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Engineering Probability and
Statistics - SE-205 -Chap 3
By
S. O. Duffuaa
Lecture Objectives
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Present the following:
Concept of random variables
Probability distributions
Probability mass function
Random Experiment and random
variables
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Throwing a coin
S = { H, T}.
Define a mapping X: { H, T}  R
X(H) = 1 and X(T) = 0, Also the
probability of 1 and 0 are the same as for
H and T.
Then we call X a random variable.
Random Experiment and random
variables
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In the experiment on the number of
defective parts in three parts the sample
space S = { 0, 1, 2, 3}
Find P(0), P(1), P(2) and P(3)
P(0) = 1/8, P(1) = 3/8, P(2) = 3/8
and P(3) = 1/8
Probability Mass Function

X
f(x)
o
1
2
3
1/8
3/8
3/8
1/8
Properties of f(x)
f(x)  0
 f(x) = 1
Give many examples in class
Probability Mass Function

Build the probability mass functions for
the following random variables:
• Number of traffic accidents per month on
campus.
• Class grade distribution
• Number of “ F” in SE 205 class per semester
• Number of students that register for SE 205
every semester.
Cumulative Distribution Function
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It is a function that provide the cumulative
probability up to a point for a random
variable (r.v). Defined as follows for a
discrete r.v:
P( X  x) = F(x) =  f(t)
t x
Cumulative Distribution Function
(CDF)
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Example of a cumulative distribution
function
F(x) = 0 x  -2
= 0.2 -2  x  0
= 0.7 0  x  2
= 1.0 2  x
What is the density function for the
above F(x). Note you need to subtract
Probability Mass Function Corresponding
to Previous CDF

X
f(x)
-2
0
2
0.2
0.5
0.3
The above density function is the one corresponding to
the previous CDF is
Mean /Expected Value of a Discrete
Random Variable (r.v)
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The mean of a discrete r.v denoted as E(X)
also called the expected value is given as:
E(X) = μ =  x f(x)
x

The expected value provides a good idea a bout the
center of the r.v.
 compute the mean of the r.v in previous slide:
 E(X) = (-2) (0.2) + (0) (0.5) + (2)(0.3) = 0.2
Variance of A Random Variable

The variance is a measure of variability.
 What is variability?
 The variance is defined as:
 V(X) =σ2 = E(X-μ)2 =  (x-μ)2f(x)
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Compute the variance of the r.v in the slide before the previous one.

σ2 = (-2-0.2)2 (0.2) + (0-0.2)2(0.5) + (2-0.2)2(0.3)
=
Also see example 3-9 and 3-11in the text.
Expected Value of a Function of a r.v
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Let X be a r.v with p.m.f f(x) and let h(X)
is a function of X. Then the expected value
of h(X) is given as:
E(H(X)) = x h(x) f(x)
Compute the expected value of h(X) = X2 - X for the
r.v in the previous slides.
See example 3-12 in text book.
Discrete Random Variables
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In this section will study several discrete
distributions. For each distribution the
student must be familiar with the following
about each distribution:
Range and probability mass function
Cumulative distribution function
Mean and variance
2-3 applications
Discrete Random Variables
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The following distributions will be studied
•
•
•
•
•
•
Discrete uniform
Bernoulli
Binomial
Hyper-geometric
Geometric
Poisson
Discrete Uniform
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A random variable is discrete uniform if every
point in its range has the same probability. If there
are n points in the range, then the probability of
each point is
 f(x) = 1/n
 An alternative way of defining uniform as
follows: Suppose the rang is a, a+1, a+2, … b
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The number of points is (b-a+1)
f(x) = 1/(b-a+1) for
x = a, a+1, a+2, …, b
Discrete Uniform
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The CDF F(x) you just multiply by the
number of points less than or equal to x
The mean of the uniform is (b+a)/2
The variance of it is [(b-a+1)2 – 1]/12
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Applied to following situations:
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• Random number generation
• Drawing a random sample
• Situation where vales have equal probabilities.
Bernoulli Trials
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A trial with only two possible outcomes is used so
frequently as a building block of a random
experiment that it is called a Bernoulli trial. It is
usually assumed that the trials that constitute the
random experiment are independent. This implies
that the outcome from one trial has no effect on
the outcome to be obtained from any other trial.
Furthermore, it is often reasonable to assume that
the probability of a success in each trial is
constant.
Bernoulli Trials
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If we denote success by 1 an failure by 0, then the
probability mass function f(x) is given as:
f(1) = p and f(0) = 1-p = q, as you see the range
is 0 and 1
F(x) simple
Mean =E(X) = p
Variance = σ2 = p(1-p) = pq
Applications:
• Building block for other distributions
• Experiment with two outcomes
Binomial Random Variable
A random experiment consisting of n repeated trails such that
1) the trials are independent,
2) each trial results in only two possible outcomes, labeled as
“success” and “failure”, and
y
3) the probability of a success in each trial, denoted as p, remeins
constant
(1) F ( x) 
x
is called a binomial experiment.
The random variable X that equals the number of trials that result in a
success has binomial distribution with parameters p and n = 1, 2, ….
The probability mass function of X is
 n x
f ( x)    p (1  p) n x ,
 x
x  0,1, ...., n
Binomial Random Variable
0.4
0.18
n
0.15
p
n
• 20 0.5
0.3
0.12
p
• 10
0.1
• 10
0.9
y
x
f(x) 0.09
kjhkjhkjh
f(x) 0.2
0.06
0.1
0.03
0
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1617 1819 20
x
0
1
2
3
4
5
x
Figure 4-6 Binomial distributions for selected values of n and p
6
7
8
9
10
Binomial Random Variable
If X is a binomial random variable with parameters p and n, then
  E ( X )  np
y
and
x
 2 V ( Xkjhkjhkjh
)  np(1  p)
Applications:
•
Design of sampling plans for quality control
• Estimation of product defects.
Geometric Random Variable
In a series of independent Bernoulli trials, with constant probability p of
a success, let the random variable X denote the number of trials until the
first success. Then X has a geometric distribution with parameters p
y
and
f ( x)  (1  p) x 1 p ,
x 1, 2, .....
x
kjhkjhkjh
Geometric Distribution
If X is a geometric random variable with parameters p, then the mean and
variance
  E ( X ) 1 / p
y
x
and
kjhkjhkjh
 V ( X )  (1  p) / p 2
2
Applications:
• Quality control, design of control charts
• Estimation
Hyper-Geometric Distribution
A set of N objects contains
K objects classified as successes and
N – K objects classified as failures

A sample of size of n objects is selected, at random
(without replacement)
from the N objects, where K  N and n  N
y
x
Let the random variable X denote the number of successes in the sample.
Then X has a hypergeometric distribution and
 K  N  K 
 

x  n  x 

f ( x) 
N
 
n
x  max 0, n  K  N  to min K , n
Hyper-Geometric Distribution
If X is a hypergeometric random variable with parameters N, K and n, then
the mean and variance of X are
  E ( X )  np
y
x
and
 N n
  V ( X )  np(1  p)

 N 1 
2
kjhkjhkjh
where p = K/N
Applications:
• Design of inspection plans for quality control
• Design of control charts
Poisson Random Variable
Given an interval of real numbers, assume counts occur at random throughout
the interval. If the interval can be partitioned into subintervals of small enough
length such that
1) the probability of more than one count inay subintervals is zero,
2) the probability of one count in a subinterval is the same for all
subintervals and proportional
to the length of the subinterval, and
(1) F ( x) 
3) the count in each subinterval is independent of other subintervals,
x
then the random experiment is called a Poisson process
If the mean number of counts in the interval is  > 0, the random variable X
that equals the number of counts in the interval has a Poisson distribution
with parameters , and the probability mass function of X is
e   x
f ( x) 
,
x
x  0,1, 2, ....
Poisson Random Variable
If X is a Poisson random variable with parameters , then the mean and
variance of X are
  E (X )  
y
x
and
kjhkjhkjh
 V ( X )  
2
Applications:
• Model number of arrivals to a service facility
• Model number of accidents per month
• Demand for spare parts per month