Transcript Document

• 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
Summary Measures
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
3.
Standard Deviation
  2
Summary Measures
Calculation Table
x
p(x)
Total
x p(x)
x p(x)
x–
(x – 
(x – p(x)
(x  p(x)
Thinking Challenge
• Sample space={G,BG,BBG,BBB}
• X= number of children
x
1
2
3
P(x)
1/2
1/4
2/8
• 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
Thinking Challenge
• Sample space={G,BG,BBG,BBB}
• X= number of boys they will have
x
0
1
2
3
P(x)
1/2
1/4
1/8
1/8
• 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
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*
x
p(x)
x p(x)
x–
0
.25
0
–1.00
1.00
.25
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
 =1.0
(x –   (x –  p(x)
 .50
 .71
P(-≤x≤+)
P(-2≤x≤+2)
P(-3≤x≤+3)
≥
≥
≥
0
3/4
8/4
0.68
0.95
0.997
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:
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 Distribution
Number 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 Probability
Characteristics 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
Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3
n!
tails?
p ( x) 
p x (1  p) n  x
x !(n  x)!
•© 1984-1994 T/Maker Co.
5!
p (3) 
.53 (1  .5)53
3!(5  3)!
 .3125
Binomial Probability
Distribution
 n  x n x
n!
x
n x
p( x)    p q 
p (1  p )
x ! ( n  x )!
 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)
•p
n=5
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
n = 5 p = 0.1
•Mean
  E ( x)  np
P(X)
1.0
.5
.0
X
•Standard Deviation
  npq
0
1
2
3
4
5
n = 5 p = 0.5
.6
.4
.2
.0
P(X)
X
0
1
2
3
4
5
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)
c) 0.35
d) 0.2522
e) 0.2616
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
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)