left( x \right) =-2{ \left( x+1 \right) }^{ -2-1 }\)
\(=-2{ \left( x+1 \right) }^{ -3' }\)
\(=\frac { -2 }{ { \left( x+1 \right) }^{ 3 } } \)
\(\therefore I=\int { { e }^{ x }\left[ f\left( x \right) +f^{ ' }\left( x \right) \right] } dx\)
\(={ e }^{ x }\quad f\left( x \right) +c\)
\(={ e }^{ x }\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } +c\)
\(=\frac { { e }^{ x } }{ { \left( x+1 \right) }^{ 2 } } +c\)

\({ y }^{ 2 }=8x\) (1)
Comparing this with the standard form \({ y }^{ 2 }=4x\),
4a = 8
a = 2
Equation of latus rectum is x = 2
Since equation (1) is symmetrical about x- axis
Required Area = 2[Area in the first quadrant between the limits x = 0 and x = 2]
\(=2\int _{ 0 }^{ 2 }{ y } dx\)
\(2\int _{ 0 }^{ 2 }{ \sqrt { 8xdx } } =2(2\sqrt { 2 } )\int _{ 0 }^{ 2 }{ { x }^{ 1/2 } } dx\)
\(=4\sqrt { 2 } { \left[ \frac { { 2x }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 2 }=4\sqrt { 2 } \times 2 \times \frac { { 2 }^{ \frac { 3 }{ 2 } } }{ 3 } \)
\(=\frac { 32 }{ 3 } \) sq. units.

Let the weight assigned to the three varieties be Rs. x, Rs. y and Rs. z respectively By the given data,
x + 2y + 3z = 11
2x + 4y + 5z = 21
3x + 5y + 6z = 27
\(\Delta =\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{matrix} \right| =1\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(24 - 25) -2(12 - 15) + 3(10 - 12)
= 1(-1) -2 (-3) + 3(-2)
= -1+6 - 6 = -1\(\neq \) 0.
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied.
\(\Delta x=\left| \begin{matrix} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{matrix} \right| \)
\(=11\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| +3\left| \begin{matrix} 21 & 4 \\ 27 & 5 \end{matrix} \right| \)
= 11(24 - 25) - 2(126 - 135) + 3(105 - 108)
= 11(-1) - 2(-9) + 3 (-3)
= 11+18-9
= -2
\(\Delta y=\left| \begin{matrix} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{matrix} \right| \)
= \(\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| -11\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| \)
= 1(126 - 135) - 11(12 -15) + 3(54 - 63)
= - 9 - 11(-3) + 3(-9)
= - 9 + 33 - 27
= 3
\(\Delta z=\left| \begin{matrix} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{matrix} \right| =1\left| \begin{matrix} 4 & 21 \\ 5 & 27 \end{matrix} \right| -2\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| +11\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(108 - 105) - 2(54 - 63) + 11(10 - 12)
= 1(3) - 2(-9) + 11(-2)
= 3 + 18 - 22
= -1
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -2 }{ -1 } =2\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
and \(z=\cfrac { \Delta z }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
Hence, the weights assigned to the three varieties are 2, 3 and 1 respectively

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.

\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
Control limits for mean chart
UCL = \(\bar{\bar{X}}\) +  A_{2^\(\bar{R}\)}
= 10.66 + .577(6.3) = 14.295
CL = \(\bar{\bar{X}}\) = 10.66
Control limits for R-chart
UCL =   D_{2^\(\bar{R}\)} = 2.115 \(\times\) 6.3
= 13.324
CL = _^\(\bar{R}\) = 6.3
LCL = D₃\(\bar{R}\) = 0

Given mean μ = 60 and S.D. σ = 5

To find P(X > 60)
When X = 60, Z =\(\frac { X-\mu }{ \sigma } =\frac { 60-60 }{ 5 } \) = 0
∴ P(X > 60) = P(Z > 0) = P (0 < Z < ∞)
= 0.5
∴ 50% of students scored more than 60 marks

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 curve is y = 4x - X²
The limits are from x = 0 to x = 3
∴ Area \(\int _{ 0 }^{ 3 }{ y } dx=\int _{ 0 }^{ 3 }{ (4x-{ x }^{ 2 }) } dx\)
\({ \left[ \frac { 4{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }=2(9)-\frac { 27 }{ 3 } \)
=18 - 9
Area = 9 sq.units

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%

The highest derivative is of third order and its power is 1.
Order is 3 and degree is 1.

The objective function is minimize Z = \(\overset { m }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the constraints
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = a_i, i=1,2,.....m\) (Supply constraints)
\(\overset { m }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = b_j, j=1,2,.....n\) (demand constraints)
x_ij ≥, 0 for all i, j (non-negative restrictions)

Any sample statistic which is used to estimate an unknown population parameter is called an estimator (i.e.,). an estimator is a sample statistic used to estimate a population parameter.

\(\int _{ 0 }^{ \frac { 1 }{ 4 } }{ { \sqrt { 1-4x } } } dx=\int _{ 0 }^{ \frac { 1 }{ 4 } }{ { \left( 1-4x \right) }^{ \frac { 1 }{ 2 } } } dx\)
\(={ { { \left[ \frac { { \left( 1-4x \right) }^{ \frac { 1 }{ 2 } +1 } }{ -4\left( \frac { 1 }{ 2 } +1 \right) } \right] }_{ 0 } } }^{ \frac { 1 }{ 4 } }={ { \left[ \frac { { \left( 1-4x \right) }^{ \frac { 3 }{ 2 } } }{ -4\left( \frac { 3 }{ 2 } \right) } \right] }_{ 0 } }^{ \frac { 1 }{ 4 } }\)
\(={ { \left[ \frac { { \left( 1-4x \right) }^{ \frac { 3 }{ 2 } } }{ -6 } \right] }_{ 0 } }^{ \frac { 1 }{ 4 } }\)
\(=\frac { -1 }{ 6 } \left[ { \left( 1-4\left( \frac { 1 }{ 4 } \right) \right) }^{ \frac { 3 }{ 2 } }-{ \left( 1-4(0) \right) }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { -1 }{ 6 } \left[ { 0 }^{ \frac { 3 }{ 2 } }-{ 1 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { -1 }{ 6 } (0-1)=\frac { 1 }{ 6 } \)

Let X be a discrete random variable with probability mass function p(x), then its expected value is defined by
E(X) =\(\sum _{ x }^{ }{ x.p(x) } \)

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)

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

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

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 slope is 1+\(\frac { y }{ x } \)
⇒ \(\frac { dy }{ dx } =1+\frac { y }{ x } \Rightarrow \frac { dy }{ dx } =\frac { x+y }{ x } \)
The numerator and denominator homogeneous functions of degree 1
So put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ x\(\frac { dv }{ dx } \) = 1
Separating the variables we get,
dv = \(\frac { dx }{ x } \)
Integrating, \(\int { dv } =\int { \frac { dv }{ x } } \)
⇒ v = log x + log c
⇒ v = log x c
Replacing V by \(\frac { y }{ x } \) we get,
\(\frac { y }{ x } \) log x c ⇒ y - x log c x...(1)
Since the Curve passes through (1, 0),
0 = 1 logc ⇒ log c =0 ⇒ c = e⁰ =1
∴ c = 1
∴ (1) becomes, y = x logx.

A =\(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -1 \\ -1 & 2 & 7 \end{matrix}\begin{matrix} 4 \\ -3 \\ 6 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho (A)\le \text{minimum} \ of(3,4)\)
\(\rho (A)\le 3\)
Let us transform the matrix A to an echelon form

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

(i) Take a suitable scale for the values of x and y, and plot the various points on the graph paper for given values of x and y.
(ii) Draw a suitable curve passing through the plotted points.
(iii) Find the point corresponding to the value x = 38 on the curve and then read the corresponding value of y on the y- axis, which will be the required interpolated value.
From the graph in Figure we find that for x = 38, the value of y is equal to 35

Here p = q = 1/2
(a) probability of A beating B in exactly 3 games out of 4
\(\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 4-3 }\)
= 1/4 = 25%
(b) probability of A beating B in exactly 5 games out of 8
\(\left( \begin{matrix} 8 \\ 5 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 8-5 }\)
\(=\frac { 7 }{ 32 } \) = 21.875%
Clearly, the first event is more probable.

Given \(MR=\frac { 2 }{ x+3 } -\frac { 2x }{ (x+{ 3) }^{ 2 } } +5\)
\(\Rightarrow \int { MR } =\int { \left( \frac { 2 }{ x+3 } -\frac { 2x }{ (x+{ 3) }^{ 2 } } +5 \right) dx } \)
\(\Rightarrow R=2log(x+3)-2\int { \frac { x }{ { (x+3) }^{ 2 } } dx+5x+k } \)
\(\Rightarrow R=2log(x+3)-2\left[ \int { \frac { x+3-3 }{ { (x+3) }^{ 2 } } dx } \right] +5x+k\)
\(\Rightarrow R=21log(x+3)-2\left[ \int { \frac { 1 }{ x+3 } dx-3\int { \frac { dx }{ { (x+3) }^{ 2 } } } } \right] +5x+k\)

\(\Rightarrow R=\frac { -6 }{ x+3 } +5x+k\)
When x = 0, R = 0 ⇒ k = 0
\(0=\frac { -6 }{ 3 } +0+k\Rightarrow k=+2\)
\(\therefore R=\frac { -6 }{ x+3 } +5x-2\)
Demand function \(P=\frac { R }{ x } \)
\(\Rightarrow P=\frac { -6 }{ x(x+3) } +5+\frac { 2 }{ x } \)
\(\Rightarrow P=\frac { -6+2(x+3) }{ x(x+3) } +5\)
\(\Rightarrow P=\frac { -6+2x+6 }{ x(x+3) } +5\)

\(\Rightarrow P=\frac { 2 }{ x+3 } +5\)