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).

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

Alternative	States of Nature			Minimum	Maximum
	Good	Fair	Bad
A	100	60	+50	+50	100
B	80	50	+10	10	80
C	40	20	+5	5	40

(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

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

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

Given \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)= 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.

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

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ -2 & 4 & -4 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & -2 & 2 \\ 4 & -3 & 4 \\ -4 & 4 & -4 \end{matrix} \right) \)	\(R_{ 1 }\rightarrow { R }_{ 1 }\left( -1 \right) \)
\(\sim \left( \begin{matrix} 1 & -2 & 2 \\ 0 & 5 & -4 \\ -2 & 4 & -4 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-4{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -2 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 32 }+2R_{ 1 }\)

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

Given values are

x	1	2	3	4
y	-1	-1	1	5

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

\(\int _{ 1 }^{ 4 }{ f(x) } dx=\int _{ 1 }^{ 3 }{ f(x) } dx+\int _{ 3 }^{ 4 }{ f(x) } dx\)
= \(\int _{ 1 }^{ 3 }{ (7x+3) } dx+\int _{ 3 }^{ 4 }{ 8x } dx\)
= \({ \left[ \frac { { 7x }^{ 2 } }{ 2 } +3x \right] }_{ 1 }^{ 3 }{ +\left[ \frac { { 8x }^{ 2 } }{ 2 } \right] }_{ 3 }^{ 4 }\)
= \(\frac { 63 }{ 2 } +9-\frac { 13 }{ 2 } +64-36\)
= 62

In a throw of a pair of dice the doublets are (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)
Probability of getting a doublet p = 6/36 = 1/6
⇒ q = 1 – p = 5/6 and also n = 4 is given
The probability of successes
 \(=\left( \begin{matrix} 4 \\ x \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ x }\left( \frac { 5 }{ 6 } \right) ^{ 4-x }\)
Therefore the probability of 2 successes are
\(P(X=2)\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { (\frac { 1 }{ 6 } ) }^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 4-2 }\)
\(=6\times \frac { 1 }{ 36 } \times \frac { 25 }{ 36 } \)
\(=\frac { 25 }{ 216 } \)

Given marginal revenue
MR = 9-4x²
\(\Rightarrow \frac { dR }{ dx } =9-4{ x }^{ 2 }\) 
\(\Rightarrow \int { dR } =\int { (9-4{ x }^{ 2 })dx } \)
\(\Rightarrow R=9x-\frac { { 4x }^{ 3 } }{ 3 } +k\quad ...(1)\)
When no product is sold, revenue is zero
∴ When x = 0, R = 0
⇒0 = 0 - 0 + k ⇒ k = 0
∴ (1) becomes 
\(R=9x-\frac { 4{ x }^{ 3 } }{ 3 } \)
We know that R = Px where P is the demand function
\(\Rightarrow \mathrm{P} \not x=\not x\left(9-\frac{4 x^{2}}{3}\right)\)
\(\Rightarrow P=9-\frac { 4{ x }^{ 3 } }{ 3 } \)
Hence, the demand function is \(9-\frac { 4{ x }^{ 3 } }{ 3 } \)

Transition probability matrix

Where A represents the percentage of subscribers and B represents the percentage of non - subscribers.
A 40% = ·40
By the given data, 40% received the order of subscription = 60% are non-subscribers.
A 40% = ·40
and B 60% = ·60

((-40)(-45) + (·60)(-30) (-40)(.55) + (-60)(.70))
(-18+·18 ·22+42)
(·36 ·64)
\(\Rightarrow\) 36% of those receiving the current letter can be expected to order a subscription

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 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 the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 :
Select a smallest element in each row and subtract this from all the elements in its row.
∴ The cost matrix of the given assignment problem is

Here column IV has no zero. Go to step 2.
Step 2:
Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3:
Examine the row with exactly one zero. Mark the zero by \(\Box \) and draw a vertical line. After examining all the rows examine the column with one zero. mark the zero by \(\Box\) and draw a horizontal line.

Here only 4 assignments have been made.
The numbers not lying on the line are

and min. of these numbers is 1.
Now subtract 1 from all these numbers and add 1 to the numbers on the intersecting line (ie. 6, 7, 2). Other numbers remains the same.
∴ The new cost matrix is

Now, repeat Step 3.
Thus, all the 5 assignments have been made.
The optimal assignment schedule and total cost is

Person	Job	Cost
P	V	7
Q	I	6
R	III	6
S	II	9
T	IV	10
Total Cost		Rs. 38

Let I = \(\int _{ a }^{ b }{ \frac { f\left( x \right) }{ f\left( x \right) +f(a+b-x) } } dx=\frac { b-a }{ 2 } --(1)\)
By the property, \(\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f(a+b-x) } dx\)
\(I=\int _{ a }^{ b }{ \frac { f(a+b-x) }{ f(a+b-x)+f(a+b-(a+b-x)) } } dx\)
= \(\int _{ a }^{ b }{ \frac { f(a+b-x) }{ (a+b-x)+f(a+b-a-b+x) } } dx\)
= \(\int _{ a }^{ b }{ \frac { f(a+b-x) }{ f(a+b-x)+f(x) } } ---(2)\)
Adding (1) and (2) we get, 2I= \(\int _{ a }^{ b }{ \left( \frac { f(x) }{ f(x)+f(a+b-x) } +\frac { f(a+b-x) }{ f(a+b-x)+f(x) } \right) } \)

= \(\int _{ a }^{ b }{ dx } ={ \left[ x \right] }_{ a }^{ b }=b-a\)
2I = b - a
\(\Rightarrow I=\frac { b-a }{ 2 } \) Hence proved

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

x	y	\(\Delta y\)	\(\Delta ^{ 2 }y\)	\(\Delta ^{ 3 }y\)	\(\Delta ^{ 4 }y\)
1891	98,752
		33,533
1901	1,32,285		2,258
		35,791		–10,435
1911	1,68,076		-8,177		41,376
		27,614
1921	1,95,690			30,941
			22,764
		50,360
1931	2,46,050

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

Sample	1	2	3	4	5	6	7	8	9	10	Total
Mean	383	508	505	582	557	337	514	614	707	753	5460
Range	95	128	100	91	68	65	148	28	37	80	840

\(\overset { = }{ X } =\frac { \sum { \bar { X } } }{ 10 } =\frac { 5460 }{ 10 } =546\)
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
\(UCL={ D }_{ 4 }\bar { R } =2.004(84)=168.336\)
\(CL=\bar { R } =84\)
\(LCL={ D }_{ 3 }\bar { R } =0(84)=0\)

Commodities Price Weights
(V) \(P=\frac { { P }_{ 1 } }{ { P }_{ 0 } } \times 100\) PV 2007
(P₀) 2011
(P₁) A 350 400 40 114.285 4571.4 B 175 250 35 142.857 4999.99 C 100 115 15 115 1725 D 75 105 20 114 2280 E 60 80 25 133.333 3333.33     Total 135   17429 76 Value of Cost of Living Index Number\(\frac { \sum { PV } }{ \sum { V } } =\frac { 17429 76 }{ 135 } =129.1093\)
Hence, the Cost of Living Index Number for a particular class of people for the year 2011 is increased by 129.1093 % as compared to the year 2007.

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.

p = probability that a person will involve in an accident in a year
= 0.1/100 = 1/1000
given n = 10,000
so, \(\lambda\) = np = 10000\((\frac{1}{10000})\) = 10
Probability that not more than 5 will involve in such an accident in a year
P(X \(\le\) 5) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)
\(={ e }^{ -10 }[1+\frac { 10 }{ 1! } +\frac { { 10 }^{ 2 } }{ 2! } +\frac { { 10 }^{ 3 } }{ 3! } +\frac { { 10 }^{ 4 } }{ 4! } +\frac { { 10 }^{ 5 } }{ 5! } ]\)
= 0.06651

Given MC = 8 + 6x
\(C(x)=\int { (8+6x)dx } \) + k₁
= 8x + 3x² + k_{1 ...}(1)
But given when x = 0, C = 0 ⇒ k₁ = 0
ஃ C(x) = 8x + 3x²  ....(2)
Given that MR = 24
R(x) = \(\int { MR } \) dx + k₂
= \(\int { 24 } \) dx + k₂
= \(\int { 24 } \) + k₂
Revenue = 0, when x = 0 ⇒ k₂ = 0
R(x) = 24x ...(3)
Total Profit functions P(x) = R(x) – C(x)
P(x) = 24x − 8x − 3x²
= 16x − 3x²

i) Since P[X\(\le\)a₁] = P[X>a₁] 
\(P[X\le { a }_{ 1 }]=\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ f(x)dx } =\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ { 5x }^{ 4 } } dx=\frac { 1 }{ 2 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { 0 }_{ } }^{ a1 }=1/2\)
\({ a }_{ 1 }={ (0.5) }^{ \frac { 1 }{ 5 } }\)
ii) \(P[X>{ a }_{ 2 }]=0.05\)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { f(x })dx=0.05 } \)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { 5x }^{ 4 }dx=0.05 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { a }_{ 2 } }^{ 1 }=0.05\)
\({ a }_{ 2 }={ [0.95] }^{ \frac { 1 }{ 5 } }\)

Commodity	Quantity	Price(Rs.)		p1q0	p0q0
	2000(q0)	2000 (p0)	2003 (p1)
A	100	8	12	1200	800
B	25	6	7.50	187.50	150
C	10	5	5.25	52.50	50
D	20	48	52	1040.00	960
E	65	15	16.50	1072.50	975
F	30	19	27.00	810	570

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

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

Given n = 10, P(H) = \(\frac { 1 }{ 2 } \) ⇒ p =\(\frac { 1 }{ 2 } \)
q=1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
P(X = x) = nC_x p^xq^n-x
∴ P(X = 5) = 10C₅ p⁵q⁵
= \(\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } \left( \frac { 1 }{ 2 } \right) ^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 10 }\)
= \(\frac { 6\times 7\times 6 }{ 2^{ 10 } } \)
= \(\frac { 2\times 3\times 7\times 2\times 3 }{ 2^{ 10 } } =\frac { 63 }{ { 2 }^{ 8 } } \)
= \(\frac { 63 }{ 256 } \).

By data we have,
x₀ = 0, x₁ = 1, x₂ = 3
y₀ = 5, y₁ = 6, y₂ = 50 and x = 2
Using Lagrange's formula, we get
\(y=\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 0 }-{ x }_{ 1 }) } { y }_{ 2 }\)
= \(5\times \frac { (2-0)(2-3) }{ (0-1)(0-3) } +6\times \frac { (2-0)(2-3) }{ (1-0)(1-3) } +50\times \frac { (2-0)(2-1) }{ (3-0)(3-1) } \)

= \(\frac { 10 }{ 3 } -6+\frac { 50 }{ 3 } =\frac { 60 }{ 3 } -6=20-6=14\)
∴ f(2) = 14

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

Since A and B are non-singular,
|A| \(\neq \) 0, |B|\(\neq \) 0
Consider |AB| |A|·|B|
\(\neq \) 0 since |A|\(\neq \) 0 and |B|\(\neq \) 0.=? |AB| \(\neq \) 0
\(\therefore\) AB is non-singular.

Differentiating w.r.t 'x' we get, 2x + 2y\(\frac { dy }{ dx } \)= 0
Dividing by 2, we get,
x+y\(\frac { dy }{ dx } \) = 0

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

1. Systematic samples are not random samples.
2. If N is not a multiple of n, then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.

\(\int { \frac { 1 }{ 9-{ 16x }^{ 2 } } } dx\)
\(=\frac { 1 }{ 16 } \int { \frac { 1 }{ \frac { 9 }{ 16 } -{ x }^{ 2 } } } dx\)
\(=\frac { 1 }{ 16 } \int { \frac { 1 }{ { \left( \frac { 3 }{ 2 } \right) }^{ 2 }-{ x }^{ -2 } } } \)
\(=\frac { 1 }{ 16 } \left[ \frac { 1 }{ 2\left( \frac { 3 }{ 4 } \right) } \log\left| \frac { \frac { 3 }{ 4 } +x }{ \frac { 3 }{ 4 } -x } \right| \right] +c\)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } \log\left| \frac { x-a }{ x+a } \right| \right] +c\)
\(=\frac { 1 }{ 16 } \times \frac { 2 }{ 3 } \log { \left| \frac { \frac { \left( 3+4x \right) }{ 4 } }{ \frac { \left( 3-4x \right) }{ 4 } } \right| } +c\)
\(=\frac { 1 }{ 24 } \log { \left| \frac { 3+4x }{ 3-4x } \right| } +c\)

Given p.d.f. is f(x) = \(\begin{cases} { 2e }^{ -2x },>0 \\ 0,\quad otherwise \end{cases}\)
Expected life of this piece of equipment is 
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ 0 }^{ \infty }{ x.{ 2e }^{ -2x } } } \)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=2\left[ \frac { 1! }{ { 2 }^{ 2 } } \right] } \)
[\(\because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \) gamma integral here n = 1, a = 2]
\(E(X)=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)