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asked Mar 6, 2023 141k views

1 Answer

Rugolinifr · Answer 1 · 2023-03-09T18:50:44+0000

(i) The mean is

$\displaystyle E(X) = \sum_x x \, P(X = x) \\\\ E(X) = 1\cdot0.175 + 2\cdot0.315 + 3\cdot0.211 + 4\cdot0.092 + 5\cdot0.207 \\\\ \boxed{E(X) = 2.839}$

The variance is

V(X) = E((X - E(X))^2) = E(X^2) - E(X)^2

Compute the second moment
E(X^2) :

$\displaystyle E(X^2) = \sum_x x^2 \, P(X = x) \\\\ E(X) = 1^2\cdot0.175 + 2^2\cdot0.315 + 3^2*0.211 + 4^2*0.092 + 5^2*0.207 \\\\ E(X^2) = 9.997$

Then the variance is

$\boxed{V(X) \approx 1.9171}$

(ii) For a random variable
Z=aX+b , where
a,b are constants, we have

E(Z) = E(aX+b) = E(aX) + E(b) = a E(X) + b

and

$V(Z) = E((aX+b)^2) - E(aX+b)^2 \\\\ V(Z) = E(a^2 X^2 + 2ab X + b^2) - (a E(X) + b)^2 \\\\ V(Z) = a^2 (E(X^2) - E(X)^2) \\\\ V(Z) = a^2 V(X)$

Then for
$Y=\frac{X+3}2$ , we have

$E(Y) = \frac12 E(X) + \frac32 \\\\ \boxed{E(Y) = 2.918}$

$E(Y^2) = E\left(\left(\frac{X+3}2\right)^2\right) = \frac14 E(X^2) + \frac32 E(X) + \frac94 \\\\ \boxed{E(Y^2) \approx 9.0028}$