1. WARM – UP Refer to the square diagram below and let X be the x- coordinate and Y be the y-coordinate of any randomly chosen point. Find the Conditional Probability P(Y X) = ? 0 0.5 1 X Y Y > X P(Y X) = Y X) = P(Y X) P(Y > X) P(Y > X) 1 / 8 1 / 8 1 / 2 1 / 2 1/4 1/4

2. VALID WAYS OF PROVING INDEPENDENCE P(A∩B) = P(A)∙P(B) P(A) = P(A|B) = P(A|C) = P(A|D) X 2 Test of Independence INVALID WAYS OF PROVING INDEPENDENCE P(A∩B) = P(A) P(A|B) = P(B|A) P(A|C) = P(B|C) P(A) = P(B) P(A∩B) = P(AUB) P(A∩B) = P(AUC)

3. CHAPTER 16 - MEANS OF RANDOM VARIABLES The Mean of a random variable is a weighted average of the possible values of X. The Mean is also called the Expected Value and is noted by the symbol, μ x or E(X) The MEAN of a DISCRETE VARIABLE (Weighted Average) Let X = the random variable whose distribution follows: μ x = x 1 p 1 + x 2 p 2 + x 3 p 3 + … + x k p k E(X) = μ x = Σx i p i Values of Xx1x1 x2x2 x3x3 …xkxk Probabilityp1p1 p2p2 p3p3 …pkpk

4. EXAMPLE #1: An insurance company acknowledges that its payout probability for claims is as follows: Policyholder Outcome Policy x Probability P(X = x ) Death$100001 / 1000 Disability$50002 / 1000 Neither$0997 / 1000 1.What can the company expect to payout per policyholder? 2.If they insure 200 clients and each client pays $100 for the policy, how much profit should they expect? E(X) = μ x = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20 E(X) = μ x = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20 2.Profit = Revenue – Expenses = 200($100) – 200($20) = $16000 = 200($100) – 200($20) = $16000 1. E(X) = μ x = x 1 p 1 + x 2 p 2 + x 3 p 3 = 1. E(X) = μ x = x 1 p 1 + x 2 p 2 + x 3 p 3 = Σx i ·p i

5. What can the average student expect to make on the AP Exam? μ x = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = μ x = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = EXAMPLE #2: The following probability distribution represent the AP Statistics scores from previous years. AP Score 12345 Probability 0.140.240.340.210.07 μ x = E(X) = 2.83 μ x = E(X) = 2.83

6. The VARIANCE OF A RANDOM VARIABLE The Variance and the Standard Deviation are the measures of the spread of a distribution. The variance is the average of the square deviations of the variable X from its mean (X – μ x ) 2. The variance is denoted by: σ X 2. The Standard Deviation is the square root of the Variance and is denoted by: σ x. Values of Xx1x1 x2x2 x3x3 …xkxk Probabilityp1p1 p2p2 p3p3 …pkpk Let X = the random variable whose distribution follows and has mean :E(X) = μ x = Σx i ·p i The Variance of a discrete random variable is: σ x 2 = (x 1 – μ x ) 2 p 1 + (x 2 – μ x ) 2 p 2 + … + (x k – μ x ) 2 p k Var(X) = σ x 2 = Σ (x i – μ x ) 2 p i

7. What is the Standard Deviation of the AP Exam results? μ x = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = μ x = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = EXAMPLE #2: The following probability distribution represent s the AP Statistics scores from previous years. AP Score 12345 Probability 0.140.240.340.210.07 μ x = E(X) = 2.83 μ x = E(X) = 2.83 σ x 2 = (x 1 – μ x ) 2 p 1 + (x 2 – μ x ) 2 p 2 + … + (x k – μ x ) 2 p k = (1 – 2.83) 2 ·(.14) + (2 – 2.83) 2 ·(.24) + (3 – 2.83) 2 ·(.34) + (4 – 2.83) 2 ·(.21) + (5 – 2.83) 2 ·(.07) = = (1 – 2.83) 2 ·(.14) + (2 – 2.83) 2 ·(.24) + (3 – 2.83) 2 · (.34) + (4 – 2.83) 2 ·(.21) + (5 – 2.83) 2 · (.07) = Σ (x i – μ x ) 2 p i σ x 2 = 1.2611 σ x = 1.123

8. 2. σ x 2 = (x 1 – μ x ) 2 p 1 + (x 2 – μ x ) 2 p 2 + … + (x k – μ x ) 2 p k = (10000 – 20) 2 ·(1/1000) + (5000 – 20) 2 ·(2/1000) + (0 – 20) 2 ·(997/1000) = (10000 – 20) 2 ·(1/1000) + (5000 – 20) 2 ·(2/1000) + (0 – 20) 2 · (997/1000) EXAMPLE #1: An insurance company acknowledges that its payout probability for claims is as follows: Policyholder Outcome Policy x Probability P(X = x ) Death$100001 / 1000 Disability$50002 / 1000 Neither$0997 / 1000 1.What can the company expect to payout per policyholder? 2.What is the Standard Deviation of payouts for the distribution of Policyholders? 1. E(X) = μ x = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20 σ x 2 = 149600 σ x 2 = 149600 σ x = √149600 = $386.78 σ x = √149600 = $386.78

10. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet E(X) = μ x = E(X) = μ x = x WIN p WIN + x LOSE p LOSE E(RED) = μ = $ E(RED) = μ = $1(18/38) + -$1(20/38) μ = μ = $–0.05 E(25) = μ = $ E(25) = μ = $35(1/38) + -$1(37/38) μ = μ = $–0.05 PAYOUTS Color1 to 1 Single # 35 to 1 E(RED) E(25)

11. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet E(X) = μ x = E(X) = μ x = x WIN p WIN + x LOSE p LOSE E(Corner) = μ = $ E(Corner) = μ = $8(4/38) + -$1(34/38) μ = μ = $–0.05 E(Split) = μ = $ E(Split) = μ = $17(2/38) + -$1(36/38) μ = μ = $–0.05 PAYOUTS Corner8 to 1 Split 17 to 1 E(Corner) E(Split)

12. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet E(X) = μ x = E(X) = μ x = x WIN p WIN + x LOSE p LOSE E(Street) = μ = $ E(Street) = μ = $11(3/38) + -$1(35/38) μ = μ = $–0.05 E(Dozen) = μ = $ E(Dozen) = μ = $2(12/38) + -$1(26/38) μ = μ = $–0.05 PAYOUTS Street 11 to 1 Dozen 2 to 1 E(Street) E(Dozen) =