Chapter 5
Nutan S. Mishra
Department of Mathematics and Statistics
University of South Alabama
Discrete Variables
A random variable is a variable whose value is
determined by the outcome of a random
experiment.
A random variable assigns a numerical value to an
event in S
Example: In tossing a coin S = {H,T}
Define random variable as follows:
X = 1 when H occurs
X = 0 when T occurs.
Here X is a discrete random variable.
Discrete Random Variable
In some cases the outcomes of the experiment are
themselves numerical values, in such a case we
may not have to define a random variable
separately.
Example: rolling a die
S ={1,2,3,4,5,6}. All outcomes are numerical thus
we do not have to define the random variable
separately.
Define X = # dots showed up
X takes values 1,2,3,4,5,6
Discrete Random variables
Toss three coins and note the number of
heads showed up.
Let X = # heads occurred. Then x can take
the values 0,1,2,3
Number of vehicles owned by a family
Number of dependents in a family
Number of goals scored by a player
Continuous Random Variable
• When outcomes of an experiment are numerical
values in an interval then X is continuous
• Example: Recording the GPA of students
• X = GPA of a student
• Then X  [0,4] and X can take any value in this
interval.
• X= highest temperature on a given day
• X= Time taken by a runner to complete a race.
Probability Distribution
Probability distribution of a discrete random
variable is the set of all values of X and the set of
corresponding probabilities P(X=x).
Example: toss a single coin and observe the
number of heads occurred. Define X = # heads
Then X can take two values 0 or 1
X
P(X=x)
0
.5
1
.5
These two columns together are
called probability distribution of X
Probability distribution
Toss three coins and define
X = # heads then X=0,1,2,3
S = {TTT,TTH,THT,HTT,HHT,HTH,THH,HHH}
The probability distribution x is :
X
P(X=x)
0
1/8
1
3/8
2
3/8
3
1/8
Properties of distribution
P(X=x) is denoted by f(X=x) or just f(x) .
f(x) is called probability mass function.
Two properties
1. f(x)  0
2. f(x) = 1
X
P(X=x) = f(x)
0
.5
1
.5
X
P(X=x)=f(x)
0
1/8
1
3/8
2
3/8
3
1/8
Exercise 5.8
x
P(x)
x
P(x)
0
.10
2
.35
1
.05
3
.28
2
.45
4
.20
3
.40
5
.14
ΣP(x) = 1.00
x
P(x)
7
-.25
8
.85
9
.40
ΣP(x) = 1.00
ΣP(x) = .97
Exercise 5.10
x
0
1
2
3
4
5
6
P(x)
.11
.19
.28
.15
.12
.09
.06
a. P(x=3) = .15
b. P(x ≤ 2) = P(x=0)+P(x=1)+P(x=2) =
c. P(x ≥ 4) = P(x=4)+P(x=5)+P(x=6) =
d. P(1 ≤x ≤4) = P(x=1)+P(x=2)+P(x=3)+P(x=4)
e. P(x<4) = 1- P(x ≥ 4)
f. P(x>2) = 1- P(x ≤ 2)
g. P(2 ≤x ≤5) = P(x=2)+P(x=3)+P(x=4)+p(x=50)
Exercise 5.14
X = # TV sets owned by a family
Size of the dataset is 2500 (This is a population)
Use classical approach to find probability distribution of x.
# TV sets owned (x)
# families (f)
P(x)
0
120
.048
1
970
.388
a. The two highlighted
columns together form
the probability distribution
of X
b. These probabilities are
exact because this is a
3
410
.164
population data and we
4
270
.108
are using classical
approach to compute
sum
2500
1.00
probabilties.
C. P(x=1) = .388, P(x≥ 3) = P(x=3)+P(x=4) = .064+.108
2
730
.292
P(2 ≤ x ≤ 4) = P(x=2)+P(x=3)+P(x=4) = .292+.064+.108
P(x<4) = 1-P(x≥4) = 1-P(x=4) = 1- .108
Mean of a discrete random variable
Mean of a discrete random variable x is the value
that is expected to occur per repetition of the
experiment.
   xP(x)
# TV sets owned (x)
P(x)
xP(x)
0
.048
0
1
.388
.388
2
.292
.584
3
.164
.492
4
.108
.432
sum
1.00
ΣxP(x)=1.896
Mean = 1.896 TV sets
Interpretation : If we repeat
this experiment number of
times then on the average a
family owns 1.896 TV sets.
Variance of discrete random variable
 2   x 2 P( x)   2

 x P( x)  
2
2
# TV sets owned (x)
P(x)
xP(x)
x2
x2P(x)
0
.048
0
0
0
1
.388
.388
1
.388
2
.292
.584
4
1.168
3
.164
.492
9
1.476
4
.108
.432
16
1.728
sum
1.00 ΣxP(x)=1.896
σ2 = 4.76 – (1.896)2 = 1.165184
σ = 1.0794
4.76
Exercise 5.28
Machines
sold/day (x)
P(x)
xP(x)
x2
x2P(x)
4
.08
0.32
16
1.28
5
.11
0.55
25
2.75
6
.14
0.84
36
5.04
7
.19
1.33
49
9.31
8
.20
1.6
64
12.8
9
.16
1.44
81
12.96
10
.12
1.2
100
12
7.28
Mean of x = 7.28
56.14
σ (Standard deviation) = 7.4927
Interpretation: if experiment of collecting data on machines sales is collected
for a large number of days then the on the average 7.28 machines would be
sold per day.
Counting revisited
Useful links:
http://www.unc.edu/~knhighto/econ70/lec4
/lec4.htm
http://pavlov.psyc.queensu.ca/~flanagan/202
_1999/lecture9/lecture9.html
Factorials
Example: three students with names A B C and
three chairs red blue and yellow.
Question: in how many ways students can be
allocated to the chairs?
A
B C
A C B
B A C
B C A
C A
B
C B A
In general n distinct objects can be arranged in n
places in n! (factorial n) ways
n! = n*(n-1)*…*1
3! = 3*2*1 = 6 ways
Permutations
Example: four students A B C D and two chairs
Question: How many ways a team of two students can
occupy the chairs?
A B
A C
A D
B C
B D
C D
B A
C A
D A
C B
D B
D C
# ways we can arrange n items in r places
Is called permutation
nPr = n*(n-1)*…(n-r+1)
4P2 = 4*3 = 12
Combinations
Example: four students A B C D and two identical chairs.
Question: In how many ways we can allocate students to
the chairs?
A B
A C
A D
B C
B D
C D
Here order does not matter (that is it does not matter
which student occupies which chair because chairs are
identical)
# ways choosing r items out of n distinct items is
n!
n Cr 
r!(n  r )!
4C 2
= 6
Table III on page C7 lists the values of combinations
Bernoulli trials
An experiment which has only two possible
outcomes is a Bernoulli trial with
probability of one outcome p and that of
the other as 1-p.
Example: toss a coin : has only two
outcomes H and T if P(H) =.5 then P(T) =
1-.5 = .5
If the coin is not fair and P(H) = .7 then P(T)
= 1-.7 = .3
Bernoulli distribution
More examples of Bernoulli trials:
Inspecting a car at an assembly plant and
declaring it as lemon or good.
If P(L) = .001 then P(G) = .999
People entering into a football stadium.
Classifying them into one of the genders.
M or F if P(M) = .62 then P(F) = .38
Binomial Experiment
A binomial experiment consists of n independent
Bernoulli trials.
That is repeating the Bernoulli experiment n times.
Each repetition is independent of the other.
Example: Toss three fair coins.
n=3 Each coin is a Bernoulli trial. Outcome of one
toss does not affect that of the others i.e. all
tosses are independent. And p=.5 for all the
three tosses.
Binomial experiment
Definition:
1. The experiment consists of n identical
trials.
2. Each trial has only two possible outcomes
3. The probability of outcomes remain
constant at each trial.
4. The trials are independent.
Binomial probability distribution
In a Binomial experiment define a random variable x as
X =# heads occurred in n tosses or
X = # successes in n trials.
Then x takes values 0, 1,2 …,n
Such an X is called Binomial random variable with n and p
specified.
If we repeat the binomial experiment a large number of
times then we are interested probabilities of x assuming
different values
i.e in tossing n coins what is P(X=0) or P(X=1) and so on
In other words we are interested in probability distribution
of x.
Binomial distribution
Consider a binomial experiment with n trials and p
as probability of success in a single trial then
X = # successes in n independent trials is a
binomial variable and its probability distribution
is called Binomial distribution with parameters n
and p
P(x=k) = nCk pk (1-p)n-k for x = 0,1, …n
Alternatively replace 1-p =q
P(x=k) = nCk pk qn-k for x = 0,1, …n
Exercise 5.51
a.
b.
c.
Rolling a die many times and observing # spots is not
a binomial experiment because there are more than
two possible outcomes of a roll
Rolling a die many times and observing whether the
number is even or odd is a binomial experiment
because at each roll there are only two possible
outcomes: even or odd. Besides all the rolls are
independent.
Yes this a binomial experiment because we are
selecting a few people, each person has only two
possible answers: in favor or not in favor. The
probability of an individual being in favor is known i.e.
.54 and all persons answers are independent.
Exercise 5.53
a. x= binomial vairate with n=8 and p=.7
P(x=5) = 8C5 *p5 (1-p)8-5 = 8C5 *p5 (1-p)3 = 8C5 *(.7)5 (.3)3
= .2441 (from table iv)
b. Given n=4, p= .40
P(x=3) = 4C3 *p3 (1-p)4-5 = 4C3 *(.4)3 (.6) =
Mean and Variance of binomial
Let x be a binomial variable with parameters
n and p. then mean of x
µ = Σx. nCk pk (1-p)n-k = np
µ = np
Variance σ2 = npq
Exercise 5.58
Question asked: Do you eat home cooked food three or more times a
week?
Yes: 85% No: 15%
P(Y) = p = .85 P(N) = 1-p =q = .15
a.
n=12 (a random sample of 12 Americans) is selected
X = # Americans from the sample of 12 who say Yes
Then X has binomial distribution with parameters n=12 and p= .85
X may take values between 0 to 12
That is there may be o Americans who say yes or may be 2 .. 0r at the
most 12 will say yes.
b. What is the probability that 10 out of 12 Americans say yes to the
posed question?
P(X=10) = 12C10 *p10(1-p)12-10 = 12C10 *p10(1-p)2