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.

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
= \(\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } \left( \frac { 1 }{ 2 } \right) ^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 10 }\)
= \(\frac { 6\times 7\times 6 }{ 2^{ 10 } } \)
= \(\frac { 2\times 3\times 7\times 2\times 3 }{ 2^{ 10 } } =\frac { 63 }{ { 2 }^{ 8 } } \)
= \(\frac { 63 }{ 256 } \).

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

When a die is rolled, sample space 
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
∴P(X = 0) = Probability of getting an odd number = \(\frac{3}{6}\)[∵ Their are 3 favourable events]
= \(\frac{1}{2}\)
Thus, the probability distribution of the random variable X is given by

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

Given f'(x) = 8x³ -2x²
∴ ∫ f'(x) dx = ∫ (8x³ - 2x²) dx
⇒ f(x) = \(\frac { { 8x }^{ 4 } }{ 4 } -\frac { { { 2x }^{ 3 } } }{ 3 } +c\)
⇒ f (x) = 2x⁴ - \(\frac { { { 2x }^{ 3 } } }{ 3 } +c\) ...(1)
Also, f(2) = 1
⇒ 1 = 2(2⁴) - \(\frac { { 2\left( { { 2 }^{ 3 } } \right) } }{ 3 } +c\)
⇒ 1 = 32 - \(\frac { { 16 } }{ 3 } +c\)
⇒ 1 - 32 + \(\frac { { 16 } }{ 3 } \) = c ⇒ -31 + \(\frac { { 16 } }{ 3 } \) =c
⇒ \(\frac { { -93+16 } }{ 3 } =c\)
⇒ c = \(\frac { { -77 } }{ 3 } \)
∴ (1) ⟶ f(x) = 2x⁴ - \(\frac { { 2x }^{ 3 } }{ 3 } -\frac { { -77 } }{ 3 } \)

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

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

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right| =12-12=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\therefore \Delta =\Delta x=\Delta y=0\)
\(\therefore \) The system is consistent with infinite number of solutions
let y = k, \(k\epsilon R\)
\(\therefore 2x+3k=4\Rightarrow 2x=4-3k\)
\(\Rightarrow x=\cfrac { 1 }{ 2 } \left( 4-3k \right) ,k\epsilon R\)
\(\therefore \) Solution set is \(\left\{ \cfrac { 4-3k }{ 2 } ,k \right\} ,k\epsilon R\)