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• 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)