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.

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.

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

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.

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

∴ 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 probability mass function is

\(E(X)=\sum _{ x=3 }^{ 4,5 }{ xp(x) } \)
= 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.

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)

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

Thus, we have

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

These values are summarized in the following probability table.

(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.

From the given data, you first calculate the probability distribution of the random variable. Then using it you calculate mean and variance.
X p(x)
1  F(1) = \(\frac{1}{6}\)
2  F(2)-F(1) = \(\frac{2}{6}\)-\(\frac{1}{6}\) = \(\frac{1}{6}\)
3  F(3)-F(2) = \(\frac{3}{6}\)-\(\frac{2}{6}\) = \(\frac{1}{6}\)
4  F(4)-F(3) = \(\frac{4}{6}\)-\(\frac{3}{6}\) = \(\frac{1}{6}\)
5  F(5)-F(4) = \(\frac{5}{6}\)-\(\frac{4}{6}\) = \(\frac{1}{6}\)
6  F(6)-F(5) = 1-\(\frac{5}{6}\) = \(\frac{1}{6}\)
The probability mass function is

Mean of the random variable X = E(X)\(\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left( 1\times \frac { 1 }{ 6 } \right) +\left( 2\times \frac { 1 }{ 6 } \right) +\left( 3\times \frac { 1 }{ 6 } \right) +\left( 4\times \frac { 1 }{ 6 } \right) +\left( 5\times \frac { 1 }{ 6 } \right) +\left( 6\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } (1+2+3+4+5+6)\)
= 7/2
\(E({ X }^{ 2 })=\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
\(=\left( { 1 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 2 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 3 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 4 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 5 }^{ 2 }\times \frac { 1 }{ 6 } \right) +\left( { 6 }^{ 2 }\times \frac { 1 }{ 6 } \right) \)
\(=\frac { 1 }{ 6 } ({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 6 }^{ 2 })\)
\(=\frac { 91 }{ 6 } \)
Variance of the Random Variable \(X=V(X)=E\left( { X }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 91 }{ 6 } -{ \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 35 }{ 12 } \)

Given \( f(x)\begin{cases} { Ae }^{ -x/5 },\quad \text{for}\quad x\ge 0 \\ 0 \quad ,\quad \text{otherwise }\end{cases}\)
a) since f(x) is a p.d.f.,
\(\int _{ 0 }^{ \infty }{ { Ae }^{ \frac { -x }{ 5 } }dx=1\Rightarrow A.\frac { { \left[ { e }^{ \frac { -x }{ 5 } } \right] }_{ 0 }^{ \infty } }{ \frac { - }{ 5 } } =1 } \)
\(\Rightarrow -5A[{ e }^{ -\infty }-{ e }^{ 0 }]=1\)
\(\Rightarrow -5A[0-1]=1\quad [\because { e }^{ -\infty }=0\quad and\quad { e }^{ 0 }=1]\)
\(\Rightarrow 5A=1\Rightarrow A=\frac { 1 }{ 5 } \)
\(\therefore A=\frac { 1 }{ 5 } \)
b) i) Probability that the person will talk over the phone more than 10 minutes is P(X >10)
∴ P(X>10) =\(\\ \int _{ 10 }^{ \infty }{ \frac { 1 }{ 5 } { e }^{ \frac { -x }{ 5 } }dx } [\because A=\frac { 1 }{ 5 } ]\)
\(={ \frac { 1 }{ 5 } \left[ \frac { { e }^{ \frac { -x }{ 5 } } }{ \frac { -1 }{ 5 } } \right] }_{ 10 }^{ \infty }\)
\(=-\left[ { e }^{ -\infty }-{ e }^{ \frac { -10 }{ 5 } } \right] =-\left[ 0-{ e }^{ -2 } \right] \)
\(={ e }^{ -2 }=\frac { 1 }{ { e }^{ 2 } } \left[ \because { e }^{ -\infty }=0 \right] \)
ii) Probability that the person will take over the phone less that 5 minutes is P(X < 5)
\(\therefore P(X<5)=\int _{ 0 }^{ 5 }{ { e }^{ \frac { -x }{ 5 } }dx } [\because A=\frac { 1 }{ 5 } ]\)
\(=\frac { 1 }{ 5 } { \left[ \frac { { e }^{ \frac { -x }{ 4 } } }{ \frac { -1 }{ 5 } } \right] }_{ 0 }^{ 5 }=-{ e }^{ \frac { -5 }{ 5 } }{ -e }^{ 0 }\)
\(=-\left( { e }^{ -1 }-1 \right) \left[ \because { e }^{ 0 }=1 \right] \)
\(=1-{ e }^{ -1 }-\frac { 1 }{ e } =\frac { e-1 }{ e } \)
\(\therefore P(X<5)=\frac { e-1 }{ e } \)
iii) The probability that the person will take over the phone between 5 and 10 minutes is P(5 < X < 10)
\(=-\left[ { e }^{ \frac { -10 }{ 5 } }-{ e }^{ \frac { -5 }{ 5 } } \right] \)
\(=-\left[ { e }^{ -2 }-{ e }^{ -1 } \right] \)
\(={ e }^{ -1 }-{ e }^{ -2 }=\frac { 1 }{ e } -\frac { 1 }{ { e }^{ 2 } } \)

i) Since P[X\(\le\)a₁] = P[X>a₁] 
\(P[X\le { a }_{ 1 }]=\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ f(x)dx } =\frac { 1 }{ 2 } \)
\(i.e., \int _{ 0 }^{ { a }_{ 1 } }{ { 5x }^{ 4 } } dx=\frac { 1 }{ 2 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { 0 }_{ } }^{ a1 }=1/2\)
\({ a }_{ 1 }={ (0.5) }^{ \frac { 1 }{ 5 } }\)
ii) \(P[X>{ a }_{ 2 }]=0.05\)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { f(x })dx=0.05 } \)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { 5x }^{ 4 }dx=0.05 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { a }_{ 2 } }^{ 1 }=0.05\)
\({ a }_{ 2 }={ [0.95] }^{ \frac { 1 }{ 5 } }\)