(i) The mean is
[tex]\displaystyle E(X) = \sum_x x \, P(X = x) \\\\ E(X) = 1\cdot0.175 + 2\cdot0.315 + 3\cdot0.211 + 4\cdot0.092 + 5\cdot0.207 \\\\ \boxed{E(X) = 2.839}[/tex]
The variance is
[tex]V(X) = E((X - E(X))^2) = E(X^2) - E(X)^2[/tex]
Compute the second moment [tex]E(X^2)[/tex] :
[tex]\displaystyle E(X^2) = \sum_x x^2 \, P(X = x) \\\\ E(X) = 1^2\cdot0.175 + 2^2\cdot0.315 + 3^2\times0.211 + 4^2\times0.092 + 5^2\times0.207 \\\\ E(X^2) = 9.997[/tex]
Then the variance is
[tex]\boxed{V(X) \approx 1.9171}[/tex]
(ii) For a random variable [tex]Z=aX+b[/tex], where [tex]a,b[/tex] are constants, we have
[tex]E(Z) = E(aX+b) = E(aX) + E(b) = a E(X) + b[/tex]
and
[tex]V(Z) = E((aX+b)^2) - E(aX+b)^2 \\\\ V(Z) = E(a^2 X^2 + 2ab X + b^2) - (a E(X) + b)^2 \\\\ V(Z) = a^2 (E(X^2) - E(X)^2) \\\\ V(Z) = a^2 V(X)[/tex]
Then for [tex]Y=\frac{X+3}2[/tex], we have
[tex]E(Y) = \dfrac12 E(X) + \dfrac32 \\\\ \boxed{E(Y) = 2.918}[/tex]
[tex]E(Y^2) = E\left(\left(\dfrac{X+3}2\right)^2\right) = \dfrac14 E(X^2) + \dfrac32 E(X) + \dfrac94 \\\\ \boxed{E(Y^2) \approx 9.0028}[/tex]
