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 denote the height of the student
Given μ = 64.5 inches and σ = 3.3 inches
Given that P(-∞ < Z < C) = 0.99
⇒ P( -∞ < Z < 0) + P (0 < Z < C) = 0.99
⇒ 0.5 + P (0 < Z < C) = 0.99
⇒ P(0 < Z < C) = 0.99 - 0.5 = 0.49...(1)
From the standard normal distribution table
P(0 < Z < 2.33) = 0.49...(2)
From (1) & (2), C = 2.33
we know that Z = \(\frac { X-\mu }{ \sigma } \)
⇒ 2.33 = \(\frac { X-64.5 }{ 3.3 } \)
⇒ X = (2.33) (3.3) + 64.5
⇒ X = 72.19 inches
Hence, the height below which 99% of the student lie is 72.19 inches.

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 \(\frac { dc }{ dq } =\frac { { c }^{ 2 }+2cq }{ { q }^{ 2 } } \)
This is a homogeneous equation in e and q of order 2
∴ Put c = vq and \(\frac { dc }{ dq } =v+q\frac { dv }{ dq } \)
∴ v + q\(\frac { dv }{ dq } \) = v²a² + 2vq₀q =\(\frac { { q }^{ 2 }({ v }^{ 2 }+2v) }{ { q }^{ 2 } } \)v²+2vv
q\(\frac { dv }{ dq } \) = v²+2v-v = v²+v
Separating the variables we get,
\(\frac { dv }{ v+v } =\frac { dq }{ q } \)
Integrating \(\int { \frac { dv }{ v(v+1) } } =\int { \frac { dq }{ q } } \)
[ \(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
⇒ 1 = A(v+1)+B
put v = -1
1 = -B ⇒ B = -1
put v = 0
⇒ 1 = A ]
\(\int { \left( \frac { 1 }{ v } -\frac { 1 }{ v+1 } \right) dv } =\int { \frac { dq }{ q } } \)
⇒ log v -log (v + 1) = log q + log k
⇒ log\(\frac { v }{ v+1 } \) = logq.k
⇒ \(\frac { v }{ v+1 } \) = q.k
Replacing v by \(\frac { c }{ q } \), we get
\(\frac { c/q }{ c/q+1 } \) = q.k
⇒ \(\frac { c }{ c+q } \) = kq
⇒ c = kq(c+q)  .....(1)
Given when c = 1 and q = 1
⇒ 1 = k(1) (1+1) ⇒ 1 = 2 k ⇒ k = \(\frac { 1 }{ 2 } \)
∴ (1) ⇒ c = \(\frac{q}{2}\)(c+q)
∴ ⇒ 2c = q(c+q)

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)
To find y at x = 1946
\(\therefore\) x₀ + nh = 1946, x₀ = 1941, h = 10
1941 + n(10) = 1946 \(\Rightarrow\) n = 0.5

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\(​​{ y }_{ \left( x=1946 \right) }=20+\frac { 0.5 }{ 1! } (4)+\frac { 0.5(0.5-1) }{ 2! } (1)+\frac { 0.5(0.5-1)(0.5-2) }{ 3! } (1)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3) }{ 4! } (0)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3)(0.5-4) }{ 5! } (-9)\)
= 20+2-0.125+0.0625-0.24609
= 21.69 lakhs

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 price index number
\(P^{L}_{01}\) = \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{0}}} \times 100\) 
= \(\frac {1150}{425} \times {100}\) = 270.58
Paasche's index number
\(P^{L}_{01}\) =  \({\frac {\sum p_{1}q_{0}}{\sum p_{0}q_{1}}} \times 100\)
=  \(\frac {1450}{550} \times {100}\) = 263.63

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 5460 }{ 10 } =546\)
The control limits for \(\overset{-}{X}\) chart is
\(UCL=\overset { = }{ X } +{ A }_{ 2 }\bar { R } =546+0.483(84)=586.57\)
\(CL=\overset { = }{ X } =546\)
\(LCL=\overset { = }{ X } -{ A }_{ 2 }\bar { R } =546-0.483(84)=505.43\)
\(\bar { R } =\frac { \sum { R } }{ n } =\frac { 840 }{ 10 } =84\)
The control limits for Range chart is
\(UCL={ D }_{ 4 }\bar { R } =2.004(84)=168.336\)
\(CL=\bar { R } =84\)
\(LCL={ D }_{ 3 }\bar { R } =0(84)=0\)

Sample size n = 50,
Sample mean \(\bar { X } \) = 1850
Population mean μ = 1800
Population standard deviation σ = 100
Null Hypotheses H₀:
μ = 1800 (i.e., To claim that the breaking strength of the cables have increased)
Alternative Hypotheses H₁:
μ≠1800(To claim that the breaking strength of the cables have not increased)
The level of significance ∝ = 1% =.001
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 1850-1800 }{ \frac { 100 }{ \sqrt { 50 } } } =\frac { 50 }{ \frac { 100 }{ 7.07 } } =\frac { 50 }{ 14.144 } =3.535\)
\(\Rightarrow \therefore Z=3.535\)
The Significant value \({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Here \(Z<{ Z }_{ \frac { \alpha }{ 2 } }i.e.,3.535<2.58\)
Inference: Since \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level of significance, the null hypothesis Ho is rejected.
Hence, we conclude that μ≠ 1800 and we cannot support the claim that the breaking strength of the cables have increased.

Let I = \(\int _{ 0 }^{ 1 }{ \frac { x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
\(=-\int _{ 0 }^{ 1 }{ \frac { -x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
[Multiply and divide by -1]
\(=-\int _{ 0 }^{ 1 }{ \frac { 1-x-1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
[Adding and subtracting 1 in the numberator]
\(=\int _{ 1 }^{ 0 }{ \frac { 1-x-1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } } dx\)
\(\left[ \because \int _{ a }^{ b }{ f\left( x \right) } dx=-\int _{ b }^{ a }{ f\left( x \right) dx } \right] \)
\(=\int _{ 1 }^{ 0 }{ \left( \frac { 1-x }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } -\frac { 1 }{ { \left( 1-x \right) }^{ \frac { 3 }{ 4 } } } \right) } dx\)
\(=\int _{ 1 }^{ 0 }{ \left( { \left( 1-x \right) }^{ 1-\frac { 3 }{ 4 } }-{ \left( 1-x \right) }^{ -\frac { 3 }{ 4 } } \right) } dx\)
\(=\int _{ 1 }^{ 0 }{ { \left( 1-x \right) }^{ \frac { 1 }{ 4 } }dx-\int _{ 1 }^{ 0 }{ { \left( 1-x \right) }^{ -\frac { 3 }{ 4 } } } } dx\)
\(={ \left[ \frac { { \left( 1-x \right) }^{ \frac { 1 }{ 4 } +1 } }{ -1\left( \frac { 1 }{ 4 } +1 \right) } -\frac { { \left( 1-x \right) }^{ -\frac { 3 }{ 4 } +1 } }{ -1\left( -\frac { 3 }{ 4 } +1 \right) } \right] }_{ 1 }^{ 0 }\)
\(={ \left[ -\frac { { { \left( 1-x \right) }^{ \frac { 5 }{ 4 } } } }{ \frac { 5 }{ 4 } } +\frac { { \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ \frac { 1 }{ 4 } } \right] }_{ 1 }^{ 0 }\)
\(={ \left[ -\frac { 4 }{ 5 } { \left( 1-x \right) }^{ \frac { 5 }{ 4 } }+4{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } \right] }_{ 1 }^{ 0 }\)
\(=-\frac { 4 }{ 5 } \left( { 1 }^{ \frac { 5 }{ 4 } } \right) +4\left( { 1 }^{ \frac { 1 }{ 4 } } \right) -0\)
\(=-\frac { 4 }{ 5 } (1)+4(1)=-\frac { 4 }{ 5 } +4\)
\(=\frac { -4+20 }{ 5 } =\frac { 16 }{ 5 } \)

Let p be the probability of having a laptop
Given p = \(\frac { 40 }{ 100 } \) = 0.4 ⇒ q = 1-p = 1-0.4 = 0.6
n = 15
(i) P(X = 3)
=¹⁵C₃ (0.4)³ (0.6)¹²  ∵ p(x) =ⁿC_x p^xq^n-x, n = 15, x = 3

= 455 (0.064) (0.002176) = 0.0634
(ii) P (12 travellers not having laptop)
= P (3 travellers having laptop)
[∵ n = 15 and 15 - 12 = 3]
= P(X = 3)
= 0.0634.
(iii) P (atleast 3 travellers have laptop)
= P (X ≥ 3) = 1 - P(X < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
=1-[¹⁵C₀ (0.4)⁰ (0.6)¹⁵ + ¹⁵C₁ (0.4)² (0.6)¹³ (0.6)¹⁴]

= 1 - (0.6)¹³ [(0.6)² + 6(0.6)¹ + 105(0.16)]
= 1 - (0.6)¹³ [0.36 + 3.6 + 16.8]
= 1 - (0.0013) (20.76) = 1 - 0.0270
= 0.9730.

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

From the values of p(x) given in the probability distribution, we obtain
F(1) = P(x\(\le\)1) = P(1) = 0.10
F(2) = P(x\(\le\)2) = P(1) + P(2)
= 0.10+0.12 = 0.22
F(3) = P(x\(\le\)3) = P(1)+P(2)+P(3)
= F(2)+P(3)
= 0.22+0.20
= 0.42
F(4) = F(3)+P(4)
= 0.42+0.30
= 0.72
F(5) = F(4)+P(5)
= 0.72+0.15
= 0.87
F(6) = F(5)+P(6)
= 0.87+0.08
= 0.95
F(7) = F(6)+P(7)
= 0.95+0.05
= 1.00


\(F(x) \text { is } F_{x}(x)= \begin{cases}0, & \text { if } x<1 \\ 0.10, & \text { if } x \leq 1 \\ 0.22, & \text { if } x \leq 2 \\ 0.42, & \text { if } x \leq 3 \\ 0.72, & \text { if } x \leq 4 \\ 0.87, & \text { if } x \leq 5 \\ 0.95, & \text { if } x \leq 6 \\ 1, & \text { if } x \leq 7\end{cases}\)

Given non-homogeneous equations are
5x+ 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 4 \\ 9 \\ 5 \end{matrix} \right) \)

\(\therefore\) The system is consistent with infinitely many solutions let us rewrite the above echelon form into matrix form
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \\ 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ -1 \\ 0 \end{matrix} \right) \)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
let z = k where k\(\in\) R
\((2)\Rightarrow \cfrac { -11 }{ 3 } y+\cfrac { k }{ 3 } =-1\)

\(\Rightarrow \ -11y=-3-k\) 
11y = 3 + k
\(\Rightarrow \quad y=\cfrac { 1 }{ 11 } \left( 3+k \right) \)
Substituting \(y=\cfrac { 1 }{ 11 } \left( 3+k \right) \) and z = k in (1) we get,
\(x+\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) +\cfrac { 2 }{ 3 } k=3\)
\(=\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k }{ 33 } -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 21-48k }{ 33 } =\cfrac { 3(7-16k) }{ 33 } \)
= \(\cfrac { 1 }{ 11 } (7-6k)\)
 \(\therefore\) Solution set is \(\left\{ \cfrac { 1 }{ 11 } \left( 7-16k \right), \cfrac { 1 }{ 11 } (3+k),k \right\} \)K \(\in\) R
Hence, for different values of k, we get infinitely many solutions.

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).

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)

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

\(\int { \frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 }-{ x }^{ 2 } }{ { x }^{ 2 }-x+1 } } dx=\int { \frac { { \left( { x }^{ 2 }+1 \right) }^{ 2 }-{ x }^{ 2 } }{ { x }^{ 2 }-x+1 } } \)

[∵ a² - b² = (a+b) (a-b)]
= ∫ (x² + 1 + x)dx
= \(\frac { { x }^{ 3 } }{ 3 } +x+\frac { { x }^{ 2 } }{ 2 } +c\)

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 values are

The difference table is

x₀ + nh = x ⇒ 1 + n = x ⇒ n = x - 1
∴ By Newton's forward interpolation formula,
y= y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)

[∵ Δ³y₀ = 0]
⇒ y = -1 + 0 + x² - 3x + 2
⇒ y = x² - 3x + 1
Hence, the required second degree polynomial is  y = x² - 3x + 1

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.

\(MC=125+10x-\frac { { x }^{ 2 } }{ 9 } \)
\(C=\int { MC } dx+k\)
\(=\int { \left( 125+10x-\frac { x^{ 2 } }{ 9 } \right) } dx+k\)
= 125 + 5x²- \(\frac { { x }^{ 3 } }{ 27 } \) + k
Fixed cost k = 250
C = 125x + 5x²- \(\frac { { x }^{ 3 } }{ 27 } \) + 250
When x = 15
C = 125(15) + 5(15)²- \(\frac { { (15) }^{ 3 } }{ 27 } \) + 250
= 1875 +1125 −125 + 250
C = Rs. 3,125

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. 

\(\frac { { d }^{ 2 }y }{ { dx }^{ 3 } } -3{ \left( \frac { dy }{ dx } \right) }^{ 6 }+2y={ x }^{ 2 }\)
∴ order = 3, 
∴ Degree = 1

\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
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 n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

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 MR = 9 - 6x² + 2x
⇒ഽMR=ഽ(9 - 6x² + 2x)sx
\(\Rightarrow R=9x-\frac { { 6x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } +k\)
⇒ R = 9x - 2x³ + x² + k
When x = 0, R = 0 ⇒ k = 0
∴ R = 9x - 2x³ + x²

Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
 We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)

If the total supply = total demand, then the given problem is a balanced transportation problem.

A region corresponding to a test statistic in the sample space which tends to rejection of Ho is called critical region.

Gamma integral \(\Gamma \) (n+1) = n! where n is a positive integer.
∴ \(\Gamma \) (4) = (4-1)!
= 3! = 3\(\times\)2 = 6

Given E(X) = 0
then Var (X) = E(X²) - [E(X)]²
= E(X²)-0² [∵ E(X)]²  [∵E(X) = 0]
var (X) = E(X²)
Thus, Var(X) = E(X²) hence proved.