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denpristay [2]
2 years ago
12

An electronic system has each of two different types of components in joint operation. Let X and Y denote the lengths of life, i

n 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.
Mathematics
1 answer:
AURORKA [14]2 years ago
6 0

Answer:

a

The average cost is E(C) = 66

b

The standard deviation of  cost is  \sigma  = 9.798

Step-by-step explanation:

From the question we are told that

          E(X) =  4

          E(Y) =  2

          E(X^2) =  24

           E(Y^2) =  8

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

The variance   of X is mathematically represented as

           V(X) =  E(X^2) - [E(X)]^2

Substituting values  

              V[X] =  24 - 4^2

               V[X] =8

The variance  of Y is mathematically represented as

           V(Y) =  E(Y^2) - [E(Y)]^2

Substituting values  

              V[Y] =  8 - 2^2

               V[X] =4

The average of  replacing the two component is  

        E(C) =  2 * E(X)  + 4* E(Y)

substituting value

         E(C) =  50 + 2 * (4)  + 4* (2)

         E(C) = 66

The variance  of replacing the two component is  

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

                                                                  and X and  Y are independent  

=>      V(C) =  2^2 * V(X) + 4^2 * V(Y)

substituting values

=>       V(C) =  4 * 8 +  16 * 4        

=>         V(C) =  32 + 64

=>         V(C) = 96

The standard deviation is

         \sigma  =  \sqrt{V(C)}  

substituting values

           \sigma  =  \sqrt{96}

           \sigma  = 9.798

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