Let x be the weight and y be the number of students.
Difference table of cumulative frequencies are given below

Let us calculate the number of students whose weight is below 70. For this we use forward difference 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 = 70
\(\therefore\) x₀+nh = 70, x₀ = 40, h = 20
40+n(20) = 70 \(\Rightarrow\) n = 1.5
\({ y }_{ \left( x=70 \right) }=250+1.5\left( 120 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) }{ 2! } \left( -20 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) }{ 3! } \left( -10 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) \left( -1.5 \right) }{ 4! } \left( 20 \right) \)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59
\(\cong \) 424.
Number of students whose weight is between
60 and 70 = y(70)−y(60) = 424−370 = 54

Here the number of rows and columns are equal.
\(\therefore\) The given assignment problem is balanced.
Now let us find the solution.
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

Column 3 contains 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: (Assignment):
Examine the rows with exactly one zero. Row B contains exactly one zero. Mark that zero by \(\square\) (i.e) Person B is assigned to Job 1. Mark other zeros in its column by ×.

Now, Row C contains exactly one zero. Mark that zero by \(\square\). Mark other zeros in its column by × .

Now, Row D contains exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its column by × .

Row E contains more than one zero, now proceed column wise. In column 1, there is an assignment. Go to column 2. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

There is an assignment in Column 3 and column 4. Go to Column 5. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by × .

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 9

Let P be the principal at time ‘t’
\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P=0.05P\)
⇒ ഽ\(\frac { dP }{ P } \) = ഽ0.05 dt + c
log_e P = 0.05t + c
P = e^0.05te^c
P = c₁e^0.05t   (1)
Given P = 2000 when t = 0
⇒  c₁ = 2000
∴ (1) ⇒ P = 2000e^0.05t
To find t , when P = 4000
(2) ⇒ 4000 = 2,000e^0.05t
2 = e^0.05t
0.05t = log2
t = \(\frac { 0.0931 }{ 0.05 } \) = 14 years (approximately)

Laspeyre’s price index number
\({ P }_{ 01 }^{ L }=\frac { \sum { { p }_{ 1 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \times 100=\frac { 1110 }{ 952 } \times 100=116.60\)
On an average, there is an increase of 16.60 % in the price of the commodities when the year 2000 compared with the year 2010.
Paasche’s price index number
\({ P }_{ 01 }^{ P }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 1 } } } \times 100=\frac { 1460 }{ 1236 } \times 100=118.12\)
On an average, there is an increase of 18.12 % in the price of the commodities when the year 2000 compared with the year 2010.
Fisher’s price index number
\({ P }_{ 01 }^{ F }=\sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \times 100=\sqrt { \frac { 1110\times 1460 }{ 952\times 1236 } } \times 100=117.36\)
On an average, there is an increase of 17.36 % in the price of the commodities when the year 2000 compared with the year 2010.

Here φ is the mean length of the components in the population.
The formula for the confidence interval is
\(\bar{x}-Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\)
\({ Here } \ \sigma=1.6, Z_{\alpha / 2}=1.96, \bar{x}=90 \text { and } \mathrm{n}=64\)
Then \(S.E=\frac { \sigma }{ \sqrt { n } } =\frac { 1.6 }{ \sqrt { 64 } } =0.2\)
Therefore, 90 - (1.96 x 0.2)\(\le φ \le\) 90 + (1.96 x 0.2)
\(\text { i.e. } \ (89.61 \leq \varphi \leq 90.39)\)
This implies that the probability that the true value of the population mean length of the components will fall in this interval (89.61,90.39) at 95% . Hence we concluded that 95% confidence interval ensures acceptance of the component by the consumer.

Consider a 2,000 individuals getting injection of a given serum , n = 2000
Let X be the number of individuals suffering a bad reaction.
Let p be the probability that an individual suffers a bad reaction = 0.001
and q = 1– p = 1– 0.001 = 0.999
Since n is large and p is small, Binomial Distribtuion approximated to poisson distribution
So, λ = np = 2000 × 0.001 = 2
(i) Probability out of 2000, exactly 3 will suffer a bad reaction is
\(P(X=3)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -2 }{ 2 }^{ 3 } }{ 3! } =0.1804\)
(ii) Probability out of 2000, more than 2 individuals will suffer a bad reaction
= P(X > 2)
1-[P(X\(\le\)2)]
= 1 – [P(x = 0) + P(x = 1) + P(x = 2)]
\(=1-\left[ \frac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } +\frac { { e }^{ -2 }{ 2 }^{ 1 } }{ 1! } +\frac { { e }^{ -2 }{ 2 }^{ 2 } }{ 2! } \right] \)
\(=1-{ e }^{ 2 }\left( \frac { { 2 }^{ 0 } }{ 0! } +\frac { { 2 }^{ 1 } }{ 1! } +\frac { { 2 }^{ 2 } }{ 2! } \right) \)
= 0.323

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) \(P[X>{ a }_{ 2 }]=0.05\)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { f(x })dx=0.05 } \)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { 5x }^{ 4 }dx=0.05 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { a }_{ 2 } }^{ 1 }=0.05\)
\({ a }_{ 2 }={ [0.95] }^{ \frac { 1 }{ 5 } }\)

The equation y² = 16x represents a parabola (Open rightward)
Required Area = 2\(\int _{ a }^{ b }{ y } dx\)
\(=2\int _{ 0 }^{ 4 }{ \sqrt { 16x } } \ dx\)
\(=8\int _{ 0 }^{ 4 }{ { x }^{ \frac { 1 }{ 2 } } } dx=8{ \left[ { \frac { { x }^{ { \frac { 3 }{ 2 } } } }{ \frac { 3 }{ 2 } } } \right] }_{ 0 }^{ 4 }=\frac { 16 }{ 3 } \left( { \left( 4 \right) }^{ \frac { 3 }{ 2 } } \right) =\frac { 128 }{ 3 } \) sq.units

\(\int { \frac { { 3x }^{ 2 }-2x56 }{ \left( x-1 \right) -\left( { x }^{ 2 }+5 \right) } } dx\)
⇒ 3x²-2x+5 = A (x²+5) + (Bx+c) (x-1)
Putting x =1,
3 - 2+5 = A (1+ 5)
⇒ 6 = A (6)
⇒ A = 1
Putting x = 0,
5 = 5 A - C
⇒ 5 = 5 - C    [∵ A = 1]
⇒ C = 5 - 5 ⇒ C = 0
Putting x = -1,
3 + 2+ 5 = A (6) + (C-B) (-2)
⇒ 10 = 6A + 2B - 2C
⇒ 10 = 6 + 2B + 0
⇒ 10 - 6 = 2B ⇒ 4 = 2B
⇒ B = 2
\(=\int { \left( \frac { A }{ x-1 } +\frac { Bx+c }{ { x }^{ 2 }+5 } \right) } dx\)
\(=\int { \left( \frac { 1 }{ x-1 } +\frac { 2x+0 }{ { x }^{ 2 }+5 } \right) } dx\)
\(=\int { \frac { 1 }{ x-1 } } dx+\int { \frac { 2x }{ { x }^{ 2 }+5 } } dx\)
\(=\log { \left| x-1 \right| } +\log { \left| { x }^{ 2 }+5 \right| } +c\)
\(\left[ \because \int { \frac { { f }^{ 1 }(x) }{ f(x) } dx=\log { \left| f\left( x \right) \right| +c } } \right] \)
\(=\log { \left| \left( { x }^{ 2 }+5 \right) \left( x-1 \right) \right| } +c\)
[∵ log m + log n = log mn]
\(=\log { \left| { x }^{ 3 }-{ x }^{ 2 }+5x-5 \right| } +c\)
\(=\frac { { 3x }^{ 2 }-2x+5 }{ \left( x-1 \right) \left( { x }^{ 2 }+5 \right) } =\frac { A }{ (x-1) } +\frac { Bx+C }{ \left( { x }^{ 2 }+5 \right) } \)

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
 AX = B

For the equations to be inconsistent
\(\rho ([A,B])\neq \rho (A)\)
It is possible if k − 2 = 0.
\(\therefore \)  k = 2