An electronic system has each of two different types of components in joint operation. Let X and Y denote the lengths of life, in hundreds of hours, for components of type I and type II, respectively. The performance of a component is independent for each other. Let E(X) = 4, E(Y) = 2, E(X2) = 24, E(Y2) = 8.
The cost of replacing the two components depends upon their length of life at failure and it is given by C = 50 + 2X + 4Y.
(i) Compute the average cost of replacing the two components. Your final answer must be a number.
(ii) Compute the standard deviation of cost of replacing the two components. Your final answer must be a number.

Respuesta :

Answer:

a

The average cost is [tex]E(C) = 66[/tex]

b

The standard deviation of  cost is  [tex]\sigma = 9.798[/tex]

Step-by-step explanation:

From the question we are told that

          [tex]E(X) = 4[/tex]

          [tex]E(Y) = 2[/tex]

          [tex]E(X^2) = 24[/tex]

           [tex]E(Y^2) = 8[/tex]

The cost of replacing the two component is  C =  50 + 2 X + 4 Y

The variance   of X is mathematically represented as

           V(X) =  [tex]E(X^2) - [E(X)]^2[/tex]

Substituting values  

              [tex]V[X] = 24 - 4^2[/tex]

               [tex]V[X] =8[/tex]

The variance  of Y is mathematically represented as

           V(Y) =  [tex]E(Y^2) - [E(Y)]^2[/tex]

Substituting values  

              [tex]V[Y] = 8 - 2^2[/tex]

               [tex]V[X] =4[/tex]

The average of  replacing the two component is  

        [tex]E(C) = 2 * E(X) + 4* E(Y)[/tex]

substituting value

         [tex]E(C) = 50 + 2 * (4) + 4* (2)[/tex]

         [tex]E(C) = 66[/tex]

The variance  of replacing the two component is  

        [tex]V(C) = V(50 + 2X +4Y)[/tex]      Note: The variance of constant is zero

                                                                  and X and  Y are independent  

=>      [tex]V(C) = 2^2 * V(X) + 4^2 * V(Y)[/tex]

substituting values

=>       [tex]V(C) = 4 * 8 + 16 * 4[/tex]        

=>         [tex]V(C) = 32 + 64[/tex]

=>         [tex]V(C) = 96[/tex]

The standard deviation is

         [tex]\sigma = \sqrt{V(C)}[/tex]  

substituting values

           [tex]\sigma = \sqrt{96}[/tex]

           [tex]\sigma = 9.798[/tex]