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 < ½ | Y > X) = ? 0 0.5 1 Y P(Y < ½ | Y > X) = P(Y < ½ ∩ Y > X) P(Y > X) 1 / 8 1 / 2 1/4 X

VALID WAYS OF PROVING INDEPENDENCE P(A∩B) = P(A)∙P(B) P(A) = P(A|B) = P(A|C) = P(A|D) X2 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)

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, μx or E(X) Values of X x1 x2 x3 … xk Probability p1 p2 p3 pk The MEAN of a DISCRETE Random VARIABLE (Weighted Average) Let X = the random variable whose distribution follows: μx = x1p1 + x2p2 + x3p3 + … + xkpk E(X) = μx = Σxi pi

What can the average student expect to make on the AP Exam? EXAMPLE #2: The following probability distribution represent the AP Statistics scores from previous years. AP Score 1 2 3 4 5 Probability 0.14 0.24 0.34 0.21 0.07 What can the average student expect to make on the AP Exam? μx = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = μx = E(X) = 2.83

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 – μx)2 . The variance is denoted by: σX2. The Standard Deviation is the square root of the Variance and is denoted by: σx. Let X = the random variable whose distribution follows and has mean :E(X) = μx = Σxi·pi Values of X x1 x2 x3 … xk Probability p1 p2 p3 pk The Variance of a discrete random variable is: σx2 = (x1 – μx)2 p1 + (x2 – μx)2 p2 + … + (xk – μx)2 pk Var(X) = σx2 = Σ (xi – μx)2 pi

What is the Standard Deviation of the AP Exam results? EXAMPLE #2: The following probability distribution represent s the AP Statistics scores from previous years. AP Score 1 2 3 4 5 Probability 0.14 0.24 0.34 0.21 0.07 What is the Standard Deviation of the AP Exam results? μx = E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = μx = E(X) = 2.83 σx2 = (x1 – μx)2 p1 + (x2 – μx)2 p2 + … + (xk – μx)2 pk = Σ (xi – μx)2 pi σx2 = 1.2611 σx = 1.123

2. σx2 = (x1 – μx)2 p1 + (x2 – μx)2 p2 + … + (xk – μx)2 pk EXAMPLE #1: An insurance company acknowledges that its payout probability for claims is as follows: What can the company expect to payout per policyholder? What is the Standard Deviation of payouts for the distribution of Policyholders? Policyholder Outcome Policy x Probability P(X = x) Death $10000 1 / 1000 Disability $5000 2 / 1000 Neither $0 997 / 1000 1. E(X) = μx = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20 2. σx2 = (x1 – μx)2 p1 + (x2 – μx)2 p2 + … + (xk – μx)2 pk = (10000 – 20)2·(1/1000) + (5000 – 20)2·(2/1000) + (0 – 20)2· (997/1000) σx2 = 149600 σx = √149600 = $386.78

Chapter 16 - Probability Models HW: Page 381: #1-2(find µ and σ) , #3 - 7

Chapter 16

Chapter 16 - Probability Models HW: Page 381: #1-2(Find µ and σ) , #3 - 7

The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet Color 1 to 1 Single # 35 to 1 E(X) = μx = xWINpWIN + xLOSEpLOSE E(RED) E(RED) = μ = $1(18/38) + -$1(20/38) μ = $–0.05 E(25) E(25) = μ = $35(1/38) + -$1(37/38) μ = $–0.05

The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet Corner 8 to 1 Split 17 to 1 E(X) = μx = xWINpWIN + xLOSEpLOSE E(Corner) E(Corner) = μ = $8(4/38) + -$1(34/38) μ = $–0.05 E(Split) E(Split) = μ = $17(2/38) + -$1(36/38) μ = $–0.05

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