• Your 3rd quiz will cover sections 4.1-4.3

a) {HHH,HTT,THT,TTH,THH,HTH,HHT,TTT}

{0,1,2,3}

b) {1/8,3/8,3/8,1/8}

d) P(x=2 or x=3)= P(x=2)+P(x=3)=3/8+1/8=1/2

Thinking Challenge

• A couple plans to have children until they get a girl, but they agree that they will not have more than three children even if they are all boys. (Assume boys and girls are equally likely)

• Create a probability model for the number of children they will have.

Thinking Challenge solution

• Sample space={G,BG,BBG,BBB}• X= number of children

x 1 2 3

P(x) 1/2 1/4 2/8

1. Expected Value (Mean of probability distribution)• Weighted average of all possible values

• = E(x) = x p(x)

2. Variance• Weighted average of squared deviation about mean

• 2 = E[(x (x p(x)=x2p(x)-2

Summary Measures

3. Standard Deviation2

Summary Measures Calculation Table

x p(x) x p(x) x –

Total (x p(x)

(x – (x – p(x)

xp(x)

Thinking Challenge

• Sample space={G,BG,BBG,BBB}• X= number of children

• Find the expected number of children • E(X)=1*1/2+2*1/4+3*2/8=7/4=1.75• Find the standard deviation of children they will have 2 = 1*1/2+4*1/4+9*2/8-(7/4)2 = .6875 = (0.6875)1/2=0.83

x 1 2 3

P(x) 1/2 1/4 2/8

Thinking Challenge

• Sample space={G,BG,BBG,BBB}• X= number of boys they will have

• Find the expected number of boys they will have• E(X)=1*1/4+2*1/8+3*1/8=7/8=0.875• Find the standard deviation of children they will have 2 = 1*1/4+4*1/8+9*1/8-(7/8)2 = 1.109 = (1.109)1/2=1.053

x 0 1 2 3

P(x) 1/2 1/4 1/8 1/8

Thinking Challenge

You toss 2 coins. You’re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails?

•© 1984-1994 T/Maker Co.

Expected Value & Variance Solution*

0 .25 –1.00 1.00

1 .50 0 0

2 .25 1.00 1.00

0

.50

.50

=1.0

x p(x) x p(x) x – (x – (x – p(x)

.25

0

.25

Probability Rules for Discrete Random Variables

Let x be a discrete random variable with probability distribution p(x), mean µ, and standard deviation . Then, depending on the shape of p(x), the following probability statements can be made:

• Chebshev's Rule Empirical RuleP(-≤x≤+) ≥ 0 0.68P(-2≤x≤+2) ≥ 3/4 0.95P(-3≤x≤+3) ≥ 8/4 0.997

• Chebshev's Rule Empirical RuleP(-≤x≤+) ≥ 0 0.68P(-2≤x≤+2) ≥ 3/4 0.95P(-3≤x≤+3) ≥ 8/4 0.997

a) 34.5, 174.75,13.219

c) 34.5 ±2*13.219=8.062,60.938

The probability that x will fall in this interval is 1.00.

Under the empirical rule we expect that this value would be 0.95.

4.3

The Binomial Distribution

Binomial DistributionNumber of ‘successes’ in a sample of n observations (trials)

• Number of reds in 15 spins of roulette wheel

• Number of defective items in a batch of 5 items

• Number correct on a 33 question exam

• Number of customers who purchase out of 100 customers who enter store (each customer is equally likely to purchase)

Binomial ProbabilityCharacteristics of a Binomial Experiment

1.The experiment consists of n identical trials.

2.There are only two possible outcomes on each trial. We will denote one outcome by S (for success) and the other by F (for failure).

3.The probability of S remains the same from trial to trial. This probability is denoted by p, and the probability of F is denoted by q. Note that q = 1 – p.

4.The trials are independent.

5.The binomial random variable x is the number of S’s in n trials.

Binomial Probability Distribution Example

3 5 3

!( ) (1 )

!( )!

5!(3) .5 (1 .5)

3!(5 3)!

.3125

x n xnp x p p

x n x

p

Experiment: Toss 1 coin 5 times in a row. Note number of tails. What’s the probability of 3 tails?

•© 1984-1994 T/Maker Co.

Binomial Probability Distribution

!( ) (1 )

! ( )!x n x x n xn n

p x p q p px x n x

p(x) = Probability of x ‘Successes’

p = Probability of a ‘Success’ on a single trial

q = 1 – p

n = Number of trials

x = Number of ‘Successes’ in n trials (x = 0, 1, 2, ..., n)

n – x = Number of failures in n trials

Binomial Probability Table (Portion)

n = 5 •p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

Cumulative Probabilities

P(x=3)=P(x ≤ 3) – P(x ≤ 2) = .812 – .500 = .312

Binomial Distribution Characteristics

.0

.5

1.0

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

E (x ) np•Mean

•Standard Deviation

npq

Thinking Challenge

• You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right?

• If you guessed on all 33 questions, what would be your grade? Would you pass?

X= number of correct answers

•Binomial random variable since we just guess the answer.

•Total number of trials=33

•p= 0.25

•Expected number of correct answers=33*0.25=8.25

Binomial Distribution Thinking Challenge

You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability of

A. No sales?

B. Exactly 2 sales?

C. At most 2 sales?

D. At least 2 sales?

Binomial Distribution Solution*

n = 12, p = .20

E(X)=n*p=12*0.2=2.4

=(np(1-p))1/2=(12*0.2*0.8)1/2 =1.38

A. p(0) = .0687

B. p(2) = .2835

C. p(at most 2) = p(0) + p(1) + p(2)= .0687 + .2062 + .2835= .5584

D. p(at least 2) = p(2) + p(3)...+ p(12)= 1 – [p(0) + p(1)] = 1 – .0687 – .2062= .7251

Using binomial table n = 10, p = .20

•A. P(X = 0) = p(0) = P(X ≤ 0) = .107 (Table I)

Using binomial table n = 10, p = .20

•B.P(X = 2) = p(2) = P(X ≤ 2) – P(X ≤ 1)

= .678 - .376 = .302

Using binomial table n = 10, p = .20

•C. P(at most 2) = P(X ≤ 2) = .678

Using binomial table n = 10, p = .20

• D.P(at least 2) = P(X ≥ 2) = 1 – P(X ≤ 1)

= 1 – .376 = .724

By using TI-84:

• B. P(X = 2) = p(2) = P(X ≤ 2) – P(X ≤ 1)

• binomcdf(10,.20,2) - binomcdf(10,.20,1)

a) Adult not working during summer vacation.

b)

•The experiment consist of 10 identical trials. A trial for this experiment is an individual.

•There are only two possible outcomes: work or do not work

•The probability remains same for each individual (trial)

•Individuals are independent

c) 0.35

d) 0.2522

e) 0.2616

Thinking Challenge

• The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages.

• What are the mean and standard deviation of the number of real messages you should expect to find in your inbox?

• What is the probability that you will find only 1 or 2 real messages?

Thinking Challenge

• The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages.

• What are the mean and standard deviaiton of the number of real messages you should expect to find in your inbox? (2.25, 1.43)

• What is the probability that you will find only 1 or 2 real messages? (0.5117)