HW 19 Mean and Variance There is a chance that a bit transmi

HW 1.9. (Mean and Variance.) There is a chance that a bit transmitted through a digital transmission channel is received in error. Let X equal the number of bits in error in the next four bits transmitted. The possible values for X are {0, 1, 2,3, 4}. Based on a model for the errors that is presented in the following section, probabilities for these values will be determined. P(X = 0) = 0.6561, P(X = 1) = 0.2916, P(X = 2) = 0.0186, P(X = 3) = 0.0036, P(X = 1) = 0.0001. The probability distribution of X is specified by the possible values along with the probability of each. Then the pmf of X is: 1. E(X)= 2. V(X) = 3. E(10X+5)= 4. V(10X+5)= 5. E(root X) Bonus! V(root X) =

Solution

1)

E(X) = 0.6561 * 0 + 0.2916*1 + 0.0486*2 + 0.0036*3 + 0.0001*4 = 0.4 Answer

2)

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

E(X^2) = 0.6561 * 0 + 0.2916*1 + 0.0486*4 + 0.0036*9+ 0.0001*16 =0.52

V(X) = 0.52 - (0.4)^2 = 0.36 Answer

3)

E(10X+5) = 0.6561 * 5 + 0.2916*15 + 0.0486*25 + 0.0036*35 + 0.0001*45 = 9 Answer

4)

V(10X+5) = E((10X+5)^2) - (E(10X+5))^2

E((10X+5)^2) = 0.6561 * 5^2 + 0.2916*15^2 + 0.0486*25^2 + 0.0036*35^2+ 0.0001*45^2 = 117

V(10X+5) = 117 - (9)^2 = 36 Answer

5)

E(sqrt(X)) = 0.6561 * 0 + 0.2916*1 + 0.0486*sqrt(2) + 0.0036*sqrt(3) + 0.0001*2 = 0.367 Answer

Bonus:

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

= 0.4 - (0.367)^2

= 0.265 Answer