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

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).

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

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 = \(\frac{1}{18}\)

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
\(P\left[ x\le \frac { 1 }{ 2 } \right] =\int _{ -\infty }^{ \frac { 1 }{ 2 } }{ f(x)dx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ axdx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 2xdx } \)
\(=\frac { 1 }{ 4 } \)

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

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

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

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 = \(\frac{1}{4}\)+\(\frac{1}{2}+\frac{1}{4}=1\)

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

We know that,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \)
\(k\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \) 
\(2k\int _{ -\infty }^{ \infty }{ { e }^{ -|x| }dx=1 } \)   (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an function)
\(2k\int _{ 0 }^{ \infty }{ { \left[ \frac { { e }^{ -x } }{ -1 } \right] }_{ 0 }^{ \infty } } =1\)
\(k=\frac { 1 }{ 2 } \)
Mean of the random variable is
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(E(X)=\int _{ -\infty }^{ \infty }{ xk{ e }^{ -|x| }dx } \)    (\(\because { xe }^{ -|x| }\) is an odd function of x)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { xe }^{ -|x| } } \)
= 0
\(E\left( { x }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } f(x)dx\)
\(=\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { ke }^{ -|x| }dx\)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { e }^{ -|x| }dx\)
\(=\int _{ 0 }^{ \infty }{ { x }^{ 2 } } { e }^{ -x }\)  (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an even function)
\(=\Gamma 3\left( \because \Gamma \left( \alpha \right) =\int _{ 0 }^{ \infty }{ { x }^{ \alpha -1 } } { e }^{ -x }dx,\alpha >0;\Gamma n=(n-1)! \right) \)
= 2
\(V(X)=E\left( { x }^{ 2 } \right) -{ \left[ E(X) \right] }^{ 2 }\)
\(=2-{ \left[ 0 \right] }^{ 2 }\)
= 2

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 probability function is

i) Since the given function is a probability function, each ρ_i>0 and Σρ_i=1
⇒ 0+k+2k+2k+3k+k²+2k²+7k²+k = 1
⇒ 10k²+9k = 1
⇒10k²+9k-1 = 0
on factoring we get
(k+1)(10k-1) = 0
⇒k = -1 or k = \(\frac{1}{10}\)
k = -1 is not possible [each ρ_i>0]
we get k = \(\frac{1}{10}\)
ii) P(X<6) = P(X = 0)+P(X = 1)+P(X=2+P(X=3)+P(X=4)+P(X=5)
P(X<6) = 0 + k + 2k + 2k + 3k + k²
= 8k+k²
= \(8\left( \frac { 1 }{ 10 } \right) { \left( \frac { 1 }{ 10 } \right) }^{ 2 }\)
\(=\frac { 8 }{ 10 } +\frac { 1 }{ 100 } =\frac { 80+1 }{ 100 } =\frac { 81 }{ 100 } \)
\(\therefore P(X<6)=\frac { 81 }{ 100 } \)
Now P(X≥6) = P(X=6)+P(X=7)
= 2k²+7k²+k
= 9k²+k
\(=9{ \left( \frac { 1 }{ 10 } \right) }^{ 2 }+\frac { 1 }{ 10 } \)
\(=\frac { 9 }{ 100 } +\frac { 1 }{ 10 } =\frac { 9+10 }{ 100 } =\frac { 19 }{ 100 } \)
\(\therefore P(X\ge 6)=\frac { 19 }{ 100 } \)
And P(0X<5) = P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= k+2k+2k+3k
\(8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } \)
\( \therefore P(0\))
iii) Given P(X ≤ x) ≥ \(\frac{1}{2}\)
⇒P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)
= 0+k+2k+2k+3k
\(=8k=8\left( \frac { 1 }{ 10 } \right) =\frac { 8 }{ 10 } =\frac { 4 }{ 5 } >\frac { 1 }{ 2 } \)
\(\therefore P(X\le 4)\frac { 1 }{ 2 } \le x=4\)
∴ The minimum value of x is 4.

(a) We know that
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\)
\(\int _{ 0 }^{ 10 }{ Axdx } +\int _{ 10 }^{ 20 }{ A(20-x)dx=1 } \)
\(A\left\{ { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }+{ \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 } \right\} =1\)
A[(50-0)+(400-200)-(200-50)] = 1
\(A=\frac{1}{100}\)
(b) (i) The probability that the number of pounds of bread that will be sold tomorrow is more than 10 pounds is given by
\(P(10\le X\le 20)=\int _{ 10 }^{ 20 }{ \frac { 1 }{ 100 } (20-x) } dx\)
\(=\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 }\)
\(=\frac { 1 }{ 100 } [(400-200)-(200-50)]\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is less than 10 pounds, is given by
\(P(0\le X\le 20)=\int _{ 0 }^{ 10 }{ \frac { 1 }{ 100 } } xdx\)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }\)
\(=\frac { 1 }{ 100 } (50-0)\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is between 5 and 15 pounds is
\(P(5\le X \le15)=\int _{ 5 }^{ 10 }{ \frac { 1 }{ 100 } xdx } +\int _{ 10 }^{ 15 }{ \frac { 1 }{ 100 } (20-x)dx } \)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 5 }^{ 10 }+\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 15 }\)
= 0.75