**Unformatted text preview: **Business Statistics (BUSA 3101)
Dr. Lari H. Arjomand
[email protected] Slide 1 Chapter 6 Discrete Probability Distributions
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s
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s Random Variables
Discrete Probability Distributions
Expected Value and Variance
Binomial Distribution
Poisson Distribution (Optional Reading)
Hypergeometric Distribution (Optional Reading) .40
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.10 0 1 2 3 4 Slide 2 Random Variables A random variable is a numerical description of the
A random variable is a numerical description of the outcome of an experiment.
outcome of an experiment. A discrete random variable may assume either a
A discrete random variable may assume either a finite number of values or an infinite sequence of
finite number of values or an infinite sequence of values.
values. A continuous random variable may assume any
A continuous random variable may assume any numerical value in an interval or collection of
numerical value in an interval or collection of intervals.
intervals. Slide 3 Example: JSL Appliances
s Discrete random variable with a finite number of values Let x = number of TVs sold at the store in one day,
Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4) where x can take on 5 values (0, 1, 2, 3, 4) Slide 4 Example: JSL Appliances
s Discrete random variable with an infinite sequence of values Let x = number of customers arriving in one day,
Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . . where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no
finite upper limit on the number that might arrive. Slide 5 Random Variables
Examples
Random Variable x Type Family
size x = Number of dependents reported on tax return Discrete Distance from
home to store x = Distance in miles from home to the store site Continuous Own dog
or cat x = 1 if own no pet; = 2 if own dog(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s) Discrete Question Slide 6 Random Variables
Definition & Example Definition: A random variable is a quantity resulting from a random experiment that, by chance, can assume different values.
Example: Consider a random experiment in which a coin is tossed three times. Let X be the number of heads. Let H represent the outcome of a head and T the outcome of a tail. Slide 7 Example (Continued) The sample space for such an experiment will be: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH. Thus the possible values of X (number of heads) are X = 0, 1, 2, 3. This association is shown in the next slide.
Note: In this experiment, there are 8 possible outcomes in the sample space. Since they are all equally likely to occur, each outcome has a probability of 1/8 of occurring. Slide 8 Example (Continued)
Example TTT
TTT
TTT
TTT
TTH
TTH 11 THT
THT 11 THH
THH 22 HTT
HTT 11 HTH
HTH
Sample
Space 00 22 HHT
HHT 22 HHH
HHH 33 X Slide 9 Example (Continued) The outcome of zero heads occurred only once.
The outcome of one head occurred three times.
The outcome of two heads occurred three times.
The outcome of three heads occurred only once.
From the definition of a random variable, X as defined in this experiment, is a random variable. X values are determined by the outcomes of the experiment. Slide 10 Probability Distribution: Definition Definition: A probability distribution is a listing of all the outcomes of an experiment and their associated probabilities.
The probability distribution for the random variable X (number of heads) in tossing a coin three times is shown next. Slide 11 Probability Distribution for Three Tosses of a Coin Slide 12 Data Types
Data Numerical
(Quantitative) Discrete Categorical
(Qualitative) Continuous Slide 13 Discrete Random Variable Examples
Experiment Random
Variable Possible
Values Make 100 sales calls # Sales Inspect 70 radios # Defective 0, 1, 2, ..., 70 Answer 33 questions # Correct Count cars at toll
# Cars
between 11:00 & 1:00 arriving 0, 1, 2, ..., 100 0, 1, 2, ..., 33
0, 1, 2, ..., ∞ Slide 14 Discrete Probability Distributions The probability distribution for a random variable
The probability distribution for a random variable describes how probabilities are distributed over
describes how probabilities are distributed over the values of the random variable.
the values of the random variable. We can describe a discrete probability distribution
We can describe a discrete probability distribution with a table, graph, or equation.
with a table, graph, or equation. Slide 15 Discrete Probability Distributions The probability distribution is defined by a
The probability distribution is defined by a probability function, denoted by ff(x), which provides
probability function , denoted by (x), which provides the probability for each value of the random variable.
the probability for each value of the random variable. The required conditions for a discrete probability
The required conditions for a discrete probability function are:
function are:
ff(x) > 0
Σf(x) = 1 P(X) ≥ 0
ΣP(X) = 1 Slide 16 Discrete Probability Distributions
Example
s Using past data on radio sales, …
s a tabular representation of the probability distribution for radio sales was developed. Number Units Sold of Days x f(x)
0 6 0 .03 1 40 1 .2 2 100 2 .5 3 40 3 .2 4 10 4 .05 5 4 5 .02 200 1.00 6/200 Slide 17 Discrete Probability Distributions
Graphical Representation of Probability Distribution
.50
Probability s .40 .30
.20
.10
0 1 2 3 4 Values of Random Variable x (TV sales) Slide 18 Discrete Probability Distributions As we said, the probability distribution of a discrete random variable is a table, graph, or formula that gives the probability associated with each possible value that the variable can assume.
Example : Number of Radios Sold at Sound City in a Week
x, Radios
p(x), Probability 0
p(0) = 0.03 1
p(1) = 0.20 2
p(2) = 0.50 3
p(3) = 0.20 4
p(4) = 0.05 5
p(5) = 0.02 Slide 19 Expected Value of a Discrete Random Variable The mean or expected value of a discrete random variable is: µX = ∑xp( x)
All x Example: Expected Number of Radios Sold in a Week
x, Radios
p(x), Probability
x p(x)
0
p(0) = 0.03
0(0.03) = 0.00
1
p(1) = 0.20
1(0.20) = 0.20
2
p(2) = 0.50
2(0.50) = 1.00
3
p(3) = 0.20
3(0.20) = 0.60
4
p(4) = 0.05
4(0.05) = 0.20
5
p(5) = 0.02
5(0.02) = 0.10 1.00
2.10 20
Slide Variance and Standard Deviation The variance of a discrete random variable is: σ = ∑ ( x − µ X ) p( x)
2
X 2 All x The standard deviation is the square root of the variance. σX = σ 2
X Slide 21 Variance and Standard Deviation
Example: Variance and Standard Deviation of the Number of
Radios Sold in a Week x, Radios
0
1
2
3
4
5 p(x), Probability
p(0) = 0.03
p(1) = 0.20
p(2) = 0.50
p(3) = 0.20
p(4) = 0.05
p(5) = 0.02
1.00 µx = 2.10 (x - µX)2 p(x)
(0 – 2.1)2 (0.03) = 0.1323
(1 – 2.1)2 (0.20) = 0.2420
(2 – 2.1)2 (0.50) = 0.0050
(3 – 2.1)2 (0.20) = 0.1620
(4 – 2.1)2 (0.05) = 0.1805
(5 – 2.1)2 (0.02) = 0.1682
0.8900 Variance σ = 0.89
2
X Standard deviation Slide 22 σ X = 0.89 = 0.9434 Expected Value and Variance (Summary) The expected value, or mean, of a random variable
The expected value, or mean, of a random variable is a measure of its central location.
is a measure of its central location.
E(x) = µ = Σxf(x) The variance summarizes the variability in the
The variance summarizes the variability in the values of a random variable.
values of a random variable.
Var(x) = σ 2 = Σ(x µ)2f(x) The standard deviation, σ, is defined as the positive
The standard deviation, σ, is defined as the positive square root of the variance.
square root of the variance. Slide 23 Discrete Probability Distribution Models
Discrete
Probability
Distribution Binomial HyperGeometric Negative
Binomial Poisson Slide 24 Binomial Distribution
s Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
1. The experiment consists of a sequence of n identical trials. identical trials.
2. Two outcomes, success and failure, are possible
2. Two outcomes, success and failure, are possible on each trial. on each trial.
3. The probability of a success, denoted by p, does
3. The probability of a success, denoted by p, does not change from trial to trial. not change from trial to trial.
stationarity
assumption
4. The trials are independent.
4. The trials are independent. Slide 25 Binomial Distribution Our interest is in the number of successes
Our interest is in the number of successes occurring in the n trials.
occurring in the n trials. We let x denote the number of successes
We let x denote the number of successes occurring in the n trials.
occurring in the n trials. Slide 26 Binomial Distribution
s Binomial Probability Function n!
f (x ) = p x (1 − p )( n − x )
x !(n − x )! where: f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial Slide 27 Binomial Distribution
s Binomial Probability Function n!
f (x ) = p x (1 − p )( n − x )
x !(n − x )! n!
x !(n − x )!
Number of experimental
Number of experimental outcomes providing exactly
outcomes providing exactly
x successes in n trials
x successes in n trials p x (1 − p )( n − x )
Probability of a particular
Probability of a particular sequence of trial outcomes
sequence of trial outcomes with x successes in n trials
with x successes in n trials Slide 28 Thinking Challenge Example
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? Slide 29 Thinking Challenge Solutions
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 Slide 30 Using SWStat+ P (X=0) = ?
(Statistics; Probability Distributions) Slide 31 Using SWStat+ P(X≥2) = ? & P(X≤2)=? & P(X=2)=?
(Statistics; Probability Distributions) Slide 32 Using SWStat+ to Graph Slide 33 Thinking Challenge Example The Department of Labor Statistics for the state of Kentucky reports that 2% of the workforce in Treble County is unemployed. A sample of 15 workers is obtained from the county. Compute the following probabilities (Hint Binomial):
three are unemployed. Note: (n = 15, p = 0.02).
P(x= 3) = 0.0029 (from Binomial Table). three or more are unemployed.
P(x ≥ 3) = 1 [0.7386 +0.2261 + 0.0323] = 0.0031. Slide 34 Using SWStat+ P (X=3) = ? & P(X≥3) = ?
(Statistics; Probability Distributions) Slide 35 Thinking Challenge Example (Continued) at least one of the sampled workers is unemployed. P(x ≥ 1) = 1 P(x = 0) = 1 – 0.7386 = 0.2614 at most two of the sampled workers are unemployed. P(x ≤ 2) = 0.7386+ 0.2261 + 0.0323 = 0.997 2) = Slide 36 Another Example A city engineer claims that 50% of the bridges in the county needs repair. A sample of 10 bridges in the county was selected at random. What is the probability that exactly 6 of the bridges need repair? This situation meets the binomial requirements. Why? VERIFY. n = 10, p = 0.5, P(x = 6) = 0.2051. Use Binomial Table
OR
Use SWStat Slide 37 Example Continued What is the probability that 7 or fewer of the bridges need repair? We need P(x ≤ 7) = P(x = 0) + P(x = 1) + ... + P(x = 7) = 0.001 + 7) = 0.0098 + ... + 0.1172 = 0.9454 OR P(x ≤ 7) = 1 – P(x=8) – P(x=9) – P(x=10) = 1 – 7) = (.0439+.0098+.0010) = 0.9454 Use Binomial Table
OR
Use SWStat Slide 38 Using SWStat+ Find P(X=6)=? Slide 39 Using SWStat+ Find P(X≤7)=? Slide 40 Binomial Distribution
s More Example: Evans Electronics Wendy is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Slide 41 Binomial Distribution
Example (Continued) Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year?
Useing the equation. Let: p = .10, n = 3, x = 1 n!
f ( x) =
p x (1 − p ) ( n − x )
x !( n − x ) !
3!
f (1) =
(0.1)1 (0.9)2 = 3(.1)(.81) = .243
1!(3 − 1)! Slide 42 Using SWStat+ Find P(X =1) = ? Also, use Tree Diagram to Solve This Problem Slide 43 Binomial Distribution
s Tree Diagram Leaves (.1) 3 Prob.
.0010 2 .0090 L (.1) 2 .0090 1 .0810 L (.1) 2 .0090 S (.9) Leaves (.1) x S (.9) 2nd Worker 3rd Worker
L (.1) S (.9) 1st Worker 1 .0810 1 .0810 0 .7290 Stays (.9) Leaves (.1)
Stays (.9) L (.1) Stays (.9)
S (.9) Slide 44 Binomial Distribution
s Expected Value (Mean)
E(x) = µ = np s Variance
Var(x) = σ 2 = np(1 − p) s Standard Deviation σ = np(1 − p ) Slide 45 Binomial Distribution: Using SWStat+
Example (Continued)
s s Evans is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? What is the mean, variance and the standard deviation? Slide 46 Binomial Distribution
s Expected Value (Mean)
E(x) = µ = 3(.1) = .3 employees out of 3 s Variance
Var(x) = σ 2 = 3(.1)(.9) = .27 s Standard Deviation σ = 3(.1)(.9) = .52 em ployees Slide 47 Using SWStat+ Find P(X=1)=? & E(X) & σ Slide 48 Poisson & Hypergeometric Distributions Optional Readings
Optional Readings Slide 49 End of Chapter 6 Slide 50 ...

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