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Chapter 5 Discrete Probability Distributions
Section 5.1 Random Variables
Random Variable
-- numerical description of the outcome of an experiment
Discrete Random Variable
-- assumes either a finite number of values or an infinite sequence of values
Discrete Random Variable (finite # of values)
Example: # of voters in a given county election
Example: JSL Appliances
Let x = # of TV’s sold at the store in 1 day
Discrete Random Variable (infinite sequence of values)
Let x = # of customers arriving in one day
In some cases, you have to convert text answers to numeric through coding
For gender -> x = 1 for male x = 2 for female
Continuous Random Variables
-- may assume any numerical value in an interval or collection of intervals
Example: mileage, temperature, time, weight
Random Variable Summary Examples
Question/Experiment
Random Variables
Family Size
x = # of dependents reported on tax return
Distance from home
to store
x = distance in miles from home to store
Own a dog or cat
x = 1 if own no pet
x = 2 if own dog(s) only
x = 3 if own cat(s) only
x = 4 if own dog(s) and cat(s)
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Type
Section 5.2 Discrete Probability Distribution
-- can be described with a table, graph or equation
Probability Distribution
-- for a random variable, the probability distribution describes how probabilities are distributed over the
values of the random variable
-- defined by a probability function, denoted as f(x), which provides the probability for each value of the
random variable
For a discrete probability distribution,
f(x) > 0
∑ f(x) = 1
Example
Graphically
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Discrete Uniform Probability Distribution
-- simplest example of a discrete probability distribution, given by the formula:
f(x) = 1/n  values of random variable are equally likely
where n = # of values the random vaiable may assume
Example: Experiment is rolling a die
Section 5.3 Expected Value, Variance & Standard Deviation of Discrete
Random Variable
Expected Value (Mean)
-- measure representing the central location of a random variable
-- known as the mean of a random variable
E(x) = μ = ∑x f(x)
Variance
-- summarizes the variability/dispersion in the values of a random variable
Var(x) =
 2 = ∑ (x – μ)2 f(x)
Standard Deviation
  
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Example: Expected Values of TV’s sold in one day
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Section 5.4 Binomial Probability Distribution
4 Properties of a Binomial Experiment
1) Experiment consists of a sequence of n identical trials
2) Two outcomes, success and failure, are possible on each trial
3) Probability of a success, denoted by p does not change from trail to trial ( 1- p is the probability of a
failure)
4) Trails are independent
Our interest is in the # of successes occurring in the n trials
Let x = # of successes occurring in n trials
n = # trials
p = probability of success
1-p = probability of failure
Example: Evans Electronics
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?
Check Properties:
1) 3 identical trails
2) 2 outcomes possible (Leave/Stay)
3) Probability of success (leave) = .10
4) Trials are indept
-- All 4 are satisfied
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Using Tables showing the probability of x successes in n trials (starting on p 978 in text)
p
p
n
n
xx
.05
.05
.10
.10
.15
.15
.20
.20
.25
.25
.30
.30
.35
.35
.40
.40
.45
.45
.50
.50
33
00
11
22
33
.8574
.8574
.1354
.1354
.0071
.0071
.0001
.0001
.7290
.7290
.2430
.2430
.0270
.0270
.0010
.0010
.6141
.6141
.3251
.3251
.0574
.0574
.0034
.0034
.5120
.5120
.3840
.3840
.0960
.0960
.0080
.0080
.4219
.4219
.4219
.4219
.1406
.1406
.0156
.0156
.3430
.3430
.4410
.4410
.1890
.1890
.0270
.0270
.2746
.2746
.4436
.4436
.2389
.2389
.0429
.0429
.2160
.2160
.4320
.4320
.2880
.2880
.0640
.0640
.1664
.1664
.4084
.4084
.3341
.3341
.0911
.0911
.1250
.1250
.3750
.3750
.3750
.3750
.1250
.1250
Expected Values/Variance/Standard Deviation of Binomial Distribution
Expected Value
E(x) =  = np
Variance
Var(x) =  2 = np(1 - p)
Standard Deviation
  np(1  p )
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