\(\bar{\bar{X}}\) = \(\frac{11.2 + 11.8 + 10.8 + 11.6 + 11.0+ 9.6 + 10.4 + 9.6 + 10.6 + 10.0}{10}\)
\(=\frac{106.6}{10}=10.66\)
\(\bar{R}=\frac{7+4+8+5+7+4+8+4+7+9}{10}\)
\(=\frac{63}{10}=6.3\)
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 sample size n = 400
Sample mean \(\bar { x } \) = 171.38
Population mean μ = 171.17
Population Standard deviation σ = 3.3
Null hypotheses: H₀ : μ = 171.17
Alternative hypotheses: H₁ : μ ≠ 171.17
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 171.38-171.17 }{ \frac { 3.3 }{ \sqrt { 400 } } } \)
=\(\frac { 0.21 }{ 0.165 } \) = 1.273
As the level of significance is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z < \(Z_{ \frac { \alpha }{ 2 } }\) as 1.273 < 1.96
Inference: since z < \(Z_{ \frac { \alpha }{ 2 } }\), we accept the null hypotheses at 5% level of significance.
Hence, we can conclude that the sample of 400 has taken from the population with mean height of 171.17 cm.

Let p be the probability of winning the prize
Given p = \(\\ \frac { 1 }{ 100 } \) and n = 50
∴ Mean = np = \(\frac { 1 }{ 100 } \times 50=\frac { 1 }{ 2 } \)
∴ λ = 0.5
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P(winning the prize at least once) = P(X ≥ 1)
= 1-P(X < 1)
= 1-P(X = 0)
= 1-\(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \) = 1-e^-0.5
= 1-0.6066
∴ P(X≥1) = 0.3934

Since f(x) is a probability density function,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ \infty }{ { ce }^{ -ax }dx=1 } \Rightarrow C\int _{ 0 }^{ \infty }{ { e }^{ -ax }dx=1 } \)
\(\Rightarrow c{ \left[ \frac { { e }^{ -ax } }{ -a } \right] }_{ 0 }^{ \infty }=1\Rightarrow \frac { -c }{ a } [{ e }^{ -\infty }-{ e }^{ 0 }]\)
\(\Rightarrow \frac { -c }{ a } [0-1]=1\quad [\because { e }^{ -\infty }=0,{ e }^{ 0 }=1]\)
\(\Rightarrow \frac { c }{ a } =1\Rightarrow C=a\quad \therefore C=a\)

Max (9, 8, 5) = 9
∴ d₁ is the best strategy using maximin principle.

\(\int _{ 0 }^{ \pi }{ { sin }^{ 2 } } x\ dx\) = \(\int _{ 0 }^{ \pi }{ \frac { sinx-sin3x }{ 4 } } dx\)
[∵ sin 3x = 3sin x - 4 sin³x]
= \(\frac { 1 }{ 4 } \int _{ 0 }^{ \pi }{ \left( 3sinx-sin3x \right) } dx\)
= \(\frac { 1 }{ 4 } { \left[ -3cosx+\frac { 1 }{ 3 } cos3x \right] }_{ 0 }^{ \pi }\)
= \(\frac { 1 }{ 4 } \left[ \left( -3cos\pi +\frac { 1 }{ 3 } cos3\pi \right) -\left( -3cos0+\frac { 1 }{ 3 } cos3(0) \right) \right] \)
= \(\frac { 1 }{ 4 } \left[ \left( 3-\frac { 1 }{ 3 } \right) -\left( -3+\frac { 1 }{ 3 } \right) \right] \)
= [∵ cos π = -1, cos3 π = -1, and cos0 = 1]
= \(\frac { 1 }{ 4 } \left[ \frac { 8 }{ 3 } -\left( \frac { 8 }{ 3 } \right) \right] \)
 

The given equation is of the form \(\frac { dy }{ dx } \)+ py = Q
where P = a and Q = e^x
\(\int { P } dx=\int { a } dx\) = ax
Integrating factor (L.F.) = \(e^{ \int { P } dx }\) = eax
∴ The solution is y \(e^{ \int { P } dx }=\int { Q } e^{ \int { P } dx }dx\)+C
⇒ y.e^ax =\(\int { { e }^{ x }.{ e }^{ ax } } dx+C\)
⇒ y.e^ax = \(\int { e^{ (a+1)x }dx+C } \)
⇒ y.e^ax = \(\frac { { e }^{ (a+1)x } }{ a+1 } \)+C

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 demand function is p = 25 - x - X²
and p₀ = 19
⇒ 19 = 25-x-x²
⇒ x² + x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units

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.