Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(0≤X≤10)=P(X=0)+P(X=10)
\(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(|X|≤2)=P(-2 =P(X=-2)+(P(X=0)
\(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X<0)=P(X=-2)
= 1/4

Given p.d.f is = \(f(x)=\left\{\begin{array}{l} a+b x^{2}, 0 \leq x \leq 1 \\ 0, \text { otherwise } \end{array}\right.\)
Since f(x) is a p.d.f.\(\\ \int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (a+{ bx }^{ 2 })dx=1 } \)
\(\Rightarrow { \left[ ax+\frac { { bx }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }=1\Rightarrow a+\frac { b }{ 3 } =1\)
3a+b = 3  [multiplied by 3] ...(1)
Also it is given that E(X) = \(\frac{3}{5}\)
\(\Rightarrow \int _{ 0 }^{ 1 }{ x.f(x)dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (a+{ bx }^{ 2 })dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow \int _{ 0 }^{ 1 }{ (ax+{ bx }^{ 3 })dx=\frac { 3 }{ 5 } } \)
\(\Rightarrow { \left[ \frac { { ax }^{ 2 } }{ 2 } +\frac { { bx }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 1 }=\frac { 3 }{ 5 } \Rightarrow \frac { a }{ 2 } +\frac { b }{ 4 } =\frac { 3 }{ 5 } \)
\(\Rightarrow 2a+b=\frac { 12 }{ 5 } \)  [Multiplies by 4] ...(2)

\(a=3-\frac { 12 }{ 5 } =\frac { 15-12 }{ 5 } =\frac { 3 }{ 5 } \)
Substituting a=\(\frac{3}{5}\) in(2) we get,
\(2(\frac { 3 }{ 5 } )+b=\frac { 12 }{ 5 } \Rightarrow \frac { 6 }{ 5 } +b=\frac { 12 }{ 5 } \)
\(\Rightarrow b=\frac { 12 }{ 5 } -\frac { 6 }{ 5 } =\frac { 6 }{ 5 } \)
\(\therefore a=\frac { 3 }{ 5 } ,b=\frac { 6 }{ 5 } \)
ii) \(E({ X }^{ 2 })=\int _{ 0 }^{ 1 }{ { x }^{ 2 }f(x)dx=\int _{ 0 }^{ 1 }{ { x }^{ 2 }\left( \frac { 3 }{ 5 } +\frac { 6 }{ 5 } { x }^{ 2 } \right) dx } } \)
\(\left[ \because a=\frac { 3 }{ 5 } ,b=\frac { 6 }{ 5 } \right] \)
\(=\int _{ 0 }^{ 1 }{ \left( \frac { 3 }{ 5 } { x }^{ 2 }+\frac { 6 }{ 5 } { x }^{ 4 } \right) dx } \)

\(=\frac { 1 }{ 5 } (1-0)+\frac { 6 }{ 25 } (1-0)\)
\(=\frac { 1 }{ 5 } +\frac { 6 }{ 25 } =\frac { 5+6 }{ 25 } =\frac { 11 }{ 25 } \)
\(\therefore\) Var (X) = E(X²)-[E(X)]²
\(=\frac { 11 }{ 25 } -{ \left( \frac { 3 }{ 5 } \right) }^{ 2 }\)
\(=\frac { 11 }{ 25 } -\frac { 9 }{ 25 } =\frac { 2 }{ 25 } \)
\(\therefore\) Var (X) =\(\frac{2}{25}\)

Given probability distribution function is 
∴ \(F(x)=\left\{\begin{array}{l} 0, \text { if } x<0 \\ \frac{x}{8}, \text { if } 0 \leq x<1 \\ \frac{1}{4}+\frac{x}{8}, \text { if } 1 \leq x<2 \\ \frac{3}{4}+\frac{x}{12}, \text { if } 2 \leq x<3 \\ 1, \text { for } 3 \leq x \end{array}\right.\)
a) i) P(1≤X≤2)
\(\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ \frac { 1 }{ 8 } } dx{ |\frac { 1 }{ 8 } x| }_{ 1 }^{ 2 } } \)
\(=\frac { 1 }{ 8 } (2-1)=\frac { 1 }{ 8 } (1)\)
ii) \(P(x=3)=0\quad if \ \int _{ 3 }^{ 3 }{ f(x)dx=0 } \)
b) X is not a discrete random variable since E is not a step function

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

P(X≤0)=P(X=-2)+P(X=0)
=\(\frac{1}{4}\)+\(\frac{1}{4}\)=\(\frac{1}{2}\)

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

We know that, the expected random variable 
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\)
\(=\int _{ 100 }^{ \infty }{ { xe }^{ -\frac { x }{ 100 } }dx } \)
\(=\left\{ { \left[ x\left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) \right] }_{ 100 }^{ \infty }-\int _{ 100 }^{ \infty }{ \left( \frac { { e }^{ \frac { x }{ 100 } } }{ -\frac { 1 }{ 100 } } \right) dx } \right\} (\because \int { udv=uv-\int { vdu } } )\)
\(=\left[ \left( 10000 \right) \left( { e }^{ -1 } \right) +\left( 10000 \right) \left( { e }^{ -1 } \right) \right] \)
\(=\left[ \left( 10000 \right) \left( 0.3679 \right) +\left( 10000 \right) (0.3679) \right] \)
= 7358 hours 
Therefore, the mean life of a radio tube is 7,358 hours.

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

Mean of the random variableX = E(X) = \({ \sum_x { x{ P }_{ x }(x) } } \)
= (1 × 0.15) + (2 × 0.10) + (3 × 0.10) + (4 × 0.01) + (5 × 0.08) + (6 × 0.01) +(7 × 0.05) + (8 × 0.02) + (9 × 0.28) + (10 × 0.20)
E(X) = 6.18
E(X²) = \(\sum _{ x }^{ }{ { x }^{ 2 } } { P }_{ X }(x)\)
= (1² × 0.15) + (2² × 0.10) + (3² × 0.10) + (4² × 0.01) + (5² × 0.08) + (6² × 0.01) + (7² × 0.05) + (8² × 0.02) + (9² × 0.28) + (10² × 0.20).
= 50.38
Variance of the Random Variagble X = V(X) = E(X²)-[E(X)]²
=50.38-(6.56)²
= 12.19
Therefore, the mean and variance of the given discrete distribution are 6.18 and 12.19 respectively.

Given probability distribution function
\(F(x)\begin{cases} 0,\quad ​​​​\text{for}\quad x<0 \\ \frac { 1 }{ 2 } ,\quad ​​​​\text{for}\quad 0\le x<1 \\ 0,\quad ​​​​\text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad ​​​​\text{for}\quad 2\le x<4 \\ 0,\quad ​​​​\text{for}\quad x\ge 4 \end{cases}\)
a) The distribution function F(x) is continuous since it is a step function
We know f'(x) = f(x)
∴ Probability density function
\(F(x)\begin{cases} 0,\quad \text{for}\quad x\le 0 \\ \frac { 1 }{ 2 } ,\quad \text{for}\quad 0\le x<1 \\ 0,\quad \text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad \text{for}\quad 2\le x<4 \\ 0,\quad \text{for}\quad x\ge 4 \end{cases}...(1)\)
(b) (i) Probability that a person will have to wait more than 3 minutes is P(X > 3)
∴ P(X>3) = P(3≤X<4)+P(X≥4)
\(\frac{1}{4}+0=\frac{1}{4}\) [from (1)]
ii) Probability that a person will have to wait less than 3 minutes is P(X < 3)
∴ P(X<3) = P(X≤0)+P(0≤x≤1)+P(1≤x≤2)+P(2≤x≤3)
0 + \(\frac{1}{2}+0+\frac{1}{4}\)[from (1)]
∴ P(X<3) = \(\frac{3}{4}\)
= P(1≤X<2)+P(2≤X<3)
= 0 +\(\frac{1}{4}=\frac{1}{4}\) [from (1)]

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)
\(=-\left[ { e }^{ \frac { -10 }{ 5 } }-{ e }^{ \frac { -5 }{ 5 } } \right] \)
\(=-\left[ { e }^{ -2 }-{ e }^{ -1 } \right] \)
\(={ e }^{ -1 }-{ e }^{ -2 }=\frac { 1 }{ e } -\frac { 1 }{ { e }^{ 2 } } \)

We have F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1  f(x) = 0 for x ≤ 1 
Again F(x) = 1 for x > 3 
f(x) = d/dx (1) = 0 for x > 3 
In 1 < x ≤ 3, F(x) = k(x – 1)4 
f(x) = d/dx (k(x – 1)4) = 4k(x – 1)3 
\(\therefore f(x)=4k{ (x-1) }^{ 3 }for\quad 1\le x\le 3\)
i) Since f(x) is a probability density function,
\(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ { 4k(x-1) }^{ 3 }dx=1 } \)

\(\Rightarrow k[{ (3-1) }^{ 4 }-{ (0) }^{ 4 }]=1\)
\(\Rightarrow k({ 2 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
ii) \(\therefore\)p.d.f
\(f(x)=\frac { 4\times 1 }{ 16 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)
\(f(x)=\frac { 1 }{ 4 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)

Given p.d.f is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 16 } { (3+x) }^{ 2 }, & -3\le x\le -1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } (6-2{ x }^{ 2 }), & -1\le x\le 1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 16 } { (3-x) }^{ 2 }, & 1\le x\le 3 \end{matrix} \end{cases}\)
Area under the given curve
\(\int _{ -3 }^{ 3 }{ f(c)dx=\int _{ -3 }^{ -1 }{ \frac { 1 }{ 16 } { (3+x) }^{ 2 }+\int _{ -1 }^{ 1 }{ \frac { 1 }{ 16 } (6-2{ x }^{ 2 })dx } } +\int _{ 1 }^{ 3 }{ \frac { 1 }{ 16 } { (3-x) }^{ 2 }dx } } \)
\(=\frac { 1 }{ 16 } { \left[ \frac { (3+x{ ) }^{ 3 } }{ 3 } \right] }_{ -3 }^{ -1 }+\frac { 1 }{ 16 } { \left( 6x-\frac { { 2x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 1 }+\frac { 1 }{ 16 } { \left( \frac { { (3-x) }^{ 3 } }{ -3 } \right) }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 16 } \left[ \left( \frac { { 2 }^{ 3 } }{ 3 } -0 \right) +\left( 6-\frac { 2 }{ 3 } \right) -\left( -6+\frac { 2 }{ 3 } \right) -\frac { 1 }{ 3 } \left( 0-({ 2 }^{ 3 }) \right) \right] \)
\(=\frac { 1 }{ 6 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } -\left( \frac { -16 }{ 3 } \right) +\frac { 8 }{ 3 } \right] \)
\(\\ =\frac { 1 }{ 16 } \left[ \frac { 8 }{ 3 } +\frac { 16 }{ 3 } +\frac { 16 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \left[ \frac { 8+16+16+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 16 } \times \frac { 48 }{ 3 } =\frac { 1 }{ 16 } \times 16=1\)
Hence, area under the given curve is unity.

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

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
= \(\frac{5}{10}+\frac{1}{100}\)
\(=\frac{51}{100}\)