Chapters 4
Discrete Random Variables
Random Variables:
Numerical Outcomes Of Experiments:
• Discrete
• Continuous
Discrete Random Variables:
• Result from a counting process,
e.g.
- No. of heads in n trials,
- No. of defective items in a sample,
- No. of machines operating / shift,
- No. of accidents / given intersection.
Discrete Random Variables:
y
0
1
2
3
….
A Countable Number of Values
Probability Distribution Functions:
Theoretical models of the relative
frequency distribution of discrete
random variables when n N:
Continuous Random Variables:
• Result from a measuring process
e.g.
- Diameter of shafts,
- Processing time,
- Car speed,
- % of DDT.
Continuous Random Variables:
y
0
1
2
3
….
An infinite Number of Values
Probability Density Functions:
Theoretical models of the relative
frequency distribution of continuous
random variables when n N:
Chapter 4
Discrete Random Variables
Probability Distribution:
a table, formula, or graph which
displays the probabilities for each
possible value of a random variable
y, which would be observed if an
experiment were repeated an
infinitely large number of times.
Properties:
1.The probability of y taking on any
given value:
0 p(y) 1
2. The values of y represent sets of
mutually exclusive events, therefore
p(y) = 1
For example:
A balanced coin is flipped twice
and the number of heads, y, is
observed. Find the probability
distribution of y.
Probability Distribution of y
Tabular Format:
y
0
1
2
p(y)
1/4
1/2
1/4
S p(y)=1
Probability Distribution of y
Graphical Format:
p(y)
0.5
0.4
0.3
0.2
0.1
0
0
1
2
y
Probability Distribution of y
Formula:
1 2
P( y ) =
4 y
Where,
(.) is the number of combinations of 2
taken y at a time.
Expected Value of y:
the mean or expected value of a
discrete random variable, y, with
probability distribution p(y) is:
= E( y) = y.P( y)
The variance of y, is the expected
value of the squared deviation of y
from its mean that is:
= Var ( y) = E[( y − ) ]
2
2
= E( y ) −
2
2
Example:
What is the expected value and variance
of the number of heads y, when you flip a
balanced coin twice?
y
0
1
2
p(y)
1/4
1/2
1/4
S p(y)=1
The mean, or expected value of
y is as follows:
S y p(y)= 0(1/4) + 1(1/2) + 2(1/4) = 1
Thus,
The expected value = 1
The variance of y is as follows:
2 = (0 - 1)2 (1/4) + (1-1)2 (1/2)
+ (2-1)2 (1/4)
OR,
2
+ (1)2 (1/2)
+ (2)2 (1/4)} - (1)2
= {(0)2 (1/4)
2 = 1/2 , and = 0.707
The Binomial Experiment:
Any experiment where there are 2
possible outcomes, the population N,
is large, and the sample taken from
the population n, is relatively small,
(say n < 0.10 N)
Characteristics of a Binomial
Random Variable:
1. The experiment consists of n
identical, independent trials
2. There are only 2 possible
outcomes on each trial, say S
(success) and F (failure)
3. The probability of success p, is the
same from trial to trial and the
probability of failure is denoted by q,
Note: p+q=1
4. The binomial random variable y, is
the number of Successes in n trials
The Binomial Probability Distribution:
n y n− y
P( y ) = p q
y
for y = 0,1,2,..n
where:
p = probability of a success on a
single trial, and q = 1-p
y = # of successes in n trials
n
n!
=
y y!(n − y )!
The mean = np, and
The variance 2 = npq
Example:
Suppose that a large lot of fuses
contains 10% defectives. Find the
probability that a random sample of 5
fuses contains:
A) Exactly one defective,
B) At most one defective,
C) At least one defective.
The Cumulative Binomial
Probability Table:
Appendix B, Table 2, p 901-904.
For n and p; find: P(y < k)
Multinomial Probability Distribution:
▪
▪
An extension of the binomial
experiment,
Consists of n identical independent
trials, with k possible outcomes.
Properties of the Multinomial
Experiment:
1. The experiment consists of n
identical, independent trials
2. There are only k possible
outcomes on each trial
3. The probabilities of k outcomes,
denoted by p1, p2, ... pk, remain
the same from trial to trial;
Note:
p1+p2+ ... + pk=1
4. The random variables of interest
are the counts y1, y2, ...yk in each
of the k categories.
The Multinomial Probability
Distribution:
n!
P( y1, y2 ,.., yk ) =
( p1) y1 ( p2 ) y2 ....( pk ) yk
y1! y2!..yk !
where:
pi = probability of outcome i on a single trial
▪
p1 + p2 + ... + pk = 1
▪
n= y1 + y2 + ... + yk = number of trials
▪
yi = # of occurrences of i in n trials
▪
mean i = npi, and
▪
variance i2= npi (1-pi)
Example:
For a multinomial distribution with; n=40,
k=3, p1=0.1, p2=0.3, and p3=0.6, find:
P( y1=10,y2=10,y3=20)
P( y1=10,y2=10,y3=20)
40!
= ________
10! 10! 20!
= 0.00055
(0.1)10 (0.3)10 (0.6)20
Note:
• No tabulation possible.
• The binomial is a multinomial
with k = 2.