Question

The waiting time W for accessing one record from a computer database is a random variable uniformly distributed between 0 and 28 milliseconds. The read time R for moving the information from the disk to the main memory is 7 milliseconds. The random variable X milliseconds is the total access time (waiting time + read time) to get one block of information from the disk. Before performing a certain task, the computer must access 6 different blocks of information from the disk. (Access times for different blocks are independent of one another.). Compute the followings:

a. E[X]
b. Vae[X]
c. E[A]

Answer 1

Solution :

Let

Waiting time for accessing one record = W

Read time for moving the information = R

Total time for accessing to get one block of information = X

So, the random variable X is defined as :

X = W + R

  = W + 7

a). Calculating E(X)

E(X) = E(W+7)

      = E(W) + 7

      $=\int_{0}^{28}\frac{1}{28-0}w dw +7$

       $=\frac{1}{28}\int_{0}^{28}w dw +7$

       $=\frac{1}{28}\left[\frac{w^2}{2}\right]_0^{28}+7$

      $=\frac{1}{56}\left[28^2\right]+7$

      $=\frac{28}{2}+7$

     = 14 + 7

     = 21

b). Calculating Var(X)

V(X) = V(W+7)

      = V(W)+0

      = V(W)

     = $E(W^2)-[E(W)]^2$

     $=\int_0^{28}\frac{1}{28}w^2 dw-[14]^2$

    $=\frac{1}{28}\left[\frac{w^3}{3}\right]_0^{28}-196$

   $=\frac{1}{28\times 3}\times 28^3-196$

  $=\frac{28\times 28}{3}-196$

 = 65.33

c). Considering A is the random variable that can be defined as follows:

A = 6(W+7)

   =6W + 42

So calculating E(A)

E(A) = E(6W + 42)

      = E(6W) +42

      = 6E(W) + 42

     = 6(14) + 42

     = 84 + 42

     = 126