Download Chapter 5 Sections 1,2 Discrete Probability Distributions

Chapter 5
Discrete Probability Distributions
M227
Sections 1,2
OBJECTIVES
Construct a probability distribution for a random variable.
Find the mean, variance, standard deviation, and expected value for a discrete random
variable.
• Find the exact probability for X successes in n trials of a binomial experiment.
• Find the mean, variance, and standard deviation for the variable of a binomial
distribution.
•
•
INTRODUCTION
• Many decisions in business, insurance, and other real-life situations are made by
assigning probabilities to all possible outcomes pertaining to the situation and then
evaluating the results.
• This chapter explains the concepts and applications of discrete probability
distributions. In addition, a special probability distribution, the binomial distribution,
is explained.
Random Variables
• A random variable is a variable whose values are determined by chance.
• A variable is a characteristic or attribute that can assume different values.
Examples: A die is rolled; a letter, say X, can be used to represent the outcomes. Then
the values that X can assume are 1,2,3,4,5,6.
Two coins are tossed; a letter, say Z, can be used to represent the number of
heads; thus Z can assume the values 0,1,2.
Discrete Probability Distribution
A discrete probability distribution consists of the values a random variable can assume and
the corresponding probabilities of the values. The probabilities are determined
theoretically or by observation.
Example: A coin is tossed three times. Let X represent the number of heads. X can take
the values: 0,1,2,3. Compute the probabilities for each value of X :
X = 0 No Heads TTT 1/8
X = 1 TTH, THT, HTT, 1/8 + 1/8 + 1/8 = 3/8
X = 2 HHT, HTH, THH 1/8 + 1/8 + 1/8 = 3/8
X = 3 HHH
1/8
Summarize information in a table form as follows:
Number of heads
X
0
1
2
3
Probability P(X)
1/8
3/8
3/8
1/8
Section 5-1, 5-2
Page 1
Chapter 5
Discrete Probability Distributions
M227
•
Sections 1,2
Above example illustrates a theoretical probability distribution. Probability distributions
can be constructed from empirical probabilities as well.
Example: Let X be the number of saws rented per day. During a 90 day period the data
are:
X
0
1
2
Number of days
45
30
15
90 Total
P(0) = 45/90 = 0.5
P(1) = 30/90 = 0.33
P(2) = 15/90 = 0.17
Number of saws rented
X
Probability P(X)
0
1
2
0.5
0.33
0.17
Requirements for Probability Distribution
•
•
The sum of the probabilities of all events in the sample space must equal 1.
The probability of each event in the sample space must be between or equal to 0 and 1
We can use the above criteria to determine if a given distribution van qualify for a
probability distribution.
Section 5-1, 5-2
Page 2
Chapter 5
Discrete Probability Distributions
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Section 3
Formula for the Mean of a Probability Distribution
In order to find the mean for a probability distribution, one must multiply each possible
outcome by its corresponding probability and find the sum of the products.
µ = X 1 ⋅ P( X 1 ) + X 2 ⋅ P( X 2 ) + ... + X n ⋅ P( X n ) =
∑ X ⋅ P( X ) ,
where X 1 , X 2 , .... X n are the
outcomes and P( X 1 ), P( X 2 ), ... P( X n ) are the corresponding probabilities
Rounding Rule
The mean, variance, and standard deviation should be rounded to one more decimal place
than the outcome, X.
Example: Find the mean of the number of heads that occur when three coins are tossed.
The probability distribution is
Number of heads X
Probability P(X)
0
1/8
1
8
3
8
1
3/8
3
8
1
8
µ = 0 ⋅ + 1⋅ + 2 ⋅ + 3 ⋅ =
The mean is:
2
3/8
3
1/8
12
= 1.5
8
Formula for the Variance and Standard Deviation of a Probability Distribution
σ 2 = ∑ [X 2 ⋅ P( X )] − µ 2
Variance:
Standard Deviation: σ =
∑  X 2 ⋅ P( X )  − µ 2
Example: Five balls numbered 0, 2, 4, 6, and 8 are placed in a bag. After the balls are mixed, one
is selected, its number is noted, and then it is put back in the bag. If this experiment is repeated
many times, find the mean, variance, and standard deviation.
Let X be the number on each ball. The probability distribution is:
X
P(X)
1
5
0
1/5
1
5
1
5
2
1/5
1
5
4
1/5
6
1/5
1
5
µ = 0⋅ + 2⋅ + 4⋅ + 6⋅ + 8⋅ = 4

1
1
1
1
1
σ 2 = ∑  X 2 ⋅ P( X )  − µ 2 = 02 ⋅ + 22 ⋅ + 42 ⋅ + 62 ⋅ + 82 ⋅  − 42 = 8
5
5
5
5
5

σ = 8 = 2.8
Section 5-3
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Chapter 5
Discrete Probability Distributions
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Section 3
Expected value or expectation is used in various types of games of chance, in insurance,
and in other areas, such as decision theory
• The symbol E(X) is used for the expected value.
• The expected value of a discrete random variable of a probability distribution is the
theoretical average of the variable. The formula is:
µ = E ( X ) = ∑ X ⋅ P( X )
Example: 1,000 tickets are sold at $1 each for a color television valued at $300. What is
the expected value of the gain if a person purchases one ticket?
Gain X
PX)
E ( X ) = 349 ⋅
Win
$349
1/1000
Lose
-$1
999/1000
1
999
+ ( −1) ⋅
= −0.65
1000
1000
Meaning: Average of the losses is $0.65 for each of the 1,000 ticket holders.
Section 5-3
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Chapter 5
Discrete Probability Distributions
Section 4
BINOMIAL DISTRIBUTION
• Many types of probability problems have only two possible outcomes or they can be
reduced to two outcomes.
• Examples include: when a coin is tossed it can land on heads or tails, when a baby is
born it is either a boy or girl, etc.
• Situations like that are called binomial experiments.
A binomial experiment is a probability experiment that satisfies the following four
requirements:
Each trial can have only two possible outcomes—success or failure.
There must be a fixed number of trials.
The outcomes of each trial must be independent of each other.
The probability of success must remain the same for each trial.
•
The outcomes of a binomial experiment and the corresponding probabilities of these
outcomes are called a binomial distribution.
Notation for Binomial Distributions
P( S )
P( F )
p
q
n
X
The symbol for the probability success
The symbol for the probability failure
The numerical probability of success
The numerical probability of failure
P( S ) = p
and P ( F ) = 1 − p = q
The number of trials
The number of successes in n trials
Binomial Probability Formula
In a binomial experiment, the probability of exactly X successes in n trials is
n!
P( X ) =
⋅ p X ⋅ q n− X
(n − X )! X !
Example: A survey found that 30% of teenage consumers receive their spending money
from part-time jobs. If 5 teenagers are selected at random, find the probability that at least
3 of them will have part-time jobs.
5!
P(3 have part − time jobs ) =
(0.3)3 (0.7)2 = 0.132
(5 − 3)!3!
5!
P(4 have part − time jobs ) =
(0.3)4 (0.7)1 = 0.028
(5 − 4)!4!
5!
P(5 have part − time jobs ) =
(0.3)5 (0.7)0 = 0.002
(5 − 5)!5!
P(at least 3 have part-time jobs) = 0.132 + 0.028 + 0.002 = 0.162
•
Tables of Probabilities for Binomial Distributions : Appendix C, Table B
Section 5-4
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Chapter 5
Discrete Probability Distributions
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Section 4
Mean, Variance, and Standard Deviation for the Binomial Distribution
Mean:
µ = n⋅ p
Variance:
σ2 = n⋅ p⋅q
Standard Deviation:
σ2 = n⋅ p⋅q
EXCEL: Use of Excel to create Probabilities of Binomial Distributions: BINOMDIST
build-in function.
Example: let n = 20 and p = 0.3; then each cell in the second column represents the
probability that exactly X successes will occur in these 20 trials.
The formula in cell 2 of column B looks like: =BINOMDIST(A2,20,0.3,False), A2=0;
The formula in cell 3 of column B looks like: =BINOMDIST(A3,20,0.3,False), A3=1;
And so on.
Mean
StDev
Section 5-4
Binomila Distribution p=0.3
0.25
0.2
0.15
0.1
0.05
6
2.05
Page 6
20
18
16
14
12
10
8
6
0
4
P(X)
0.00079792
0.00683934
0.02784587
0.07160367
0.13042097
0.17886305
0.19163898
0.16426199
0.11439674
0.06536957
0.03081708
0.01200665
0.00385928
0.00101783
0.00021811
3.739E-05
5.0076E-06
5.0496E-07
3.6069E-08
1.6272E-09
3.4868E-11
2
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
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