Download Learning Objective Thinking Challenge Random Variables

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WEEK5:
DISCRETE PROBABILITY
DISTRIBUTIONS AND CONTINUOUS
PROBABILITY DISTRIBUTIONS
Learning Objective
Discrete Probability Distributions
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Random Variables
Discrete Probability Distributions
 Expected Value and Variance
 Binomial Probability Distribution
 Poisson Probability Distribution
 Hypergeometric Probability
Distribution




Dr. Po-Lin Lai


.40
.30

.20
Develop the notion of a random variable
Learn that numerical data are observed values of
either discrete or continuous random variables
Study two important types of random variables
and their probability models: the binomial and
normal model
Define a sampling distribution as the probability
of a sample statistic
Learn that the sampling distribution of follows a
normal model
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Thinking Challenge
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
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?
Discrete Random Variables with a
Finite Number of Values
Random Variables
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
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A random variable is a numerical description of the
outcome of an experiment.
A discrete random variable means that random
variables that can assume a countable number
(finite or infinite) of values.
A continuous random variable may assume any
numerical value in an interval or collection of
intervals.

Example: JSL Appliances
Let x = number of TVs sold at the store in one day,
where x can take on 5 values (0, 1, 2, 3, 4)

We can count the TVs sold, and there is a finite upper
limit on the number that might be sold (which is the
number of TVs in stock).
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Discrete Random Variables with a
Infinite Number of Values
Discrete Random Variable Examples
Continuous Random Variable Examples
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
Example: JSL Appliances

We can count the customers arriving, but there is no
finite upper limit on the number that might arrive.
Discrete Probability Distributions
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Random
Variable (X)
Experiment
Let x = number of customers arriving in one day,
where x can take on the values 0, 1, 2, . . .
Possible
Values
Make 100 Sales Calls
# Sales
0, 1, 2, ..., 100
Inspect 70 Radios
# Defective
0, 1, 2, ..., 70
We can describe a discrete probability distribution
with a table, graph, or formula.
Weigh 100 People
Possible
Values
Weight
X>0
Hours
$ amount
Inter-Arrival
Time
Answer 33 Questions
# Correct
0, 1, 2, ..., 33
Count Cars at Toll
Between 11:00 & 1:00
# Cars
Arriving
0, 1, 2, ..., ∞
Measure Time
Between Arrivals
Discrete Probability Distributions
Random
Variable (X)
Measure Part Life
Amount spent on food
Salary=$2000
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The probability distribution for a random variable
describes how probabilities are distributed over
the values of the random variable.
Experiment
X>0
0<X<2000
X>0
Discrete Probability Distributions
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The probability distribution is defined by a
probability function, denoted by f(x), which provides
the probability for each value of the random variable.
The required conditions for a discrete probability
function are:
f(x) > 0
f(x) = 1
Example: JSL Appliances
Using past data on TV sales, …
a tabular representation of the probability distribution
for TV sales was developed.

Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
x
0
1
2
3
4
f(x)
.40
.25
.20
.05
.10
1.00
80/200
2
Discrete Probability Distributions
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Expected Value
Discrete Uniform Probability Distribution
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
Example: JSL Appliances
Probability
.50
.40
f(x) = 1/n
.10
where:
n = the number of values the random
variable may assume
1
2
3
4
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Variance
Weighted average of squared deviation about mean
2 = E[(x (x
p(x)
Standard Deviation
The expected value does not have to be a value the
random variable can assume.
Expected Value
Expected Value
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The expected value is a weighted average of the
values the random variable may assume.
The weights are the probabilities.
the values of the
random variable
are equally likely
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Values of Random Variable x (TV sales)

E(x) = = xf(x)
The discrete uniform probability function is
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
The expected value, or mean, of a random variable
is a measure of its central location.
The discrete uniform probability distribution is the
simplest example of a discrete probability
distribution given by a formula.
Graphical
representation
of probability
distribution
Expected Value
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
Example: JSL Appliances
x
0
1
2
3
4
f(x)
xf(x)
.40
.00
.25
.25
.20
.40
.05
.15
.10
.40
E(x) = 1.20
expected number of
TVs sold in a day

Example: JSL Appliances
x
x-
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
(x - )2
f(x)
(x - )2f(x)
1.44
0.04
0.64
3.24
7.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
Variance of daily sales =
2
TVs
squared
= 1.660
Standard deviation of daily sales = 1.2884 TVs
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Thinking Challenge
Discrete Probability Distribution Example
Expected Value & Variance Solution*
Experiment: Toss 2 coins. Count number of tails.
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?
Probability Distribution
Values, x Probabilities, p(x)
© 1984-1994 T/Maker Co.
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
x
p(x)
x p(x)
x – (x – ) (x – ) 2p(x)
0
.25
0
–1.00
1.00
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
2
= 1.0
.25
2 = .50
= .71
© 1984-1994 T/Maker Co.
Binomial Probability Distribution
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Binomial Probability Distribution
Binomial Probability Distribution
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
The binomial probability distribution is a discrete
probability distribution that provides many
applications. It is associated with a multiple-step
experiment that we can call the binomial experiment.

Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
Our interest is in the number of successes
occurring in the n trials.
2. Two outcomes, success and failure, are possible
on each trial.
We let x denote the number of successes
occurring in the n trials.
3. The probability of a success, denoted by p, does
not change from trial to trial.
stationarity
assumption
4. The trials are independent.
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Binomial Probability Distribution
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Binomial Probability Distribution
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
Binomial Probability Distribution
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Binomial Probability Function

 Example: Evans Electronics
Evans Electronics is concerned about a low retention
rate for its 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?
Binomial Probability Function
where:
x = the number of successes
p = the probability of a success on one trial
1-p = the probability of a failure on any one trial
n = the number of trials
f(x) = the probability of x successes in n trials
Number of experimental
outcomes providing exactly
x successes in n trials
Binomial Probability Distribution
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Probability of a particular
sequence of trial outcomes
with x successes in n trials
Binomial Probability Distribution
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
Example: Evans Electronics
The probability of the first employee leaving and the
second and third employees staying, denoted (S, F, F),
is given by
p(1 – p)(1 – p)
With a .10 probability of an employee leaving on
any one trial, the probability of an employee leaving
on the first trial and not on the second and third trials
is given by
(.10)(.90)(.90) = (.10)(.90)2 = .081
Binomial Probability Distribution
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
Example: Evans Electronics
Two other experimental outcomes also result in one
success and two failures. The probabilities for all three
experimental outcomes involving one success follow.
Experimental
Outcome
Probability of
Experimental Outcome
(S, F, F)
(F, S, F)
(F, F, S)
p(1 – p)(1 – p) = (.1)(.9)(.9) = .081
(1 – p)p(1 – p) = (.9)(.1)(.9) = .081
(1 – p)(1 – p)p = (.9)(.9)(.1) = .081
Total = .243

Example: Evans Electronics
Let: p = .10, n = 3, x = 1
Using the
probability
function
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Poisson Probability Distribution
Binomial Probability Distribution
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
Using a tree diagram
Example: Evans Electronics
1st Worker
2nd Worker
Leaves (.1)
Leaves
(.1)
3rd Worker
L (.1)
x
3
Prob.
.0010
S (.9)
2
.0090
L (.1)
2
.0090
S (.9)
1
.0810
2
.0090
Stays (.9)
Leaves (.1)
Stays
(.9)
L (.1)
S (.9)
L (.1)
Stays (.9)
S (.9)
1
.0810
1
.0810
0
.7290
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Poisson Probability Function
Where:
x = the number of occurrences in an interval
f(x) = the probability of x occurrences in an interval
= mean number of occurrences in an interval
e = 2.71828
Two Properties of a Poisson Experiment
1. The probability of an occurrence is the same
for any two intervals of equal length.
2. The occurrence or non-occurrence in any
It is a discrete random variable that may assume
an infinite sequence of values (x = 0, 1, 2, . . . ).
interval is independent of the occurrence or
non-occurrence in any other interval.
Poisson Probability Distribution
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

A Poisson distributed random variable is often
useful in estimating the number of occurrences
over a specified interval of time or space
Poisson Probability Distribution
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Poisson Probability Distribution
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Poisson Probability Distribution
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
Poisson Probability Function
Since there is no stated upper limit for the number of
occurrences, the probability function f(x) is applicable
for values x = 0, 1, 2, … without limit.
In practical applications, x will eventually become
large enough so that f(x) is approximately valueero and
the probability of any larger values of x becomes
negligible.
Example: Mercy Hospital
Patients arrive at the emergency room of Mercy
Hospital at the average rate of 6 per hour on weekend
evenings.
What is the probability of 4 arrivals in 30 minutes on a
weekend evening?

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Poisson Probability Distribution
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Poisson Probability Distribution
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
Example: Mercy Hospital
Poisson Probability Distribution
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
Example: Mercy Hospital
Poisson Probabilities
A property of the Poisson distribution is that
the mean and variance are equal.
0.25
Probability
 = 6/hour = 3/half-hour, x = 4
Using the
probability
function
0.20
 = 2
actually, the
sequence
continues:
11, 12, …
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
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Number of Arrivals in 30 Minutes
Poisson Probability Distribution
Thinking Challenge
Poisson Distribution Solution: Finding *
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Example: Mercy Hospital
Variance for Number of Arrivals During 30-Minute Periods

=2=3
You work in Quality Assurance
for an investment firm. A clerk
enters 75 words per minute with
6 errors per hour. What is the
probability of 0 errors in a
255-word bond transaction?


75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
6 errors/hr
= 6 errors/4500 words
= .00133 errors/word

In a 255-word transaction (interval):
 = (.00133 errors/word )(255 words)
= .34 errors/255-word transaction
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Hypergeometric Probability Distribution
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Hypergeometric Probability Distribution
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Hypergeometric Probability Distribution
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
Hypergeometric Probability Function

Hypergeometric Probability Function
The hypergeometric distribution is closely related to the
binomial distribution.
for 0 < x < r
However, for the hypergeometric distribution:
the trials are not independent, and
where: x = number of successes
n = number of trials
f(x) = probability of x successes in n trials
N = number of elements in the population
r = number of elements in the population labeled success
the probability of success changes from trial
to trial.
Hypergeometric Probability Distribution
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Hypergeometric Probability Distribution
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
Hypergeometric Probability Function
The probability function f(x) on the previous slide
is usually applicable for values of x = 0, 1, 2, … n.
However, only values of x where: 1) x < r and
2) n – x < N – r are valid.
If these two conditions do not hold for a value of x,
the corresponding f(x) equals 0.
number of ways
x successes can be selected
from a total of r successes
in the population
number of ways
n – x failures can be selected
from a total of N – r failures
in the population
number of ways
n elements can be selected
from a population of size N
Hypergeometric Probability Distribution
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Example: Neveready’s Batteries
Bob Neveready has removed two dead batteries from
a flashlight and inadvertently mingled them with the two
good batteries he intended as replacements. The four
batteries look identical.
Bob now randomly selects two of the four batteries.
What is the probability he selects the two good
batteries?


Example: Neveready’s Batteries
Using the
probability
function
where:
x = 2 = number of good batteries selected
n = 2 = number of batteries selected
N = 4 = number of batteries in total
r = 2 = number of good batteries in total
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Hypergeometric Probability Distribution
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Hypergeometric Probability Distribution
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
Example: Neveready’s Batteries
Mean

Variance

Hypergeometric Probability Distribution
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Consider a hypergeometric distribution with n trials
and let p = (r/n) denote the probability of a success
on the first trial.
When the population size is large, a hypergeometric
distribution can be approximated by a binomial distribution
with n trials and a probability of success
p = (r/N).
If the population size is large, the term (N – n)/(N – 1)
approaches 1.
The expected value and variance can be written
E(x) = np and Var(x) = np(1 – p).
Note that these are the expressions for the expected
value and variance of a binomial distribution.
continued
Continuous Probability Distributions
Continuous Probability Distributions
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



f (x)
Uniform Probability Distribution
Normal Probability Distribution
Normal Approximation of Binomial Probabilities
Exponential Probability Distribution


f (x) Exponential
Uniform
f (x)

Normal
x
x
x
Continuous Probability Distributions
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A continuous random variable can assume any value
in an interval on the real line or in a collection of
intervals.
It is not possible to talk about the probability of the
random variable assuming a particular value.
Instead, we talk about the probability of the random
variable assuming a value within a given interval.

f (x)
The probability of the random variable assuming a
value within some given interval from x1 to x2 is
defined to be the area under the graph of the
probability density function between x1 and x2.
Uniform
f (x) Exponential
f (x)
x1 x2
Normal
x
x1 x2
x
x1 xx12 x2
x
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Continuous Probability Density Function
Uniform Probability Distribution
Continuous Probability Density Function
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The graphical form of the probability distribution for a
continuous random variable x is a smooth curve

This curve, a function of x, is denoted by the symbol f(x)
and is variously called a probability density function
(pdf), a frequency function, or a probability distribution.

f (x) = 1/(b – a) for a < x < b
=0
elsewhere
The areas under a probability
distribution correspond to
probabilities for x. The area A
beneath the curve between two
points a and b is the probability
that x assumes a value betweena and b.
Uniform Probability Distribution
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
Expected Value of x
E(x) = (a + b)/2

Variance of x
Var(x) = (b - a)2/12
where: a = smallest value the variable can assume
b = largest value the variable can assume
Uniform Probability Distribution
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A random variable is uniformly distributed whenever
the probability is proportional to the interval’s length.
The uniform probability density function is:
Uniform Probability Distribution
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Example: Slater's Buffet
Slater customers are charged for the amount of
salad they take. Sampling suggests that the amount
of salad taken is uniformly distributed between 5
ounces and 15 ounces.


Uniform Probability Density Function
f(x) = 1/10 for 5 < x < 15
=0
elsewhere
Where: x = salad plate filling weight
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Uniform Probability Distribution
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Uniform Probability Distribution
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
Expected Value of x

E(x) = (a + b)/2
= (5 + 15)/2
= 10

Uniform Probability Distribution
for Salad Plate Filling Weight

f(x)
Variance of x
P(12 < x < 15) = 1/10(3) = .3
1/10
1/10
0
5
10
Salad Weight (oz.)
15
x
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
The area under the graph of f(x) and probability are
identical.
This is valid for all continuous random variables.
The probability that x takes on a value between some
lower value x1 and some higher value x2 can be
found by computing the area under the graph of f(x)
over the interval from x1 to x2.
0
Normal Probability Distribution
Area as a Measure of Probability
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
What is the probability that a customer will take
between 12 and 15 ounces of salad?
f(x)
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33

Uniform Probability Distribution
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5
10 12
Salad Weight (oz.)
x
15
Normal Probability Distribution
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


The normal probability distribution is the most important
distribution for describing a continuous random variable.
It is widely used in statistical inference.
It has been used in a wide variety of applications including:






Heights of people
Rainfall amounts
Test scores
Scientific measurements
Abraham de Moivre, a French mathematician, published The
Doctrine of Chances in 1733.
He derived the normal distribution.

Normal Probability Density Function

where:
= mean
= standard deviation
= 3.14159
e = 2.71828
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Normal Probability Distribution
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Normal Probability Distribution
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
Characteristics
Normal Probability Distribution
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
The distribution is symmetric; its skewness
measure is zero.
Characteristics
The entire family of normal probability
distributions is defined by its mean m and its
standard deviation s .

Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
Standard Deviation s
x
Normal Probability Distribution
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Normal Probability Distribution
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
Characteristics
Characteristics

The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
s = 15
x
-10
0
Normal Probability Distribution
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
The mean can be any numerical value: negative,
zero, or positive.
x
x
Mean m
s = 25
Characteristics
Probabilities for the normal random variable are
given by areas under the curve. The total area
under the curve is 1 (.5 to the left of the mean and
.5 to the right).
.5
25
x
.5
x
12
Normal Probability Distribution
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Normal Probability Distribution
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
Characteristics (basis for the empirical rule)
Standard Normal Probability Distribution
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
Characteristics (basis for the empirical rule)

99.72%
95.44%
68.26%
68.26% of values of a normal random variable
are within
of its mean.
+/- 1 standard deviation
Characteristics
A random variable having a normal distribution
with a mean of 0 and a standard deviation of 1 is
said to have a standard normal probability
distribution.
95.44% of values of a normal random variable
are within +/- 2 standard deviations
of its mean.
99.72% of values of a normal random variable
are within +/- 3 standard deviations
of its mean.
– 3s
– 1s
– 2s
Standard Normal Probability Distribution
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+ 3s
+ 1s
+ 2s
x
Standard Normal Probability Distribution
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
Characteristics
The letter z is used to designate the standard
normal random variable.
Standard Normal Probability Distribution
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Characteristics
 Converting to the Standard Normal Distribution


We can think of z as a measure of the number of
standard deviations x is from .
Example: Pep Zone
Pep Zone sells auto parts and supplies including a
popular multi-grade motor oil. When the stock of this oil
drops to 20 gallons, a replenishment order is placed.
The store manager is concerned that sales are being lost
due to stockouts while waiting for a replenishment order.

z
0
13
Standard Normal Probability Distribution
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Standard Normal Probability Distribution
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Example: Pep Zone
It has been determined that demand during replenishment
lead-time is normally distributed with a mean of 15
gallons and a standard deviation of 6 gallons.
The manager would like to know the probability of a
stockout during replenishment lead-time. In other words,
what is the probability that demand during lead-time will
exceed 20 gallons?

Standard Normal Probability Distribution
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
Solving for the Stockout Probability

Step 1: Convert x to the standard normal distribution.
z = (x - )/　= (20 - 15)/6
= .83
Step 2: Find the area under the standard normal
curve to the left of z = .83.
P(x > 20) = ?
P(z < .83)
see next slide
Standard Normal Probability Distribution
82
Standard Normal Probability Distribution
83

Solving for the Stockout Probability
Step 3: Compute the area under the standard normal
curve to the right of z = .83.

Solving for the Stockout Probability
Solving for the Stockout Probability
If the manager of Pep Zone wants the probability of a
stockout during replenishment lead-time to be no more
than .05, what should the reorder point be?
------------------------------------------------------------ (Hint: Given a probability, we can use the standard
normal table in an inverse fashion to find the
corresponding z value.)


Area = 1 - .7967
Area = .7967
= .2033
= .2033
P(x > 20)
Standard Normal Probability Distribution
84
P(z > .83) = 1 – P(z < .83)
= 1- .7967
Probability
of a stockout
Cumulative Probability Table for the Standard Normal
Distribution
0 .83
z
14
Standard Normal Probability Distribution
85
Standard Normal Probability Distribution
86

Solving for the Reorder points
Standard Normal Probability Distribution
87

Solving for the Reorder points

Step 1: Find the z-value that cuts off an area of .05
in the right tail of the standard normal
distribution.
Area = .9500
Solving for the Reorder points
Step 2: Convert z.05 to the corresponding value of x.
x = + z.05　(6)
= 15 + 1.645(6)
Area = .0500
= 24.87 or 25
0
We look up
the complement
of the tail area
(1 - .05 = .95)
z
z.05
Normal Probability Distribution
89
Solving for the Reorder points
Probability of a
stockout during
replenishment
lead-time = .05
15
90

Probability of no
stockout during
replenishment
lead-time = .95
24.87
x
A reorder point of 25 gallons will place the
probability of a stockout during leadtime at (slightly
less than) .05.
Normal Approximation of Binomial
Probabilities
Normal Probability Distribution
88



By raising the reorder point from 20 gallons to 25
gallons on hand, the probability of a stockout
decreases from about .20 to .05.
This is a significant decrease in the chance that Pep
Zone will be out of stock and unable to meet a
customer’s desire to make a purchase.
When the number of trials, n, becomes large,
evaluating the binomial probability function by hand
or with a calculator is difficult.
The normal probability distribution provides an
easy-to-use approximation of binomial probabilities
where np > 5 and n(1 - p) > 5.
In the definition of the normal curve, set
= np and
15
Normal Approximation of Binomial
Probabilities
Normal Approximation of Binomial
Probabilities
91
92
Normal Approximation of Binomial
Probabilities
93
Example
Suppose that a company has a history of making errors
in 10% of its invoices. A sample of 100 invoices has
been taken, and we want to compute the probability
that 12 invoices contain errors.
In this case, we want to find the binomial probability of
12 successes in 100 trials. So, we set:
m = np = 100(.1) = 10
= [100(.1)(.9)] ½ = 3

Add and subtract a continuity correction factor
because a continuous distribution is being used to
approximate a discrete distribution.
For example, P(x = 12) for the discrete binomial
probability distribution is approximated by
P(11.5 < x < 12.5) for the continuous normal
distribution.

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
=3
P(11.5 < x < 12.5)
(Probability
of 12 Errors)
 = 10
11.5
Normal Approximation of Binomial
Probabilities
94
Normal Approximation of Binomial
Probabilities
95

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
Normal Approximation of Binomial
Probabilities
96

Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1
P(x < 12.5) = .7967
10 12.5
x
12.5

The Normal Approximation to the Probability of 12
Successes in 100 Trials is .1052
P(x < 11.5) = .6915
x
10
x
11.5
P(x = 12)
= .7967 - .6915
= .1052
10
11.5
12.5
x
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Exponential Probability Distribution
Exponential Probability Distribution


The exponential probability distribution is useful in
describing the time it takes to complete a task.
The exponential random variables can be used to
describe:

A property of the exponential distribution is that the
mean and standard deviation are equal.

The exponential distribution is skewed to the right.
Its skewness measure is 2.
•Time between vehicle arrivals at a toll booth
•Time required to complete a questionnaire
•Distance between major defects in a highway


Density Function
for x > 0
 = expected or mean
where:
In waiting line applications, the exponential
distribution is often used for service times.
Exponential Probability Distribution

Exponential Probability Distribution
Cumulative Probabilities
where:
x0 = some specific value of x
e = 2.71828
Exponential Probability Distribution

Example: Al’s Full-Service Pump
The time between arrivals of cars at Al’s fullservice gas pump follows an exponential probability
distribution with a mean time between arrivals of 3
minutes. Al would like to know the probability that
the time between two successive arrivals will be 2
minutes or less.
Exponential Probability Distribution

Example: Al’s Full-Service Pump
f(x)
P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866
.4
.3
.2
.1
x
0
1
2
3
4
5
6
7
8
9 10
Time Between Successive Arrivals (mins.)
17
Relationship between the Poisson
and Exponential Distributions
End of Week 5
The Poisson distribution
provides an appropriate description
of the number of occurrences
per interval
The exponential distribution
provides an appropriate description
of the length of the interval
between occurrences
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