\(\bar{\bar{X}}\) = 634.5, \(\bar{R}\) = 226.4
Control limits for \(\overline { X } \) - chart are
UCL = \(\bar{\bar{X}}\) + A₂\(\bar{R}\)
= 634.5+ 0.483x 226.4
= 743.85
CL = 634.5
LCL = \(\bar{\bar{X}}\)- A₂\(\bar{R}\) = 525.15
Control limits of R-chart are
UCL = D₄\(\bar{R}\) = 2.004 x226.4
= 453.7056
CL = 226.4
LCL = D_3\(\bar{R}\) = 0
\(\overline { X } \)-chart

R - chart

Conclusion: Since one point in R-chart lie outside the control limits, the given system is not in control

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 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

Clearly P(X = 0) + P(X = 1) + P(X = 2)= 1
p + P + P(X = 2) =1
2p + P(X = 2) = 1
P(X = 2) = 1 - 2p
so, probability distribution of X is 

∴ E(X) = 0xp+1xp+2(1-2p)
= p+2-4p = 2-3p
and E(X²) = 0xp+1(p)+4(1-2p)
= p+4-8p = 4-7p
Since E(X²) = E(X)we get,
4-7p = 2-3p ⇒ 4-2 = -3p+7p
⇒2 = 4p ⇒ p =\(\frac{2}{4}=\frac{1}{2}\)
∴ p = \(\frac{1}{2}\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin } 3x\ sin\ 2x\ dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ 2sin } 3xsin2x\quad dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left[ cos(3x-2x)-cos(3x+2x) \right] } dx\)
\(\left[ \because 2sin(sinD=cos(C-D))-cos(C+D) \right] \)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (cosx-cos5x)dx } \)
= \(\frac { 1 }{ 2 } { \left[ sinx-\frac { sin5x }{ 5 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
= \(\frac { 1 }{ 2 } \left[ \left( sin\frac { \pi }{ 4 } -\frac { 1 }{ 5 } sin5\frac { \pi }{ 4 } \right) -\left( sin0-\frac { 1 }{ 5 } sin5(0) \right) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ 5 } \left( \frac { -1 }{ \sqrt { 2 } } \right) \right] \quad \quad \left[ \because sin0=0 \right] \)

= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 5\sqrt { 2 } } \right] \)
= \(\frac { 1 }{ 2 } \left( \frac { 5+1 }{ 5\sqrt { 2 } } \right) =\frac { 1 }{ 2 } \times \frac { 6 }{ 5\sqrt { 2 } } =\frac { 3 }{ 5\sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \)
= \(\frac { 3\sqrt { 2 } }{ 10 } \)

Given (6x²+2y²)dx-(x²+4xy)dy = 0
⇒ (6x²+2y²)dx = (x²+4xy)dy
⇒ \(\frac { dy }{ dx } =\frac { 6{ x }^{ 2 }+2{ y }^{ 2 } }{ { x }^{ 2 }+4xy } \)
This is a homogeneous function of degree 2.
∴ Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ v+x\(\frac { dv }{ dx } =\frac { 6+2{ v }^{ 2 } }{ 1+4v } \)
⇒ x\(\frac { dv }{ dx } =\frac { 6+2v^{ 2 } }{ 1+4v } -v=\frac { 6+2v^{ 2 }-v(1+4v) }{ 1+4v } \)
=\(\frac { 6+2{ v }^{ 2 }-v-4{ v }^{ 2 } }{ 1+4v } \) ....(1)
⇒ x\(\frac { dv }{ dx } =\frac { 6-v-2v^{ 2 } }{ 1+4v } \)
Separating the variables we get,
\(\frac { (1+4v)dv }{ 6-v-2{ v }^{ 2 } } =\frac { dx }{ x } \)
[∵ t = 6-v-2v² ⇒ dt = (-1-4v)dv, \(\int { \frac { dt }{ t } } \) = log t]
∴ -\(\int { \frac { (-1-4v)dv }{ 6-v-2v^{ 2 } } } =\int { \frac { dx }{ x } } \)
⇒ -log(6-v-2v²) = log x + log k
⇒ \(\frac { 1 }{ 6-v-2{ v }^{ 2 } } \) = kx
Replacing v by \(\frac{y}{x}\) we get, \(\frac { 1 }{ 6-\frac { y }{ x } -\frac { 2y^{ 2 } }{ { x }^{ 2 } } } \) = kx
⇒ \(\frac { { x }^{ 2 } }{ 6x^{ 2 }-xy-2y^{ 2 } } \) = kx
⇒ x = k(6x²-xy-2y²)
When x = 1, y = 2 ⇒
1 = k(6-2-8) ⇒ 1 = k(-4) ⇒ k = -\(\frac{1}{4}\)
⇒ x = \(\frac{1}{4}\)(6x²-xy-2y²)
⇒ 4x = 2y²+xy-6x²

Given x_o = 0, x₁ = 3, x₂ = 5, x₃ = 6, x₄ = 8
y_o = 276, y₁ = 460, y₂ = 414, y₃ = 343, y₄ = 110
Lagrange's formula is
y = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 })({ x }_{ 0 }-{ x }_{ 4 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 })({ x }_{ 1 }-{ x }_{ 4 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 })({ x }_{ 2 }-{ x }_{ 4 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 })({ x }_{ 3 }-{ x }_{ 4 }) } { y }_{ 3 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 4 }-{ x }_{ 0 })({ x }_{ 4 }-{ x }_{ 2 })({ x }_{ 4 }-{ x }_{ 3 })({ x }_{ 4 }-{ x }_{ 4 }) } { y }_{ 4 }\)
⇒ 276 \(\frac { (1)(-1)(-2)(-4) }{ (-3)(-5)(-6)(-8) } +460\frac { (4)(-1)(-2)(-4) }{ (3)(-2)(-3)(-5) } +414\frac { (4)(1)(-1)(-4) }{ (5)(2)(-1)(-3) } +343\frac { (4)(1)(-1)(-2) }{ (6)(3)(1)(-2) } +110\frac { (4)(1)(-1)(-2) }{ (8)(5)(3)(2) } \)
⇒ y = -3.066 + 163.555 + 441.6 - 152.44 + 3.666
⇒ y = 453.311.

Given' R'(x) = 7 + e^-0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (7+{ e }^{ -0.05x })dx } \)
\(={ \left[ 7x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
\(=700-\frac { 100 }{ 5 } ({ e }^{ -5 }-{ e }^{ 0 })\left[ 0.05=\frac { 5 }{ 100 } \right] \)
= 700-20(0.0067-1)[∵e⁰=1]
= 700-0.134
= 719.866
Since the revenue is given in thousands,
Total revenue = 719.866 x 1000
= Rs. 7,19,866

Let A represents the percent of commuters who use the transit system and B represents the percent of commuters who use their own car. Transition probability matrix

Given 50% of commuters use the transit system and 50% of the commuters use their own car this year.
(i) Percentage of commuters after one year
\(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·9+·5\(\times\).2 ·5\(\times\)·1+·5\(\times\)·8)
= (-45 + ·10 ·05 +.40)
= (-55 - 45)
A = 55% and B = 45%
(ii) Equilibrium will be reached in the long run at equilibrium, we must have
(A B)T = (A B) wher A+B = 1
\(\Rightarrow \left( \begin{matrix} A & B \end{matrix} \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
\(\left( \begin{matrix} \cdot 9A+\cdot 2B & \cdot 1A+8B \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
Equating the corresponding entries on both sides we get,
\(\cdot 9A+\cdot 2B=A\Rightarrow \cdot 9A+\cdot 2(1-A)=A\)
[Since A + B = 1, B = 1 -A]
\(\Rightarrow \cdot 9A+\cdot 2-\cdot 2A=A\)
\(\Rightarrow \cdot 2=A-\cdot 9A+\cdot 2A\)
\(\Rightarrow \cdot 2=A(1-\cdot 9+\cdot 2)\)
\(\Rightarrow \cdot 2=A=(\cdot 3)\)
\(\therefore\) 67% of the commuters will be using the transit system in the long run.