描述统计学chap05

Business Statistics B siness Statistics
Chapter 5
Some Important Discrete Probability Distributions Probability Distributions
Chapter Goals
Chapter Goals
After completing this chapter, you should be Aft l ti thi h t h ld b able to:
Interpret the mean and standard deviation for a I t t th d t d d d i ti f discrete probability distribution
Explain covariance and its application in finance E l i i d it li ti i fi
Use the binomial probability distribution to find probabilities
b biliti
Describe when to apply the binomial distribution Use Poisson discrete probability distributions to find probabilities
Definitions
Random Variables
Random Variables
A random variable represents a possible numerical value from an uncertain event. numerical value from an uncertain event.
Discrete random variables produce outcomes random variables produce outcomes that come from a counting process (e.g. number of courses you are taking this semester).
of courses you are taking this semester).
random variables produce Continuous random variables produce outcomes that come from a measurement (e.g. your annual salary, or your weight).
your annual salary,or your weight).
R d V Definitions i bl
Random Variables Random
Variables
C Discrete Continuous
Random Variable Random Variable Ch. 5Ch. 6
Can only assume a countable number of values C
Probability Distribution for a
Discrete Random Variable
Discrete Random Variable
A probability distribution (or probability mass function )(pdf) A b bilit di t ib ti(b bilit f ti)(df for a discrete random variable is a mutually exclusive listing of all possible numerical outcomes for that random variable of all possible numerical outcomes for that random variable such that a particular probability of occurrence is associated with each outcome.
Number of Classes Probability
Taken
20.2
30.4
04
40.24
50.16
016
Experiment: Toss 2 Coins. Let X = # heads.
4 possible outcomes
4possible
S M
Summary Measures
E t d V l()f di t
Expected Value (or mean)of a discrete distribution (Weighted Average)
∑
=
=
=
µ
N
1
i
i
i
)
X
(P
X
E(X)
E l T2i
X P(X) Example:Toss 2 coins,
X= # of heads,
t t d l f X
0 .25
1 .50
compute expected value of X:
E(X) = (0 x .25) + (1 x .50) + (2 x .25)
=10
2 .25
= 1.0
(continued) Variance
(continued)σ
The Covariance The Covariance
The covariance measures the strength of the
linear relationship between two variables ea e a o s p be ee o a ab es The covariance :
)
Y X (P )]Y (E Y )][(X (E X [σN
i i i i XY ∑−−=1
i =where:X = discrete variable X
=the i th outcome of X X i = the i outcome of X Y = discrete variable Y Y i = the i th outcome of Y )=probability of occurrence of the condition affecting
P(X i Y i ) probability of occurrence of the condition affecting the i th outcome of X and the i th outcome of Y
Computing the Mean for I R
Investment Returns Return per $1,000 for two types of investments
P(X i Y i )Economic condition Passive Fund X
Aggressive Fund Y
Investment
.2Recession -$ 25-$200.5Stable Economy + 50+ 60.3
Expanding Economy
+ 100
+ 350
E(X) = µX = (-25)(.2) +(50)(.5) + (100)(.3) = 50E(Y)(200)(2)(60)(5)(350)(3)95E(Y) = µY = (-200)(.2) +(60)(.5) + (350)(.3) = 95
Computing the Standard Deviation
P(X
.2
.5
.3
Computing the Covariance f I R
for Investment Returns P(X i Y i )Economic condition Passive Fund X
Aggressive Fund Y
Investment
.2Recession -$ 25-$200.5Stable Economy + 50+ 60.3
Expanding Economy
+ 100
+ 350
95)(.5)
50)(60(5095)(.2)200-50)((-25σY X ,−−−−+−−=8250
95)(.3)50)(350(100=+
Interpreting the Results for
I R
Investment Returns
The aggressive fund has a higher expected
return,but much more risk return, but much more risk
=95>µ=50
µY = 95 > µX = 50but
19321>4330
σY = 193.21 > σX = 43.30 The Covariance of 8250 indicates that the two
investments are positively related and will vary in the same direction in the same direction
Portfolio Example
Portfolio Example
Returns
Probability Stock X Stock Y
0.1-$100$50
01$100
0.30150
0.380-20
0.3150-100
You are trying to develop a strategy for investing in two different stocks. The anticipated annual return for a different stocks The anticipated annual return for a $1,000 investment in each stock has the above probability distributions.
probability distributions.
Compute the
1) Expected return for stock X
2) Expected return for stock Y
3) Standard deviation for stock X
3)Standard deviation for stock X
4) Standard deviation for stock Y
5) Covariance of stock X and stock Y
5)C i f t k X d t k Y
6) Do you think you will invest in stock X or stock Y?
Explain.
Suppose you wanted to create a portfolio that consists of stock X and stock Y. Compare the portfolio expected return and portfolio risk for each of the following proportions invested in stock X:
7) 0.1 8) 0.7
7)018)07
9)0.3 10) 0.9 11) 0.5
On the basis of the results of 7)—11), which portfolio would you recommend? Explain.
Probability Distributions Probability Distributions Probability Distributions Continuous Probability Discrete Probability Ch. 5Ch. 6
Probability
Distributions
Probability Distributions Binomial Normal
Hypergeometric Uniform
Poisson Exponential
Binomial Probability Distribution Binomial Probability Distribution
A fixed number of observations n A fixed number of observations, n
e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse Each observation is categorized as to whether or not the g
“event of interest” occurred
e.g., head or tail in each toss of a coin; defective or not defective light b lb bulb Since these two categories are mutually exclusive and collectively exhaustive(Generally called “success ” and “failure ”) When the probability of the event of interest is represented as π, then the probability of the event of interest not occurring is 1 -πConstant probability for the event of interest occurring ()for Constant probability for the event of interest occurring (π) for each observation
Probability of getting a tail is the same each time we toss the coin y g g
Binomial Probability Distribution Binomial Probability Distribution
(continued)
Observations are independent
The outcome of one observation does not affect the
outcome of the other
p g p
Two sampling methods deliver independence
Infinite population without replacement
Finite population with replacement
Possible Applications for the
pp
Binomial Distribution
A manufacturing plant labels items as either defective or acceptable
either defective or acceptable
A firm bidding for contracts will either get a contract or not
A marketing research firm receives survey A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer Ne job applicants either accept the offer or reject it
The Binomial Distribution
Counting Techniques
Counting Techniques
Suppose the event of interest is obtaining heads on the toss of a fair coin.You are to toss the coin three times. toss of a fair coin. You are to toss the coin three times. In how many ways can you get two heads?
Possible ways: HHT, HTH, THH, so there are three
P ibl HHT HTH THH th th
ways you can getting two heads.
This situation is fairly simple. We need to be able to count the number of ways for more complicated situations.
31
Binomial Distribution Formula Binomial Distribution Formula
P(X)n X !π(1-π)X n X !=−P(X)=probability of events of interest in X !
n X ()!−P(X) = probability of X events of interest in n
trials, with the probability of an “event of
interest” being πfor each trial
Example:Flip a coin four times, let x = # heads:X = number of “events of interest” in sample, (X = 0, 1, 2, ..., n )
l i (b f t i l
,n = 4π= 0.5n = sample size (number of trials or observations)
π= probability of “event of interest” 1 -π= (1 -0.5) = 0.5X = 0, 1, 2, 3, 4
p y
The Binomial Distribution E Example l
Suppose the probability of purchasing a defective computer p is 0.02. What is the probability p y of purchasing 2 defective computers in a group of 10? X = 2, n = 10, and π = .02
n! P(X = 2) = π X (1 − π ) n − X X!(n − X)! 10! = (.02) 2 (1 − .02)10 − 2 2!(10 − 2)! = (45)(.0004)(.8508) = .01531
5-31


Example
ƒ Suppose that 20% of all copies of a particular textbook fail a certain binding strength test test. Let X denote the number among 15 randomly selected copies that fail the test test.
ƒ Find the probability that at most 8 fail the test ƒ Find the probability that exactly 8 fail the test ƒ Find the probability that at least 8 fail the test
5-32


The Binomial Distribution Shape
ƒ The shape of the bi binomial i l di distribution t ib ti depends on the values of π and n „ Here, n = 5 and π = .1
P(X) .6 .4 .2 0 0 P(X) .6 .4 .2 0 0 1 2 3 4 5 X
5-33
n = 5 π = 0.1
1
2
3
4
5
X
n = 5 π = 0.5
„
Here, , n = 5 and π = .5


The Binomial Distribution Using Binomial Tables
n = 10 x 0 1 2 3 4 5 6 7 8 9 10 … … … … … … … … … … … … … π=.20 0.1074 0 2684 0.2684 0.3020 0.2013 0.0881 0 0264 0.0264 0.0055 0.0008 0.0001 0 0000 0.0000 0.0000 π=.80 π=.25 0.0563 0 1877 0.1877 0.2816 0.2503 0.1460 0 0584 0.0584 0.0162 0.0031 0.0004 0 0000 0.0000 0.0000 π=.75 π=.30 0.0282 0 1211 0.1211 0.2335 0.2668 0.2001 0 1029 0.1029 0.0368 0.0090 0.0014 0 0001 0.0001 0.0000 π=.70 π=.35 0.0135 0 0725 0.0725 0.1757 0.2522 0.2377 0 1536 0.1536 0.0689 0.0212 0.0043 0 0005 0.0005 0.0000 π=.65 π=.40 0.0060 0 0403 0.0403 0.1209 0.2150 0.2508 0 2007 0.2007 0.1115 0.0425 0.0106 0 0016 0.0016 0.0001 π=.60 π=.45 0.0025 0 0207 0.0207 0.0763 0.1665 0.2384 0 2340 0.2340 0.1596 0.0746 0.0229 0 0042 0.0042 0.0003 π=.55 π=.50 0.0010 0 0098 0.0098 0.0439 0.1172 0.2051 0 2461 0.2461 0.2051 0.1172 0.0439 0 0098 0.0098 0.0010 π=.50 10 9 8 7 6 5 4 3 2 1 0 x
Examples:
n = 10, π = .35, x = 3: n = 10, π = .75, x = 2: P(x = 3|n =10, π = .35) = .2522 P(x = 2|n =10, π = .75) = .0004
5-34


Binomial Distribution Ch Characteristics t i ti
ƒ Mean
„
µ = E(x) ( ) = nπ
σ = nπ ( (1 - π )
σ = nπ ( (1 - π )
2
Variance and Standard Deviation
Where
n = sample size π = probability of the event of interest for any trial (1 – π) = probability of no event of interest for any trial
5-35


The Binomial Distribution Characteristics
Examples
µ = nπ = (5)(.1) (5)( 1) = 0.5 05
σ = nπ (1 - π ) = (5)(.1)(1 − .1) = 0.6708
P(X) .6 .4 .2 0 0 P(X) .6 .4 .2 0 0 1 1
n = 5 π = 0.1
2
3
4
5
X
µ = nπ = (5)(.5) = 2.5
σ = nπ (1 - π ) = (5)(.5)(1 − .5) = 1.118 1 118
n = 5 π = 0.5
2
3
4
5
X
5-36


Using Excel For The Binomial Distribution
5-37


Using PHStat
ƒ Select PHStat / Probability & Prob. Distributions / Binomial…
5-38


Using PHStat
(continued)
ƒ Enter desired values in dialog box Here: n = 10 π = .35 Output for X = 0 to X = 10 will be generated by PHStat Optional check boxes for additional output
5-39


PHStat Output
P(X = 3 | n = 10, π = .35) = .2522
P(X > 5 | n = 10, 10 π = .35) 35) = .0949 0949
5-40


=Binomdist (numbers_s, trials,
probability_s, cumulative)
b bilit l ti)
The Hypergeometric Distribution The Hypergeometric Distribution
“n” trials in a sample taken from a finite
of size N
population of size N
Sample taken without replacement
Outcomes of trials are dependent
Concerned ith finding the probabilit of“X”
Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population
■Example:
Example
Five individuals from an animal population thought to be near extinction in a certain region thought to be near extinction in a certain region
have been caught, tagged, and released to
mixed into the population. After they have had
p p y
an opportunity to mix, a random sample of 10 of
these animals is selected . Let X be the number
of tagged animals in the second sample. If there
f t d i l i th d l If th
are actually 25 animals of this type in the region,
what is the probability that
what is the probability that
1) X=2?
2) X2?
2)X≤
Hypergeometric Distribution
in PHStat
i PHS
Select:
PHStat / Probability & Prob. Distributions / Hypergeometric …
Hypergeometric Distribution
in PHStat
i PHSt t
(continued)
Complete dialog box entries and get output…
N 10n 3
N=10n=3
A = 4X = 2
P(X = 2) = 0.3
The Poisson Distribution
The Poisson Distribution
Apply the Poisson Distribution when:
You wish to count the number of times an event occurs in a given area of opportunity
The probability that an event occurs in one area of t it i th f ll f t it opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events opportunity is independent of the number of events that occur in the other areas of opportunity
The probability that two or more events occur in an The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller
The average number of events per unit is λ(lambda)。