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).

We have P(X=r)∝α^r
⇒P(X = r) = λα^r,r = 0,1,2,....
Since sum of all the probabilities in a probability distribution is 1.
P(X = 0)+P(X = 1)+P(P(X = 2)+... = 1
⇒ λα0+λα1+λα2+...= 1
⇒λ(1+α+α2+.....) = 1
⇒\(\lambda(\frac{1}{1-\alpha})\) = 1
⇒ λ = 1 - α
∴ P(X = r) = (1-α)α^r, r = 0,1,2,...
Hence P(X=0) = (1-α)α⁰ = (1-α)(1) = 1-α.

(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

Let I = \(\int _{ 1 }^{ 2 }{ \frac { log\quad x }{ { x }^{ 2 } } } dx\)

u = log x and dv = \(\frac { 1 }{ { x }^{ 2 } } dx\quad ={ x }^{ -2 }dx\)
\(du=\frac { 1 }{ x } ;v=\frac { { x }^{ -2+1 } }{ { -2+1 } } =\frac { { x }^{ -1 } }{ -1 } =\frac { -1 }{ x } \)
Using integration by parts we get,
ഽu dv = uv - ഽv du
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { log\quad x }{ { x }^{ 2 } } } dx\)
= \({ \left[ -\frac { 1 }{ x } logx-\int { -\frac { 1 }{ x } .\frac { 1 }{ x } } dx \right] }_{ 1 }^{ 2 }\)
= \({ \left[ -\frac { 1 }{ x } logx+\int { \frac { 1 }{ { x }^{ 2 } } } dx \right] }_{ 1 }^{ 2 }\)
= \({ \left[ -\frac { 1 }{ x } logx-\frac { 1 }{ x } \right] }_{ 1 }^{ 2 }\)
= \(-{ \left[ \frac { 1 }{ x } logx+\frac { 1 }{ x } \right] }_{ 1 }^{ 2 }\)
= \(-\left[ \left( \frac { 1 }{ 2 } log2+\frac { 1 }{ 2 } \right) -\left( 1log1+\frac { 1 }{ { 1 }^{ 1 } } \right) \right] \)
= \(-\left[ \frac { 1 }{ 2 } log2+\frac { 1 }{ 2 } -0-1 \right] \)
\(\left[ \because log1=0 \right] \)
= \(-\left[ \frac { 1 }{ 2 } log2-\frac { 1 }{ 2 } \right] \)
= \(\frac { 1 }{ 2 } -\frac { 1 }{ 2 } log2\)
= \(\frac { 1 }{ 2 } (1-log2)\)

Given (x²-yx²)dy + (y²+xy²)dx = 0
⇒ x²(1-y)dy+y²(1+x)dx = 0
⇒ x²(1-y)dy = -y²(1+x)dx
Separating the variables we get,
\(\frac { (1-y) }{ y^{ 2 } } dy=-\frac { (1+x) }{ x^{ 2 } } \)dx
⇒ \(\frac { 1 }{ { y }^{ 2 } } dy-\frac { 1 }{ y } dy=-\frac { 1 }{ x^{ 2 } } dx-\frac { 1 }{ x } dx\)
Integrating, \(\int { { y }^{ -2 } } dy-\int { \frac { 1 }{ y } } dy=-\int { \frac { 1 }{ x^{ 2 } } } dx-\int { \frac { 1 }{ x } } \)
\(-\frac { 1 }{ y } -logy=\frac { 1 }{ x } \) -log x + C
⇒ log x - log y = \(\frac { 1 }{ x } +\frac { 1 }{ y } \)+C
⇒ log \(log\left( \frac { x }{ y } \right) =\frac { x+y }{ xy } \)+C
⇒ \(\frac { x }{ y } =e^{ \frac { x+y }{ xy } +C }\)
⇒ \(\frac { x }{ y } =K.e^{ \frac { x+y }{ xy } }\)  [where e^C = K]

Given h = 1
LHS = (E⁻¹Δ) x³
= Δ(E^-1(x³))
= Δ(x - h)³  [∵ E^-1f(x) = f(x - nh)]
= Δ(1 - h)³ [∵ h = 1]
= (x - 1+ 1)³ - (x - 1)³ [∵ Δf(x) =f(x + h) - f(x)]
= x³ - (x - 1)³
= x³ - (x³ - 3x² + 3x - 1)
[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= x³ - x³ + 3x² - 3x + 1
= 3x² - 3x + 1
= RHS
Hence proved

Tendency movements are due to nature, which repeat themselves periodically in every seasons. These variations repeat themselves in less than one year time. It is measured in an interval of time.
Seasonal variations may be influenced by natural force, social customs and traditions.

Given that \(\int _{ a }^{ b }{ dx } =1\)
\({ \left[ x \right] }_{ a }^{ b }\) = 1
b − a = 1 … (1)
Now, \(\int _{ a }^{ b }{ xdx } =1\)
\({ \left[ \frac { { x }^{ 2 } }{ a } \right] }_{ a }^{ b }\) = 1
b² − a² = 2
(b + a)(b − a) = 2
b + a = 2 … (2) [∵ b -a =1]
(1) + (2) ⇒ 2b = 3
ஃ \(b=\frac { 3 }{ 2 } \)
Now, \(\frac { 3 }{ 2 } \) - a = 1 [∵ from (1)]
 ஃ a = \(\frac { 1 }{ 2 } \)

We are given mean μ = 2.25 and standard deviation σ = 0.25.
Probability that weight of fish is less than 2 kg is P(X < 2.0)
When x = 20 \(Z=\frac { X-\mu }{ \sigma } =\frac { 2.0-2.25 }{ 0.25 } =P(Z<-1.0)=P(Z>1.0)\)
= 0.5 – 0.3413 = 0.1587
Therefore 15.87% of fishes weigh less than 2 kg.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 3 & -2 \\ 6 & -4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \)
AX = B

\(\therefore \)\(\rho ([A,B])=2\), \(\rho (A)=1\)
\(\rho (A)\neq \rho \left( [A,B] \right) \)
\(\therefore \) The given system is inconsistent and has no solution.

Grand average = \(\frac{67 + 61 + 63.33 + 63.37}{4}\)
= \(\frac{255}{4}=63.75\)
Seasonal index (S.I) = \(\frac{Quarterly average}{Grand average}\times100\)
Hence, S.I for I quarter = \(\frac{67}{63.75}\times100\) = 105.01
S.I for II quarter = \(\frac{61}{63.75}\times100\) = 95.68
S.I for III quarter = \(\frac{63.33}{63.75}\times100\) = 99.35
S.I for IV quarter = \(\frac{63.67}{63.75}\times100\) = 99.87

Let λ be the parameter
Given 3. P(X = 2) = P(X = 4)
⇒ 3. \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \)
\(\frac { 3{ \lambda }^{ 2 } }{ 2 } =\frac { { \lambda }^{ 4 } }{ 4\times 3\times 2 } \)
⇒ 36λ² = λ⁴
λ⁴-36λ² = 0
⇒ λ²(λ²-36) = 0
⇒ λ² = 0 or λ² = 36
⇒ λ = 6 since λ > 0

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

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

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions

(2 - y'')² = y''² + 2y'

⇒ 4-y'' = 2y
The highest derivative is of second order and its poweris 1.
∴ Order is 2 and degree is 1.

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

The two branches of statistical inference are
(i) Estimation and
(ii) Testing of hypotheses.

S = {1, 2, 3, 4, 5, 6}⇒ n(s) = 6
∴ X takes the values 1, 2, 3, 4, 5, 6
P(X = 1) = \(\frac{1}{6}\)
P(X = 2) = \(\frac{1}{6}\)
P(X = 3) = \(\frac{1}{6}\)
P(X = 4) = \(\frac{1}{6}\)
P(X = 5) = \(\frac{1}{6}\)
P(X = 6) = \(\frac{1}{6}\)
[Since in all the cases, only one favourable event and total no of events is 6]
∴ The probability mass function is

∴ Expected value for the random value of an unbiased die
\(E(X)=\sum _{ i=1 }^{ 6 }{ xp(x) } \)
\(=1(\frac { 1 }{ 6 } )+2(\frac { 1 }{ 6 } )+3(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 6 } )+5(\frac { 1 }{ 6 } )+6(\frac { 1 }{ 6 } )\)
\(=\frac { 1+2+3+4+5+6 }{ 6 } =\frac { 21 }{ 6 } =\frac { 7 }{ 2 } \)
\(\\ \therefore E(X)=3.5\)

\( \int \sqrt{2 x+1} \ d x=\int(2 x+1)^{\frac{1}{2}} d x\)
\(=\frac { { \left( 2x+1 \right) }^{ \frac { 3 }{ 2 } } }{ 3 } +c\)

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.

Since the number of rows is less then the number of columns, given assignment problem is unbalanced one.
To balance it, introduce a dummy row with all the entries zero.
The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1 :
Select the smallest element in each row and subtract it will all the elements in its row.

Here each row and column has atleast one zero.
Step 2:
Examine the row with only one zero, mark that zero by \(\Box\) and draw a vertical line.
After examining all the rows, examine the column with single zero, mark that zero by \(\Box\) and draw a horizontal line.
Step 3:
Only two assignment have been made.
The elements not lying on the line are
\(\begin{matrix} 6 & 10 & 14 \\ 5 & 9 & 11 \\ 5 & 5 & 12 \end{matrix}\)
and minimum is 5.
Subtract 5 from all these numbers. Other numbers remains the same.
A new cost matrix will be found .and repeat step 2.
∴ The new cost matrix is

Thus, 3 assignments have been made.
The optical assignment schedule and total cost is

Let I =  ഽ x³ sin (x⁴) dx
Put t = x⁴
⇒ dt = 4x³ dfx
\(\Rightarrow \frac { dt }{ 4 } ={ x }^{ 3 }dx\)
\(\therefore I=\int { sin } t.\frac { dt }{ 4 } =\frac { 1 }{ 4 } sint\quad dt\)
= \(-\frac { 1 }{ 4 } cos\quad t+c\)
= \(-\frac { 1 }{ 4 } cos\left( { x }^{ 4 } \right) +c\)      \(\left[ \because t={ x }^{ 4 } \right] \)

(y-x)\(\frac { dy }{ dx } \) = a²
⇒ \(\frac { dy }{ dx } =\frac { { a }^{ 2 } }{ y-x } \Rightarrow \frac { dx }{ dy } =\frac { y-x }{ { a }^{ 2 } } \)
⇒ \(\frac { dx }{ dy } =\frac { y }{ { a }^{ 2 } } -\frac { x }{ a^{ 2 } } \)
⇒ \(\frac { dx }{ dy } +\frac { x }{ { a }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } \)
This is of the form \(\frac { dx }{ dy } \)+Px = Q
where P = \(\frac { 1 }{ { a }^{ 2 } } \) and Q = \(\frac { 1 }{ { a }^{ 2 } } \)y
∴ \(\int { P } dy=\int { \frac { 1 }{ { a }^{ 2 } } dy } =\frac { 1 }{ { a }^{ 2 } } \)
I.F = \(e^{ \int { pdy } }=e^{ y/{ a }^{ 2 } }\)
∴ The solution is \(\int { x. } e^{ \int { pdy } }=\int { Q.e^{ \int { pdy } } } dy\)       
⇒ x.e^y/a2 =\(\int { \frac { 1 }{ { a }^{ 2 } } y } \) e^y/a2dy+C....(1)
Put \(\frac { 1 }{ { a }^{ 2 } } \)y = t ⇒ dy = a²dt
[∵ u = t; d = e^t]
u¹= 1; v = e^t
v₁ = e^t
\(\int { u } dv\) = uv-u¹v₁]
∴ (1) ⇒ x.e^y/a2 = a²\(\int { te^{ t } } \)dt
= a²[te^t-e^t]+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }\) = a².e^t(t-1)+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }=a^{ 2 }.e^{ \frac { y }{ { a }^{ 2 } } }\left( \frac { y }{ { a }^{ 3 } } -1 \right) \)   [∵ t = \(\frac { y }{ { a }^{ 2 } } \)]

\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2 (-3+2) - 1 (-3 -1) -1 (-2-1)
= 2(-1) -1 (-4) -1 (-3)
= -2 + 4 + 3 = 5.
Since\(\Delta \neq 0\),
we can apply Cramer's rule and the system is consistent with unique solution.
\(x=\left| \begin{matrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & 3 \end{matrix} \right| =3\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -2 \end{matrix} \right| \)
= 3(-3 + 2) -1(-3 -4) -1(-2 -4)
= 3 (-1) -1 (-7) -1 (-6)
= -3 + 7 + 6 = 10.
\(\Delta y=\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| \)
= 2(-3 -4) -3 (-3 -1) -1 (4-1)
= 2 (-7) -3 (-4) -1(3)
= 14 + 12 - 3 = -5
\(\Delta z=\left| \begin{matrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & 1 \\ -2 & 4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2(4 + 2) -1(4 -1) + 3(-2 -1)
= 2(6) -1(3) + 3(-3)
= 12 - 3 - 9
= 0

\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 0 }{ 5 } =0\)
\(\therefore\)Solution set is (2, -1, 0)
 

Given
The difference table is

To findy when x = x ⇒ x₀+ nh = x ⇒ 1 + n (1) = x ⇒ n = x-1
Newton's forward interpolation formula is
y_(x=x) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ .....
y_(x=x) = 1 + (x-1)(-2) + \(\frac { (x-1)(x-2) }{ 2 } (4)+\frac { (x-1)(x-2)(x-3) }{ 6 } (-8)+\frac { (x-1)(x-2)(x-3)(x-4) }{ 24 } (16)\)
⇒ y = 1 - 2x + 2(x² - 3x + 2) - \(\frac{4}{3}\) (x - 1) (x - 2) (x - 3) + \(\frac{2}{3}\) (x - 1) (x - 2) (x - 3) (x -4)
⇒ y = 3 - 2x + 2x² - 6x + 4 - \(\frac{4}{3}\) [(x² - 3x + 2) (x- 3)] + \(\frac{2}{3}\) [(x² - 3x + 2)(x² - 7x + 12)]
⇒ y = 2x² - 8x+ 7- \(\frac{4}{3}\) [x³- 3x²+ 2x- 3x² + 9x- 6] + \(\frac{2}{3}\) [x⁴ - 3x³ + 2x² - 7x³ + 21x² - 14x + 12x² - 36x + 24]
⇒ y = 2x²-8x+7- \(\frac{4}{3}\) x³ + 4x² - \(\frac{8}{3}\) x + 4x² - 12x + 8 + \(\frac { { 12x }^{ 4 } }{ 3 } -\frac { 20 }{ 3 } { x }^{ 3 }+\frac { 70 }{ 3 } { x }^{ 2 }-\frac { 100x }{ 3 } +\frac { 48 }{ 3 } \)
⇒ y = \(\frac{2}{3}\) x⁴ + x³ \(\left( \frac { -4 }{ 3 } \frac { -20 }{ 3 } \right) \) + x²\(\left( 2+4+4+\frac { 70 }{ 3 } \right) \) + x \(\left( -8-\frac { 8 }{ 3 } -12-\frac { 100 }{ 3 } \right) \) + 31
⇒ y =  \(\frac{2}{3}\) x⁴ - 8x³ + \(\frac{100}{3}\) x² - 56x + 31 which is the required polynomial.

Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1: Select a minimum 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 c, d and e have no zeros. Go to step 2.
Step 2 : Select the minimum element in each column and subtract this from all the elements in its column.

Since each row and column contains exactly atleast one zero, assignments can be made.
Step 3: (Assignment)
Examine the rows with exactly one zero. Mark that zero by ׈ and draw a vertical line.

Also examine the columns with exactly one zero. Mark that zero by ׈ and draw a horizontal line.
Here, numbers not lying on the line are and minimum of these number is 15.


Now, subtract 15 from all these numbers and numbers lying on the line remains the same.
Hence, the new cost matrix is as follows.
Again repeat step 3 (Assignment)

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

\(\overline {\overline{X}} = \frac {125.1}{10}\) = 12.51
\(\overline{R} = \frac {3.7}{10}\) = 0.37
Control limits for \(\overline {X}\) - chart
UCL =  \(\overline {\overline{X}} + A_{2} \overline {R}\)
= [when n = 4, A₂ = .729]
= 12.51 + .729 (0.37)
= 12.51 + 0.27 = 12.78
CL = \(\overline {\overline{X}}\) = 12.51
LCL = \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 12.51 - 0.27 = 12.24
Control limits for R-chart
UCL = D₄ \(\overline{R}\)
= 2.282(0.37) = 0.84
 [when n = 4, D₄ = 2.282]
CL = \(\overline{R}\) = 0.37
LCL = D₃ \(\overline{R}\) = 0
[When n = 4, D₃ = 0]
\(\overline {X}\) - chart

\(\overline {R}\) - chart

Conclusion: The above diagram shows all the control lines with the data points plotted.
Since all the points lie within the control limits, we can say that the process is in control.

Let x_i, (i = 1,2,3....n) is a ran dom sample of size n from a normal population with mean μ and variance σ², then the sample mean is distributed
normally with mean μ and variance \(\frac { { \sigma }^{ 2 } }{ n } \)
Thus for large samples, the standard normal variate corresponding to \(\bar { X } \) is:
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
Under the null hypothesis, that the sample has been drawn from a population with mean μ and variance σ², the test statistic is \(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)

Given mean = λ = 1.5
X follows poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) ∴ P(neither car is used)
= P(X = 0) = \(\frac { e^{ -1.5 }.(1.5)^{ 0 } }{ 0! } \) =e^-1.5
= 0.2231    [∵ e^-1.5 =0.2231]
(ii) P (Some demand is refused)
The demand may be either 0 car, 1 car or 2 cars
∴ P (Some demand is refused) = 1 - P(X ≤ 2)
= 1 - [P(X = 0) + P (X = 1) + P(X = 2)]
= 1-\(\left[ \frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { e^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } \right] \)
= 1-e^-λ (1+λ+\(\frac { { \lambda }^{ 2 } }{ 2 } \))
= 1-e^-1.5 (1+1.5+\(\frac { (1.5)^{ 2 } }{ 2 } \))
= 1 - 0.2231 (3.625) = 1 - 0.8087 = 0.1912
∴ Probability of some demand is refused = 0.1912.

Given elasticity of demand = \(\frac{p}{x^3}\)
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =\frac { p }{ { x }^{ 3 } } \)
\(\Rightarrow \frac { { -x }^{ 3 }dx }{ x } =p.\frac { dp }{ p } \)
\(\Rightarrow -{ x }^{ 2 }dx=dp\)
\(\Rightarrow -\int { { x }^{ 2 }dx } =\int { dp } \)
\(\Rightarrow -\frac { { x }^{ 3 } }{ 3 } +k=p\)
When p = 2, x = 3
\(\Rightarrow -\frac { { 3 }^{ 3 } }{ 3 } +k=2\)
k = 2 + 9 ⇒ k = 11
∴ (1) becomes
\(P=-\frac { { x }^{ 3 } }{ 3 } +11\)
= 11−\(\frac { { x }^{ 3 } }{ 3 } \)

Given C'(x) = 5 + 0.13x
R'(x) = 18
Fixed cost is Rs. 120
C'(x) = 5 + 0.13x
⇒ ∫C'(x) = ∫(5+0.13x)dx
⇒ C(x) = 5x \(+\frac { 0.13{ x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
Since fixed cost is Rs.120 ⇒ k₁ = 120
\(\therefore C(x)=5x+\frac { 0.13{ x }^{ 2 } }{ 2 } +120\) ...(1)
Also R'(x) = 18
⇒ ∫R'(x) = ∫18dx
⇒ R(x) = 18x+k₂
When x = 0, R = 0 ⇒ k₂= 0
∴ R(x) = 18x   ...(2)
Profit function ⇒ P(x) = R(x) - C(x)
\(=18x-5x-\frac { 0.13{ x }^{ 2 } }{ 2 } -120\)
[from (1) & (2)]
P(x) = 13x-0.065x²-120

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