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 
\(f(x)= \begin{cases}k & \text { for } 0< x \leq 4 \\ 0, & \text { otherwise }\end{cases}\)
Since f(x) is a p.d.f.\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ 4 }{ kdx=1\Rightarrow k\int _{ 0 }^{ 4 }{ dx=1\Rightarrow k{ [x] }_{ 0 }^{ 4 }=1 } } \)
\(\Rightarrow k(4-0)=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \)
Also P(2≤x≤4) =\(\int _{ 0 }^{ 4 }{ f(x)dx } \)
\(=2\int _{ 2 }^{ 4 }{ kdx=\int _{ 2 }^{ 4 }{ \frac { 1 }{ 4 } dx=\frac { 1 }{ 4 } \int _{ 2 }^{ 4 }{ dx } } } \)
\(=\frac { 1 }{ 4 } { [x] }_{ 2 }^{ 4 }=\frac { 1 }{ 4 } (4-2)=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\\ \therefore P(2\le X\le 4)=\frac { 1 }{ 2 } \)

When a coin is tossed, sample space S = {H, T}
Since he is receiving Rs. 4 for a head and pays Rs. 2 for a tail,
∴ X take values 4 and -2.
∴ Probability for getting a head is \(\frac{1}{2}\) and probability for getting a tail is \(\frac{1}{2}\).
The probability mass function is

X = x	4	-2
P(X = x)	\(\frac{1}{2}\)	\(\frac{1}{2}\)

∴ Expectation E(X) = Σxp(x)
= 4(\(\frac{1}{2}\)) - 2(\(\frac{1}{2}\))
= 2-1 = 1
∴ His expectation is Rs. 1
E(x²) = ∑x²p(x)
= 4²(\(\frac{1}{2}\)) + (-2)²(\(\frac{1}{2}\))
= 16(\(\frac{1}{2}\)) + 4(\(\frac{1}{2}\))
= 8 + 2 = 10
Variance (X) = E(X²)-[E(X)]²
= 10-1² = 9
∴ Variance of his gains Rs. 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.

Since the profit is Rs. 2000 and the loss is Rs. 1000
X can take values 2000 and -1000.
∴ The probability mass function is

X = x	2000	-1000
P(X = x)	0.4	0.6

∴ Expected profit E(X) = Σxp(x)
= 2000(0.4)-1000(0.6)
= 800-600
E(X) = Rs. 200
E(X²) = Σx²p(x)
= 2000²(0.4)-1000²(0.6)
= 1600000+600000
= 2200000
∴ Var(X) = E(X²)-[E(X)]²
= 2200000-(200)²
= 2200000-40000
= 2160000.
Standard deviation = \(\sqrt{variance}=\sqrt{2160000}\)
= 1469.69.

Given probability density function is
\(f(x)=\left\{\begin{array}{l} \frac{3}{x^{4}}, x \geq 1 \\ 0, \text { otherwise } \end{array}\right.\)
Here E(X) = \(\int _{ 1 }^{ \infty }{ dx\quad x.f(x)dx } \)
\(=\int _{ 1 }^{ \infty }{ x.\frac { 3 }{ { x }^{ 4 }dx } } dx=3\int _{ 1 }^{ \infty }{ \frac { 1 }{ { x }^{ 3 } } dx=3\int _{ 1 }^{ \infty }{ { x }^{ -3 }dx } } \)
\(=3{ \left[ \frac { { x }^{ -3+1 } }{ -3+1 } \right] }_{ 1 }^{ \infty }={ \left( \frac { { x }^{ -2 } }{ -2 } \right) }_{ 1 }^{ \infty }=\frac { -3 }{ 2 } { \left( \frac { 1 }{ { x }^{ 2 } } \right) }_{ 1 }^{ \infty }\)
\(=\frac { -3 }{ 2 } \left( \frac { 1 }{ \infty } -\frac { 1 }{ 1 } \right) =\frac { -3 }{ 2 } (0-1)=\frac { 3 }{ 2 } \)
\([\because \frac { 1 }{ \infty } =0]\) 
\(E(X)=\frac { 3 }{ 2 } \)
\(E({ X }^{ 2 })=\int _{ 1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 } } dx=3\int _{ 1 }^{ \infty }{ { x }^{ -2 }dx } } \)
\(=3{ \left[ \frac { { x }^{ -1 } }{ -1 } \right] }_{ 1 }^{ \infty }=-3{ \left[ \frac { 1 }{ x } \right] }_{ 1 }^{ \infty }=-3\left[ \frac { 1 }{ \infty } -1 \right] =3\)
Var(X) = E(X²)-[E(X)]²
\(\\ =3-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 }=3-\frac { 9 }{ 4 } =\frac { 12-9 }{ 4 } \)
\(Var(X)=\frac { 3 }{ 4 } \)

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

Expected value of the random variable is
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\)
\(=\int _{ 0 }^{ 25 }{ x \frac{1}{25}dx } \)
\(=\frac { 1 }{ 25 } \int _{ 0 }^{ 25 }{ xdx } \)
\(=\frac { 1 }{ 25 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 25 }\)
= 12.5
Therefore, the expected waiting time of the commuter is 12.5 minutes.

We know that,
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -3x } } dx\)
\(=3\int _{ 0 }^{ \infty }{ { xe }^{ -3x }dx } \)
\(=3\left\{ { \left[ \frac { { xe }^{ -3x } }{ -3 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -3x } }{ -3 } dx } \right\} \)( ∵ ∫udv = uv - ∫ vdu)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -3x }dx } \)
\(=\frac{1}{3}\)
Therefore, the expected life of the piece of equipment is \(=\frac{1}{3}\)hrs (in thousands).

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \),since f(x) is a density function
\(\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k{ \left[ \frac { { e }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }=1\)
⇒ k = 2
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -2x } } dx\)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx } \)
\(=2\left\{ { \left[ \frac { { xe }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -2x } }{ -2 } dx } \right\} \)
\((\because \int { udv=uv-\int { udv } } )\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -2x }dx } \) \(=\frac { 1 }{ 2 } \)

We know that,
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ 1 }{ x{ 4x }^{ 3 } } dx\)
\(=4{ \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }\)
\(E(X)=\frac { 4 }{ 5 } \)
\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } f(x)dx\)
\(=\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ 4x }^{ 3 } } dx\)
\(=4{ \left[ \frac { { x }^{ 6 } }{ 6 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
\(V(X)=\left( { x }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 4 }{ 6 } -{ \left[ \frac { 4 }{ 5 } \right] }^{ 2 }\)
\(=\frac { 2 }{ 75 } \)

\(E(X)=\sum _{ x }^{ }{ x } { P }_{ x }(x)\)
= (0 × 0.2) + (1 × 0.1) + (2 × 0.4) + (3 × 0.3)
= 1.8
\(E\left( { x }^{ 2 } \right) =\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ x }(x)\)
= (0² × 0.2) + (1² × 0.1) + (2² × 0.4) + (3² × 0.3)
= 4.4
E(3X + 2X²⁾ = 3E(X) + 2E(X²)
= (3 × 1.8) + (2 × 4.4)
= 14.2

If the random variable X is the top face of a tossed, fair, six sided die, then the probability mass function of X is
P_x (x) = \(\frac{1}{6}\), for x = 1, 2, 3, 4, 5 and 6
The average toss, that is, the expected value of X is
\(E(X)=\sum _{ x }^{ }{ { P }_{ x }(x) } \)
\(E(X)=\left( 1\times \frac { 1 }{ 6 } \right) +\left( 2\times \frac { 1 }{ 6 } \right) +\left( 3\times \frac { 1 }{ 6 } \right) +\left( 4\times \frac { 1 }{ 6 } \right) +\left( 5\times \frac { 1 }{ 6 } \right) +\left( 6\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } (1+2+3+4+5+6)\)
\(=\frac { 7 }{ 2 } \)
= 3.5
Therefore, the expected toss of a fair six sided die is 3.5.

Probability of getting a blue ball = (p) =\(\frac{1}{4}\) = 0.25
Total experiments (N) = 30
Expected value = Number of experiments × Probability
= N x p
= 30 x 0.25
= 7.50
Therefore, the expected value of getting blue ball is approximately 8.

1) 0≤F(x)≤1, -∞
2) F(-∞) = 0 and F(∞) = 1
3) is a non-decreasing function, F(a) ≤F(b) for a
4) \(\underset { h\rightarrow 0 }{ lim } \) F(x+h) = F(x), since F(x) is continuous from the right
5) F'(x) = f(x)≥0
6) P(a≤x≤b) = F(b)-F(a)

For discrete random variable:
p(x_i)≥0∀i and \(\sum _{ i=1 }^{ n }{ p({ x }_{ i }) } =1\) and X takes only
infinite number of values
For Continuous random variable:
f(X)≥0∀x and \(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \) and X takes
infinite number of values in the interval.

(i) Probability mass function:
If X is a discrete random variable with distinet values x₁, x₂, x_n then the function denoted by p_x (x) and defined by
P_X(x) = p(x)
= {p(x = xi) = p_i = p(xi)if x = x_i, i = 1, 2,...n
0 if x ≠ x_i
Here p(x_i)≥0∀i & \(\sum _{ i=1 }^{ n }{ p({ x }_{ i })=1 } \)
(ii) Probability density function:
The probability that a random variable X takes a value in th interval [t₁, t₂] [open or closed] is given by the integral of a function called the probability density function
\(P({ t }_{ 1 }\le X\le { t }_{ 2 })=\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ { f }_{ x }(x)dx } \)
Here f(x) ≥0∀ x and \(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
(iii) Probability distribution function:
Fora discrete distribution, the distribution function of a random variable X is defined as 
FX(x) = P(X ≤ x)for all x ∈ R.

Given probability distribution function is
\(= \begin{cases}kx & x =2, 4, 6 \\ k(x - 2), & x = 8 \\ 0, & \text { otherwise } \\ \end{cases}\)
where k is a constant

X = x	2	4	6	8
P(X=x)	2k	4k	6k	k(8-2) = 6k

Since the given function is a probability distribution function, each ρ_i>0Σ ρ_i=1
⇒ 2k+4k+6k+6k = 1
⇒ 18 k = 1
⇒ k = \(\frac{1}{18}\)

Given probability mass function is

X=x	3	5	8	10
P(X=x)	0.3	0.2	0.3	0.2

∴  The cumulative distribution function F_x(x) is
F_x(0) = 0 if x < 3
F_x(3) = P(X = 3) = 0.3, for 3 ≤x<5
F_x(5) = P(X = 3)+P(X = 5) = 0.3+0.2 = 0.5, for 5≤X<8
F_x(8) = P(X = 3)+P(X+5)+P(X = 8)
= 0.3+0.2+0.3
= 0.8,8≤x<10
F_x(10) = P(X = 3)+P(X = 5)+P(X = 8)+P(X = 10)
= 0.3+0.2+0.3+0.2
= 1 for x ≥10
\(F_{x}(x)= \begin{cases}0, & \text { if } x<3 \\ 0.3, & \text { if } 3 \leq x < 1 \\ 0.5, & \text { if } 5 \leq x < 8 \\ 0.8, & \text { if } 8 \leq x < 10 \\ 1, & \text { if } x ≥ 10 \\ \end{cases}\)

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ 0 }^{ -\infty }{ ax\quad dx } \Rightarrow a\int _{ 0 }^{ 1 }{ xdx=1 } \)
\(\Rightarrow a{ \left( \frac { { x }^{ 2 } }{ 2 } \right) }^{ 1 }=1\)
\(\Rightarrow \frac { a }{ 2 } (1-0)=1\)
\(\Rightarrow\)a = 2
\(P\left[ x\le \frac { 1 }{ 2 } \right] =\int _{ -\infty }^{ \frac { 1 }{ 2 } }{ f(x)dx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ axdx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 2xdx } \)
\(=\frac { 1 }{ 4 } \)

The sample space (S) = { (HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
X takes the values: 3, 2, 2, 1, 2, 1, 1, and 0

Range of X(Rx)	0	1	2	3
Px(x)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)
Fx(x)	\(\frac{1}{8}\)	\(\frac{4}{8}\)	\(\frac{7}{8}\)	1

Thus, we have

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

Let X is the random variable which counts the number of heads on the upturned faces. The outcomes are stated below

Outcomes	(HHH)	(HHT)	(HTH)	(THH)	(THT)	(TTH)	(HTT)	(TTT)
Values of X	3	2	2	2	1	1	1	0

These values are summarized in the following probability table.

Value of X	0	1	2	3	Total
P(xi)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)	\(\sum _{ i=0 }^{ 3 }{ p({ x }_{ i })=1 } \)

(i) p(x_i) \(\ge\)0\(\forall \) i and
(ii) \(\sum _{ i=0 }^{ 3 }{ p({ x }_{ i })=1 } \)
Hence, p(x_i) is a probability mass function.

P(X<3) = P(X = 1)+P(X = 2)
\(=0+\frac{1}{20}+\frac{2}{20}\) = \(\frac{3}{20}\)
P(2\(\le\)4) = P(X = 3)+P(X = 4)
\(=\frac{3}{20}+\frac{4}{20}\) = \(\frac{7}{20}\)