Question

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).

Answer 1

a)
`E(X) = 6.45`

b)
`E(X^(2) )= 57.25`

c)
`V(X) = 15.648`

d) E(3X + 2) = 21.35

e)
`E(3X^(2) +2) = 173.75`

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Explanation:

a)
`E(X) = \sum xP(x)`

`E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45`

b)

`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`

c)

`V(X) = E(X^(2) ) - (E(X))^(2) \\V(X) = 57.25 - 6.45^(2) \\V(X) = 15.648`

d)

`E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35`

e)

`E(3X^(2) +2) = 3E(X^(2) ) + 2\\E(3X^(2) +2) = (3*57.25) + 2 \\E(3X^(2) +2) = 173.75`

f)

`V(3X+2) = 3^(2) V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832`

g)

`E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45`

h)

`V(X+1) = 1^(2) V(X)\\V(X+1) = 15.648`