Prove that V(X+b) = V(X)
LHS = V(X+b) = E[(X+b)-E(X+b)]²
= E[(X+b)-E(X+b)]²
[∵E(b) = b]
= E[X-E(X)]2
= E[X-\(\bar { X } \)]2 [∵\(\bar { X } \) = E(X)]
= Var(X) = RHS
Hence proved.

Given p.d.f. is f(x) = \(\begin{cases} { 2e }^{ -2x },>0 \\ 0,\quad otherwise \end{cases}\)
Expected life of this piece of equipment is 
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ 0 }^{ \infty }{ x.{ 2e }^{ -2x } } } \)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=2\left[ \frac { 1! }{ { 2 }^{ 2 } } \right] } \)
[\(\because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \) gamma integral here n = 1, a = 2]
\(E(X)=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

 LHS = V[aX] = E[aX - E(aX)]2
= E(aX-a.E(X)]2[∵E(aX) = a.E(X)]
= E[aX-a.E(X)]2[∵E(aX) = a.E(X)]
= E[a{X-E(X)}]²
= a².E[X-E(X)]² = a²E(X-\(\bar { x }\))²
= a².[E(X)²-[E(X)]²]
= a².V(X) [∵V(X) = E(X²)-[E(X)]²]
= RHS Hence proved.

When a con is flipped once, sample space S = {H,T}
When tail occurs, you will get Rs. 2 - Rs. 0.50 = 1.50
When head occurs, you will get 1 - Rs. 0.50 = 0.50
∴ The probability mass function is

[∵ P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)]
∴ Expected Value = Σxp(x)
= 1.50(\(\frac{1}{2}\)) + 0.50(\(\frac{1}{2}\))
= 1

Given E(X) = 0
then Var (X) = E(X²) - [E(X)]²
= E(X²)-0² [∵ E(X)]²  [∵E(X) = 0]
var (X) = E(X²)
Thus, Var(X) = E(X²) hence proved.

Given X is a random variable
∴ Y = 2X + 1 is also a random variable
Given Var (X) = 5
∴ Var (Y) = Var (2X + 1) = 22 Var (X)
[∵ Var (aX + b) = a² Var (X)]
= 4(5)
∴ Var (Y) = 20

If X is a continuous random variable and f(x) is the value of its probability density function at x, then the expected value of X is
\(E(X)=\int _{ -\infty }^{ \infty }{ x.(x)dx } \)

Let X be a discrete random variable with probability mass function p(x), then its expected value is defined by
E(X) =\(\sum _{ x }^{ }{ x.p(x) } \)

Var (X) = E(X²) - [E(X)]² where
\(E({ X }^{ 2 })=\begin{cases} \sum { { x }^{ 2 }p(x)\text {for discrete random variable} } \\ \int _{ -\infty }^{ \infty }{ { x }^{ 2 }P(x)\text{dx for continuous random variable} } \end{cases}\)

Mathematical expectation E(X) is an average of the values, that the random variable takes on, where each value is weighted by the probability that the random variable is equal to that value. Values that are most probable receive more weight. Each value x is multiplied by the approximate probability that X equals the valuex.

(i) E(a) = a where a is a constant
(ii) E(aX) = a. E(X)
(iii) E(aX + b) = a E(X) + b
(iv) If X ≥ 0, then E(X) ≥ 0
(v) V(a) = 0
(vi) V(aX + b) = a² V(X).

Given that in an investment profit is Rs. 5000 with probability of 0.62 or a loss of Rs.8000 with a probability of 0.38.
Hence, the probability mass function is

∴ Expected gain E(X) = 5000(0.62) - 8000 (0.32)
= 3100-3040
= Rs. 60
Hence, the expected gain is = Rs. 60

Given probability density function is
\({ f }_{ x }(x)=\begin{cases} \begin{matrix} 2x, & 0\le x\le 1 \end{matrix} \\ \begin{matrix} 0, & otherwise \end{matrix} \end{cases}\)
\(E(X)=\sum _{ x=3 }^{ 4,5 }{ xp(x) } \)
\(E(X)=\int _{ 0 }^{ 1 }{ x.f(x)dx=\int _{ 0 }^{ 1 }{ x.2(xdx=2\int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } } } \)
\(=2{ \left( \frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 1 }=\frac { 2 }{ 3 } ({ 1 }^{ 3 }-{ 0 }^{ 3 })=\frac { 2 }{ 3 } \)
\(E(X)=\frac { 2 }{ 3 } \)

Given probability mass function is

\(\\ \\ E(X)=\sum _{ i=0 }^{ 3 }{ xp(x) } \)
= 0(0.2) + 1(0.1) + 2(0.4) + 3(0.3)
= 0.1+0.8+0.9
E(X) = 1.8

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
\(E(X)=\sum _{ i=1 }^{ 6 }{ xp(x) } \)
\(=1(\frac { 1 }{ 6 } )+2(\frac { 1 }{ 6 } )+3(\frac { 1 }{ 6 } )+4(\frac { 1 }{ 6 } )+5(\frac { 1 }{ 6 } )+6(\frac { 1 }{ 6 } )\)
\(=\frac { 1+2+3+4+5+6 }{ 6 } =\frac { 21 }{ 6 } =\frac { 7 }{ 2 } \)
\(\\ \therefore E(X)=3.5\)

Expected number of success is
E(X)\(E(X)=\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left( 0\times \frac { 6 }{ 11 } \right) +\left( 1\times \frac { 9 }{ 22 } \right) +\left( 2\times \frac { 1 }{ 22 } \right) \)
\(=\frac { 11 }{ 22 } \)
= 0.5
Therefore, the expected number of success is 0.5. Approximately one success.

Expected number of women in the interview pool is
\(E(X)=\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left[ \left( 0\times \frac { 2 }{ 11 } \right) +\left( 1\times \frac { 5 }{ 11 } \right) +\left( 2\times \frac { 4 }{ 11 } \right) \right] \)
\(=\frac{13}{11}\)

The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x₁,x₂, .... with corresponding probabilities p(x₁)p(x₂).... and the distribution function is defined by
F_x(x) = P(X≤x) for all x∈R
ie.F_x(x) = \(\sum _{ { x }_{ i }\le x }^{ }{ p({ x }_{ i }) } \)
For a continuous random variable with the probability density function f_x(x) then the distribution function F_x(x) is defined by
F_x(x) = P(X≤x)

When we perform any experiment, we expect an outcome. We associate a real numbers with each outcome of an experiment. In other words, we considering a function whose domain is the set of possible outcomes and whose range is subset of the set of real numbers such a function is called random variable.

Continuous random variable :
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable. For eg., height of students in a school.

Discrete random variable :
A variable which can assume finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.

Random variables are classified into two types namely discrete and continuous random variables.These are important for practical applications in the field of Mathematics and Statistics.

A random variable is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

When two coins are tossed,
Sample space S = {HH, HT, TH, TT}
⇒n(S) = 4
Since getting a head is termed as success,
X takes the values 2, 1, 1,0
∴ P(X = 2) = \(\frac{1}{4}\)[∵ only one (HH) favourable event]
P(X = 1) = \(\frac{1}{4}\)+\(\frac{1}{4}\) = \(\frac{1}{2}\)[∵ favourable events are HT, TH]
P(X = 0) = \(\frac{1}{4}\)[∵only one favourable event]
∴ Probability distribution function is

Here each ρ_i > 0 and Σρ_i = \(\frac{1}{4}\)+\(\frac{1}{2}+\frac{1}{4}=1\)

The given probability function is

Since the given function is a probability function, each ρ_i>0 and Σρ_i = 1
⇒ k + 2k + 3k + 4k = 1
⇒10k = 1 ⇒ k = \(\frac{1}{10}\)
⇒ k = 0.1

We know Fx (x) = P(X ≤ x) for all x ∈ R
∴ F(0) = P(X ≤ 0) = P(0) = 0.3
F(1) = P(X ≤ 1) = P(0)+P(l)
= 0.3 + 0.2 = 0.5
F(2) = P(X ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)
= 0.3 + 0.2 + 0.4 + 0.1 = 1
∴ Cumulative distribution function for the given probability distribution is 1

\(F(x)=\int _{ -\infty }^{ x }{ f(t)dt } \)
\(=\int _{ 100 }^{ x }{ \frac { 100 }{ { t }^{ 2 } } dt,\quad x\ge 100 } \)
\(={ \left[ \frac { 100 }{ -t } \right] }_{ 100 }^{ x },\quad x\ge 100\)
\(F(x)=\left[ 1-\frac { 100 }{ x } \right] ,\ge 100\)

Let X be the number of cars

i) P(xi)\(\ge\)0\(\forall \) i and
ii) \(\sum _{ i=1 }^{ \infty }{ P({ x }_{ i })=p(0)+p(1)+p(3)+p(4) } \)
= 0.03+0.32+0.38+0.19+0.08 = 1
Hence p(x_i) is a probability mass function.