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.

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

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)
\(=\frac{3}{20}+\frac{4}{20}\) = \(\frac{7}{20}\)

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 \(X=V(X)=E\left( { X }^{ 2 } \right) -{ [E(X)] }^{ 2 }\)
\(=\frac { 91 }{ 6 } -{ \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 35 }{ 12 } \)

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) \(P[X>{ a }_{ 2 }]=0.05\)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { f(x })dx=0.05 } \)
\(\int _{ { a }_{ 2 } }^{ 1 }{ { 5x }^{ 4 }dx=0.05 } \)
\(5{ { \left[ \frac { { x }^{ 5 } }{ 5 } \right] } }_{ { a }_{ 2 } }^{ 1 }=0.05\)
\({ a }_{ 2 }={ [0.95] }^{ \frac { 1 }{ 5 } }\)

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