Let X be a random variable with mean 6.1 and variance 2.4. Let Y be a random variable with mean 7.5 and variance 3.2. Find the variance of Z, Var(Z).

Var(z)= 8.122 if Z= z1+z2

Step-by-step explanation:

IF x and Y are independent then Cov(x,y)= 0

Var(x+y) = Var X + Var y + 2cov(x,y)

= 2.4 + 3.2 +0 = 5.6

Z1= X - E (X)/ √Var (X) and Z2 = Y - E(Y)/ √Var Y

Var (z1+z2) = Var (z1) + Var(z2) + 2Cov (z1,z2)

Cov (z1,z2)= E(z1,z2) - E(z1) E(z2)

= E(z1,z2) [as E(z1) E(z2)= 0]

= E {[ X-E(X)] [Y-E(Y)]/ √Var(X) Var(y)}

= E{[ X- 6.1][Y- 7.5]/√2.4(3.2)}

= E {xy-7.5x -6.1y +45.75]/ 2.771}

={ 6.1(7.5) - 7.5(6.1) - 6.1(7.5) + 45.75]/ 2.771}

= 0/2.771

= 0

Var (z1+z2) = Var (z1) + Var(z2)

= {E (X)/ √Var (X)} +{ Y - E(Y)/ √Var Y}

= {6.1/√2.4 } + {7.5/√3.2}

= 3.93+ 4.19

=8.122

2.83941151 would be the answer you're looking for! : )