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 hits 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) = 0.0486, P(X = 3) = 0.0036, P(X =4) = 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: f(x) = 1. E(X)= 2. V(X)= 3. E(10X+5)= 4. V(10X+5)= 5. E(sqrt X)= Bonus! V(sqrt X) =

Solution

1) E(x)= 0 * f(0) +1 *f(1) + 2*f(2) + 3*f(3) + 4* f(4)

= 0*0.6561 +1*0.2916 + 2*0.0486 +3*0.0036 + 4*0.0001

= 0.40

2) Var(x) = E(x²)- (E(x))²

Now

E(x²) = 0²*0.6561 +1²*0.2916 + 2²*0.0486 +3²*0.0036 + 4²*0.0001

=0 + 0.2916 + 0.1944 + 0.0324 + 0.0016

=0.52

Thus Var(X) = 0.52 - (0.40)²

   =0.36

3) E(10X+5) = 10*E(X) + 5

=10*0.40 + 5

=9

4) V(10X+5) = 10²*V(X) +0

=100*0.36

=36

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

= 0*0.6561 +1*0.2916 + 1.414214*0.0486 +1.732051*0.0036 + 2*0.0001

=0.366766

(NOTE: sqrt(X) = square root of X)

6) var(sqrt(X))= E(X) - [E(sqrt(X))]²

=0.40 - (0.366766)²

= 0.268483