Answer:
1. Mean is 1.75
2. The variance is 1.6042
3.
The distribution function is:
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384
Explanation:
The mean of Z is given as:
E(Z) =Σ6, k=0 Kp (Z = k)
Σ6,k=0 K 1/6 Σ6, n=k (n k) (1/2)^n
=( 0(21/128) + 1(5/16) + 2( 33/128) + 3 (1/6) + 4 (29/384) + 5 (1/48) + 6 (1/384))
=7/4
=1.75
Thus, the mean Z is 1.75
The variance of Z is given as:
Var (Z) = E (Z^2) - (E (Z)) ^2
Therefore,
E(Z^2) = Σ 6, k=0 K^2P ( Z=K)
= ( 0(21/128 + 1(5/16) + 4(33/128) + 9(1/6) + 16(29/384) + 25(1/48) + 36(1/384))
=14/3
Var (Z) = 14/7 - (7/4)^2
= 14/7 - 49/16
=77/48
=1.6042
Thus, the variance is 1.6042
The probability of mass function is given as:
P(Z=k) = 1/6 Σ 6, n=k (n k) (1/2)^n
The distribution function is
Z Z/K
0 21/128
1 5/16
2 33/128
3 1/6
4 29/384
5 1/48
6 1/384