Page 155 - DMTH404_STATISTICS

Unit 10: Approximate Expressions for Expectations and Variance

Solution. Notes

Let X denote the number obtained on the i th die. Therefore, the sum of points on n dice is S = X
i 1
+ X + ...... + X and
2 n
E(S) = E(X ) + E(X ) + ...... + E(X ).
1 2 n
Further, the number on the i th die, i.e., X follows the following distribution :
i
X : 1 2 3 4 5 6
i
1 1 1 1 1 1
( p X i ) :
6 6 6 6 6 6
1 7
E X
\ ( ) = (1 2 3 4 5 6+ + + + + ) = (i = 1, 2, .... n)
i
6 2
7 7 7 7n
E
n
Thus, ( ) S = + + .... + ( times) =
2 2 2 2
Example 15: If X and Y are two independent random variables with means 50 and 120
and variances 10 and 12 respectively, find the mean and variance of Z = 4X + 3Y.

Solution.
E(Z) = E(4X + 3Y) = 4E(X) + 3E(Y) = 4 ´ 50 + 3 ´ 120 = 560
Since X and Y are independent, we can write
Var(Z) = Var(4X + 3Y) = 16Var(X) + 9Var(Y) = 16 ´ 10 + 9 ´ 12 = 268

Example 16: It costs Rs 600 to test a machine. If a defective machine is installed, it costs Rs
12,000 to repair the damage resulting to the machine. Is it more profitable to install the machine
without testing if it is known that 3% of all the machines produced are defective? Show by
calculations.
Solution.
Here X is a random variable which takes a value 12,000 with probability 0.03 and a value 0 with
probability 0.97.
\ E(X) = 12000 ´ 0.03 + 0 ´ 0.97 = Rs 360.
Since E(X) is less than Rs 600, the cost of testing the machine, hence, it is more profitable to install
the machine without testing.

10.6 Summary

 Expected value of a constant is the constant itself, i.e., E(b) = b, where b is a constant.
 Using the above result, we can write an alternative expression for the variance of X, as
given below :
2
2
 = E(X - ) = E(X - 2X +  )
2
2
= E(X ) - 2E(X) +  = E(X ) - 2 + 2
2
2
2
2
2
= E(X ) -  = E(X ) - [E(X)] 2
2
2
= Mean of Squares - Square of the Mean
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