The probability distribution of the amount of memory X (GB) in a purchased flash drive is given below. x 1 2 4 8 16 p(x) .05 .10 .35 .40 .10 Compute the following: E(X), E(X2 ), V (X), E(3X 2), E(3X2 2), V (3X 2), E(X 1), V (X 1).

Respuesta :

Answer:

a) [tex]E(X) = 6.45[/tex]

b) [tex]E(X^{2} )= 57.25[/tex]

c) [tex]V(X) = 15.648[/tex]

d) E(3X + 2) = 21.35

e) [tex]E(3X^{2} +2) = 173.75[/tex]

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) [tex]E(X) = \sum xP(x)[/tex]

[tex]E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45[/tex]

b)

[tex]E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\ E(X^{2} )= 57.25[/tex]

c)

[tex]V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648[/tex]

d)

[tex]E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35[/tex]

e)

[tex]E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75[/tex]

f)

[tex]V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832[/tex]

g)

[tex]E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45[/tex]

h)

[tex]V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648[/tex]