WARM – UPRefer 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

0

0.

5

1

X

Y P(Y < ½ | Y > XY > X) =

P(Y < ½ Y < ½ | Y > X) =

P(Y < ½ ∩ Y > X)P(Y < ½ ∩ Y > X) P(Y > X)P(Y > X)

1 / 81 / 8 1 / 21 / 2

1/41/4

VALID WAYS OF PROVING INDEPENDENCEP(A∩B) = P(A)∙P(B)P(A∩B) = P(A)∙P(B)P(A) = P(A|B) = P(A|C) = P(A|D) P(A) = P(A|B) = P(A|C) = P(A|D) XX22 Test of Independence Test of Independence

INVALID WAYS OF PROVING INDEPENDENCE

P(A∩B) = P(A)P(A∩B) = P(A)P(A|B) = P(B|A)P(A|B) = P(B|A)P(A|C) = P(B|C)P(A|C) = P(B|C)P(A) = P(B)P(A) = P(B)P(A∩B) = P(AP(A∩B) = P(AUUB)B)P(A∩B) = P(AP(A∩B) = P(AUUC)C)

CHAPTER 16 - MEANS OF RANDOM VARIABLESThe 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)

The MEAN of a DISCRETE VARIABLE (Weighted Average)

Let X = the random variable whose distribution follows:

μx = x1p1 + x2p2 + x3p3 + … + xkpk

E(X) = μx = Σxi pi

Values of X x1 x2 x3 … xk

Probability p1 p2 p3 … pk

EXAMPLE #1: An insurance company acknowledges that its payout probability for claims is as follows:

PolicyholderOutcome

Policyx

ProbabilityP(X = x)

Death $10000 1 / 1000 Disability $5000 2 / 1000 Neither $0 997 / 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) E(X) = = μμxx = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20 = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20

2.2. Profit = Revenue – ExpensesProfit = Revenue – Expenses = 200($100) – 200($20) = $16000= 200($100) – 200($20) = $16000

1. 1. E(X) = E(X) = μμxx = x = x11pp11 + x + x22pp22 + x + x33pp3 = 3 = Σxi·pi

What can the average student expect to make on the AP Exam?

μμxx = = E(X) E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = = 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 1 2 3 4 5Probability 0.14 0.24 0.34 0.21 0.07

μμxx = = E(X) E(X) = = 2.832.83

The VARIANCE OF A RANDOM VARIABLEThe 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: σX

2. The Standard Deviation is the square root of the Variance and is denoted by: σx.

Values of X x1 x2 x3 … xk

Probability p1 p2 p3 … pk

Let X = the random variable whose distribution follows and has mean :E(X) = μx = Σxi·pi

The Variance of a discrete random variable is:σx

2 = (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?

μμxx = = E(X) E(X) = 1(.14) + 2(.24) + 3(.34) + 4(.21) + 5(.07) = = 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 1 2 3 4 5Probability 0.14 0.24 0.34 0.21 0.07

μμxx = = E(X) E(X) = = 2.832.83σσxx

22 = (x = (x1 1 – – μμxx))22 pp11 + (x + (x22 – – μμxx))22 pp22 + … + (x + … + (xkk – – μμxx))22 p pkk

= (1 – 2.83)= (1 – 2.83)22·(.14)·(.14) + (2 – 2.83)+ (2 – 2.83)22·(.24) + (3 – 2.83)·(.24) + (3 – 2.83)22·· (.34) + (4 – 2.83)(.34) + (4 – 2.83)22·(.21) ·(.21) + (5 – 2.83)+ (5 – 2.83)22·· (.07) = (.07) = Σ (xi – μx)2 pi

σσxx22 = = 1.26111.2611

σσxx = = 1.1231.123

2. 2. σσxx22 = (x = (x1 1 – – μμxx))22 pp11 + (x + (x22 – – μμxx))22 pp22 + … + (x + … + (xkk – – μμxx))22 p pkk

= (10000 – 20)= (10000 – 20)22·(1/1000)·(1/1000) + (5000 – 20)+ (5000 – 20)22·(2/1000) + (0 – 20)·(2/1000) + (0 – 20)22·· (997/1000)(997/1000)

EXAMPLE #1: An insurance company acknowledges that its payout probability for claims is as follows:

PolicyholderOutcome

Policyx

ProbabilityP(X = x)

Death $10000 1 / 1000 Disability $5000 2 / 1000 Neither $0 997 / 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. 1. E(X) E(X) = = μμxx = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = = 10000(1/1000) + 5000(2/1000) + 0(997/1000) = $20$20

σσxx22 = 149600 = 149600

σσxx = √149600 = = √149600 = $386.78$386.78

The reason why you SHOULD NOT gamble!

Expected Payouts on a $1.00 Bet

E(X) = E(X) = μμxx = = xWINpWIN + xLOSEpLOSE

E(RED) = E(RED) = μμ = $= $1(18/38) + -$1(20/38) μμ = = $–0.05 E(25) = E(25) = μμ = $= $35(1/38) + -$1(37/38) μμ = = $–0.05

PAYOUTSColor 1 to 1Single # 35 to 1

E(RED)E(RED)

E(25)E(25)

The reason why you SHOULD NOT gamble!

Expected Payouts on a $1.00 Bet

E(X) = E(X) = μμxx = = xWINpWIN + xLOSEpLOSE

E(Corner) = E(Corner) = μμ = $= $8(4/38) + -$1(34/38) μμ = = $–0.05 E(Split) = E(Split) = μμ = $= $17(2/38) + -$1(36/38) μμ = = $–0.05

PAYOUTSCorner 8 to 1Split 17 to 1

E(Corner)E(Corner)

E(Split)E(Split)

The reason why you SHOULD NOT gamble!

Expected Payouts on a $1.00 Bet

E(X) = E(X) = μμxx = = xWINpWIN + xLOSEpLOSE

E(Street) = E(Street) = μμ = $= $11(3/38) + -$1(35/38) μμ = = $–0.05 E(Dozen) = E(Dozen) = μμ = $= $2(12/38) + -$1(26/38) μμ = = $–0.05

PAYOUTSStreet 11 to 1Dozen 2 to 1

E(Street)E(Street)

E(Dozen) =E(Dozen) =