E(X) = Σx_i²p_i
\(\Rightarrow E(X)=1(\frac { 1 }{ 2 } )+2\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 3 }{ 25 } \right) +2A\left( \frac { 1 }{ 10 } \right) +3A\left( \frac { 1 }{ 25 } \right) +5A\left( \frac { 1 }{ 25 } \right) \)
\(\Rightarrow \frac { 1 }{ 2 } +\frac { 2 }{ 5 } +\frac { 12 }{ 25 } +\frac { A }{ 5 } +\frac { 3A }{ 25 } +\frac { A }{ 5 } \)
\(=\frac { 69 }{ 50 } +\frac { 13A }{ 25 } \)
Since E(X) = 2.94
\(\frac { 69 }{ 50 } +\frac { 13A }{ 25 } =2.94\Rightarrow \frac { 13A }{ 25 } =2.94-\frac { 69 }{ 50 } \)
= 2.94 - 1.38 = 1.56
\(\Rightarrow A=\frac { 1.56\times 25 }{ 13 } =\frac { 39 }{ 13 } =3\)
\(E({ X }^{ 2 })=1\left( \frac { 1 }{ 2 } \right) +4\left( \frac { 1 }{ 5 } \right) +16\left( \frac { 3 }{ 25 } \right) +{ (2A) }^{ 2 }\left( \frac { 1 }{ 10 } \right) { +(3A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) +{ (5A) }^{ 2 }\left( \frac { 1 }{ 25 } \right) [\because A=3]\)
\(=\frac { 1 }{ 2 } +\frac { 4 }{ 5 } +\frac { 48 }{ 25 } +36\left( \frac { 1 }{ 10 } \right) \)
\(E({ X }^{ 2 })=\frac { 25+40+96+180+162+450 }{ 50 } \)
\(=\frac { 953 }{ 50 } =19.06\)
\(V(X)=E({ X }^{ 2 })-{ [E(X)] }^{ 2 }\)
\(=19.06-{ (2.94) }^{ 2 }[\because E(X)=2.94]\)
= 19.06 - 8.6436
V(X) = 10.4164

Clearly P(X = 0) + P(X = 1) + P(X = 2)= 1
p + P + P(X = 2) =1
2p + P(X = 2) = 1
P(X = 2) = 1 - 2p
so, probability distribution of X is 

∴ E(X) = 0xp+1xp+2(1-2p)
= p+2-4p = 2-3p
and E(X²) = 0xp+1(p)+4(1-2p)
= p+4-8p = 4-7p
Since E(X²) = E(X)we get,
4-7p = 2-3p ⇒ 4-2 = -3p+7p
⇒2 = 4p ⇒ p =\(\frac{2}{4}=\frac{1}{2}\)
∴ p = \(\frac{1}{2}\)

\(E(X)=0\times \frac { 1 }{ 5 } +1(\frac { 2 }{ 5 } )+2\left( \frac { 1 }{ 5 } \right) +3\times \frac { 1 }{ 5 } \)
\(=\frac { 2 }{ 5 } +\frac { 2 }{ 5 } +\frac { 3 }{ 25 } =\frac { 7 }{ 5 } \)
\(\\ \therefore 2E(X)=\frac { 14 }{ 5 } ...(1)\)
\(E({ Y }^{ 2 })=0\times \frac { 1 }{ 5 } +{ 1 }^{ 2 }(\frac { 3 }{ 10 } )+{ 2 }^{ 2 }(\frac { 2 }{ 5 } )+{ 3 }^{ 2 }(\frac { 1 }{ 10 } )\)
\(=\frac { 3 }{ 10 } +\frac { 8 }{ 5 } +\frac { 9 }{ 10 } =\frac { 3+16+9 }{ 10 } \)
\(=\frac { 28 }{ 10 } =\frac { 14 }{ 5 } ..(2)\)
From (1) and (2), E(Y²) = 2 E(X).

i) E(X) = Σx_ip_i
= 0(0.1)+1(0.25)+2(0.3)+3(0.2)+4.(0.15)
= 2.05
E(X²) = ∑x_i²p_i
= 0(0.1)+1(0.25)+4(0.3)+9(0.2)+16(0.15)
= 5.65
Now, V(X) = E(X²)-[E(X)]²
= 5.65 - (2.05)² = 1.4475
ii) \(V\left( \frac { X }{ 2 } \right) =\frac { 1 }{ 4 } v(X)\quad [\because V(aX)={ a }^{ 2 }V(X)]\)
\(=\frac { 1 }{ 4 } (1.4475)\)
\(V\left( \frac { X }{ 2 } \right) =0.361875\)

Since X is the random variable taking values
1, 2,....7
P(X=1)+P(X=2)+...P(X=7) = 1
⇒ c+2c+2c+3c+c²+2c²+7c²+c = 1
⇒ 10c²+ 9c +1 = 0
⇒ (c+1)(10c-1)= 0
⇒ c = \(\frac{1}{10}\)[∵ c-1 is not possible]
Now, E(X) = Σxipi
= 1(c) + 2(2c) + 3(2c) + 4(3c) + 5c² + 12c² + 7(7c² + c)
= 66c²+ 30c = 66(\(\frac{1}{10}\))² + 30(\(\frac{1}{10}\))
= \(\frac{66}{100}+3=\frac{366}{100}=3.66\)
∴ E(X) = 3.66