Page 153 - DMTH404_STATISTICS

Unit 10: Approximate Expressions for Expectations and Variance

Notes
Example 11:
A random variable X has the following probability distribution :

X : - 2 - 1 0 1 2
1 1 1
Probability : p p
6 4 6
(i) Find the value of p.
2
(ii) Calculate E(X + 2) and E(2X + 3X + 5).
Solution.
Since the total probability under a probability distribution is equal to unity, the value of p
1 1 1
should be such that + p + + p + = 1 .
6 4 6
5
This condition gives p =
24

1 5 1 5 1
Further, E ( ) = - 2. - 1. + 0. + 1. + 2. = 0
X
6 24 4 24 6
1 5 1 5 1 7
( E X 2 ) = 4. + 1. + 0. + 1. + 4. = ,
6 24 4 24 6 4
X
( E X + 2) = E ( ) 2 = 0 2 = 2
+
+
7
and E (2X + 3X + 5) = 2 (X 2 ) 3 ( ) 5 = 2. + 0 5 = 8.5
2
+
E
+
+
X
E
4
Example 12:
A dealer of ceiling fans has estimated the following probability distribution of the price of a
ceiling fan in the next summer season:
P
Price ( ) : 800 825 850 875 900
p
Probability ( ) : 0.15 0.25 0.30 0.20 0.10
If the demand (x) of his ceiling fans follows a linear relation x = 6000 - 4P, find expected demand
of fans and expected total revenue of the dealer.
Solution.

Since P is a random variable, therefore, x = 6000 - 4P, is also a random variable. Further, Total
Revenue TR = P.x = 6000P - 4P is also a random variable.
2
From the given probability distribution, we have

E(P) = 800 ´ 0.15 + 825 ´ 0.25 + 850 ´ 0.30 + 875 ´ 0.20 + 900 ´ 0.10
=Rs 846.25 and
2
2
2
E(P ) = (800) ´ 0.15 + (825) ´ 0.25 + (850) ´ 0.30 + (875) ´ 0.20
2
2
2
+ (900) ´ 0.10 = 717031.25
LOVELY PROFESSIONAL UNIVERSITY 145