Given p.d.f. is \(f(x)=\left\{\begin{array}{l} 3 x^{2}, \text { if } 0< x< 1 \\ 0, \text { otherwise } \end{array}\right.\)
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx } \)
\(=\int _{ 0 }^{ 1 }{ x.(3{ x }^{ 2 })dx } \)
\(=3\int _{ 0 }^{ 1 }{ { x }^{ 3 }dx=3. } { \left[ \frac { { x }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { 3 }{ 4 } (1-0)=\frac { 3 }{ 4 } \)
\(E({ X }^{ 2 })=\int _{ 0 }^{ 1 }{ { x }^{ 2 }.3{ x }^{ 2 }dx=3 } \int _{ 0 }^{ 1 }{ { x }^{ 4 }dx=3.{ \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 } } \)
\(=\frac { 3 }{ 5 } (1-0)=\frac { 3 }{ 5 } \)
\(Var(X)=E({ X }^{ 2 })-{ [E(X)] }^{ 2 }=\frac { 3 }{ 5 } -{ \left( \frac { 3 }{ 4 } \right) }^{ 2 }\)
\(=\frac { 3 }{ 5 } -\frac { 9 }{ 16 } =\frac { 48-45 }{ 80 } =\frac { 3 }{ 80 } \)
\(V(3X-2)={ 3 }^{ 2 }V(X)[\because V(aX+b)={ a }^{ 2 }V(X)]\)
\(\\ =9\times \frac { 3 }{ 80 } =\frac { 27 }{ 80 } \)
\(\therefore V(3X-2)=\frac { 27 }{ 80 } \)

Given probability distribution function is 
X=x 1 3 5 P(X=x) 2k 3k 4k a) Since the given function is a p.d.f. each p_i≥0 and Σp_i=1
⇒2k + 3k+4k = 1 ⇒ 9k = 1 ⇒ k = \(\frac{1}{9}\)
b) P(X>2) = P(X = 3) + P(X = 5)
= 3k + 4k = 7k = 7(\(\frac{1}{9}\))
\(\left[ \because k=\frac { 1 }{ 9 } \right] \)
\(\therefore P(X\ge 2)=\frac { 7 }{ 9 } \)

Given p.d.f is
\(f(x)= \begin{cases}\frac{1}{30} e^{-\frac{x}{30}}, & \text { for } x>0 \\ 0, & \text { for } x \leq 0\end{cases}\)
Expected number of miles
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x.\frac { 1 }{ 30 } { e }^{ \frac { -x }{ 30 } }dx=\int _{ }^{ \infty }{ { xe }^{ \frac { -x }{ 36 } }dx } } \)
\(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -axdx }=\frac { n! }{ { a }^{ n+1 } } Hence\quad n=1,a=\frac { 1 }{ 30 } } \right] \)
\(=\frac { 1 }{ 30 } \left( \frac { 1! }{ { \left( \frac { 1 }{ 30 } \right) }^{ 2 } } \right) \)
\(=\frac { 1 }{ 30 } \times \frac { 1 }{ { \left( \frac { 1 }{ 30 } \right) }^{ 2 } } =\frac { 1 }{ \frac { 1 }{ 30 } } =30\)
∴ E(X) = 30 miles (in thousands) or 30,000 miles.

Given probability mass function is

x	3	4	5
P(x)	0.1	0.1	0.2

\(E(X)=\sum _{ x=3 }^{ 4,5 }{ xp(x) } \)
= 0.6+ 1.2 +2.5
E(X²) = Σx²p(x)
= 9(0.2) + 16(0.3) +25(0.5)
= 1.8 + 4.8 + 12.5
= 19.1
Var(X) = E(X²)-[E(X)]² 
= 19.1-(4.3)²
19.1- 18.49
V(X) = 0.61
Stdard deviation = \(\sqrt{Variance}=\sqrt{0 .61}\)
= 0.78

Sample space \((s)=\left\{ \begin{matrix} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} & \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} & \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ (6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) & (1,5) & (1,6) \\ (2,4) & (2,5) & (2,6) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} & \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} & \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)

Total outcomes : n(S) = 36