1. 7.2 Mathematical Expectation
Below are the results obtained when a die is thrown 30 times .
Below are probability distribution table for X where X = number obtained from the die .
Score, x f 1 2 3 4 5 6
f x
Score, x
P( X=x ) 1 2 3 4 5 6
x P( X=x )
Experimental method
Theoretical method 1
Mean score =
3.5
Expected mean =
Expected value of X,

2. Example 1 :
The probability distribution for a random variable U is given in the following table : u 1 2 3 4 5 6 7
P( U = u )
0.05
0.10
0.15
0.40
Ex 7.2 pg. 308
Q. 1 – 3
(a) Find the E( U ) .
(b) Draw a probability distribution graph for U and deduce the E( U ) .
(c) Find P [ U > E( U ) ]
P(U = u )
0.40
Solution :
0.35
0.30
0.25
= 1(0.05) +
2(0.10) +
3(0.15) +
4(0.40) +
5(0.15) +
6(0.10) +
7(0.05)
= 4
0.20
(b) Axis of symmetry is U = 4
0.15
So, E ( U ) = 4
0.10
0.05
(c) P [ U > E( U ) ]
= P ( U > 4 )
= P( U = 5 ) + P( U = 6 ) + P( U = 7 ) u
= = 0.30 1 2 3 4 5 6 7

3. Ex 7.2 pg. 308 Q. 4 – 6 x 1 3 8 P( X = x ) Example 2 :
In a game, Jooing throws 3 fair coins. She will receive RM 8 if all the three coins show “heads”, RM x if two “heads” are obtained, RM 3 if one “head” is obtained and RM 1 if no “head” appear. State, in terms of x, the expected gain for Jooing in each game. If Jooing pays RM 3.75 to participate in each game, find
Ex 7.2 pg. 308
Q. 4 – 6
(a) The value of x so that each game is fair,
(b) The expected gain or loss for Jooing after she plays 100 games if x = 3.5.
Solution :
Let T = Event that a coin shows tail.
H = Event that a coin shows head. x 1 3 8
P( X = x )
Let X = Gain for Jooing in each game .
X = { 1 , 3 , x , 8 }
P( X = 1 )
= P ( T T T )
= RM 3.75
Expected gain for 100 games
x = RM 4
P( X = 3 )
= P ( H T T ) x
(b) If x = 3.5 and Jooing pays RM 3.75,
Expected loss after 100 games
P( X = x )
= P ( H H T ) x
Expected gain for each game
P( X = 8 )
Expected gain for each game
= P ( H H H )

4. E (X2) = E (a) = a E (aX) = a E(X) Expectation Formula E (aX + b) =
Ex 7.3 pg. 310
Q. 1 – 4
E (X2) =
E (a) = a
E (aX) =
a E(X)
Expectation Formula
E (aX + b) =
a E(X) + b
( a and b are constants )
Example 1 :
The probability distribution for a random variable X is given in the following table. Find E( 2X – 1 ) and E[ ( X + 2 )2 ]
= E(X2) + 4E(X) + 4
E( 2X – 1 ) = 2E(X) – 1 x 1 2 3 4
P( X = x )
0.4
0.3
0.2
0.1
=E(X2+4X+4 )
E[(X + 2 )2]
= 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1)
= 2
= 12(0.4) + 22(0.3) + 32(0.2) + 42(0.1)
= 5
E[ ( X + 2 )2 ]
= E ( X2 + 4X + 4 )
E( 2X – 1 ) = 2E(X) – 1
= 2(2) – 1 = 3
= E(X2) + 4E(X) + 4
= 5 + 4(2) + 4
= 17

5. Memorize Var (X) = E [ ( X -  )2 ] Var (X) = E ( X 2 ) – [ E ( X ) ]2
Definition of variance =
( Data Description )
( Discrete Random Variable )
Var (X) = E [ ( X -  )2 ]
Memorize
Var (X) = E ( X 2 ) – [ E ( X ) ]2
For Discrete Random Variable
Mean X = E ( X )
where
Variance X = Var ( X )
= E ( X 2 ) – [ E ( X ) ]2

6. Example 1 :
The probability distribution for a random variable X is given in the following table. Find the expected value of X and variance of X. x 1 2 3 4
P( X = x )
0.4
0.3
0.2
0.1
Expected value of X,
= 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1)
= 2
Variance of X , Var ( X )
= E ( X 2 ) – [ E ( X ) ]2
= 12(0.4) + 22(0.3) + 32(0.2) + 42(0.1)
= 5
Var ( X ) =
5 – ( 2 )2
Ex 7.4 pg. 314
Q. 1, 2, 3, 7, 8, 10
= 1

7. Variance Formula Var (X) = E ( X 2 ) – [ E ( X ) ]2 Var (a)
= E ( a 2 ) – [ E ( a ) ]2
( a is a constant )
= a 2
– ( a ) 2
= 0
Var (aX)
= E [ ( aX )2] – [ E ( aX ) ]2
= E ( a2 X2 )
– [ aE ( X ) ]2
pg. 374
= a2 E (X2 )
– a2 [E ( X ) ]2
= a2 { E (X2 ) – [ E ( X ) ]2 }
= a2 Var (X)
( a & b are constants )
Var (aX + b)
= E [ ( aX + b )2 ] – [ E ( aX + b ) ]2
= E ( a2 X2 + 2abX + b2 )
– [ aE ( X ) + b ]2
= E (a2 X2) + E (2abX) + E (b2)
– { a2[E (X)]2 + 2abE(X) + b2 }
= a2 E (X2) + 2abE (X) + b2
– a2[E (X)]2 – 2abE(X) – b2 }
= a2 E (X2) – a2 [E (X)]2
= a2 { E (X2) – [E (X)]2 }
Variance Formula
= a2 Var (X)

8. Var (X) = E ( X 2 ) – [ E ( X ) ]2 Ex 7.4 pg. 314
Example 1 :
The discrete random variable X has the probabilities
, x = – 3, – 1, 2
P( X = x ) = , x = – 2, 1
, x = 0
Find (a ) Var ( X )
(b ) Var ( 7X )
(c ) Var ( 3X + 2 )
(d ) Var ( 3 – X )
= (– 3)(0.1) + (– 1)(0.1) + (2)(0.1) + (– 2)(0.2) + (1)(0.2) + (0)(0.3)
= – 0.4
= (– 3)2(0.1) + (– 1)2(0.1) + (2)2(0.1) + (– 2)2(0.2) + (1)2(0.2) + (0)2(0.3)
= 2.4
Var (X) = E ( X 2 ) – [ E ( X ) ]2
= 2.4 – (– 0.4 )2
= 2.24
Ex 7.4 pg. 314
Q. 4, 5, 6, 9, 11, 12
(b ) Var ( 7X )
= 72 Var (X )
= 49 ( 2.24 ) =
(c ) Var ( 3X + 2 )
= 32 Var (X)
= 9 (2.24)
= 20.16
(d ) Var ( 3 – X )
= ( –1 )2 Var ( X )
= 1 (2.24)
= 2.24
Example 2 :
X is a discrete random variable. 20
(a) If E( X + 5 ) = 8 and Var( X + 5 ) = 11, find E( X 2 )
(b) If E( X2 ) = 29 and Var( 3X ) = 36, find E( X )
5 or – 5