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7.2 Mathematical Expectation

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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


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