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

2. 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)

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, 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

4. 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

5. 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

6. 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 =
σx = 1.123

7. 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) (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 =
σx = √ = $386.78

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

9. Chapter 16

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

13. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet
Color 1 to 1
Single # 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

14. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet
Corner 8 to 1
Split 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

15. The reason why you SHOULD NOT gamble! Expected Payouts on a $1.00 Bet
Street to 1
Dozen 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