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

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.

(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

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

From the values of p(x) given in the probability distribution, we obtain
F(1) = P(x\(\le\)1) = P(1) = 0.10
F(2) = P(x\(\le\)2) = P(1) + P(2)
= 0.10+0.12 = 0.22
F(3) = P(x\(\le\)3) = P(1)+P(2)+P(3)
= F(2)+P(3)
= 0.22+0.20
= 0.42
F(4) = F(3)+P(4)
= 0.42+0.30
= 0.72
F(5) = F(4)+P(5)
= 0.72+0.15
= 0.87
F(6) = F(5)+P(6)
= 0.87+0.08
= 0.95
F(7) = F(6)+P(7)
= 0.95+0.05
= 1.00


\(F(x) \text { is } F_{x}(x)= \begin{cases}0, & \text { if } x<1 \\ 0.10, & \text { if } x \leq 1 \\ 0.22, & \text { if } x \leq 2 \\ 0.42, & \text { if } x \leq 3 \\ 0.72, & \text { if } x \leq 4 \\ 0.87, & \text { if } x \leq 5 \\ 0.95, & \text { if } x \leq 6 \\ 1, & \text { if } x \leq 7\end{cases}\)

(i) Since the condition of probability mass function is
\(\sum _{ i=1 }^{ \infty }{ p({ x }_{ i }) } =1\)
\(\therefore\) \(\sum _{ i=0 }^{ 7 }{ p({ x }_{ i }) } =1\)
0+a+2a+2a+3a+a²+2a²+7a²+a = 1
10a²+9a–1 = 0
(10a–1)(a+1) = 0
a = \(\frac{1}{10}\)and -1
Since p(x) cannot be negative, a = – 1 is not applicable. Hence, a = \(\frac{1}{10}\)
ii) P(X<3) = P(X = 0)+P(X=1)+P(X = 2)
= 0+a+2a
= 3a
\(\\ =\frac { 3 }{ 10 } \left( \because a=\frac { 1 }{ 10 } \right) \)
(iii) P(X>2) = 1-P(X\(\le\)2)
= 1-[P(X = 0)+P(X=1)+P(X=2)
= 1-\(\frac{3}{10}\)
= \(\frac{7}{10}\)
iv) P(2< x\(\le\)5) = P(X = 3)+P(X = 4)+P(X = 5)
= 2a+3a+a²
= 5a+a²
= \(\frac{5}{10}+\frac{1}{100}\)
\(=\frac{51}{100}\)

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

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

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