Introduction to Probability and Probability Distributions

Training on Teaching Basic
Statistics for Tertiary Level
Teachers
Summer 2008
Note: Some of the Slides were taken from
Elementary Statistics: A Handbook of Slide
Presentation prepared by Z.V.J. Albacea, C.E.
Reano, R.V. Collado, L.N. Comia and N.A.
Tandang in 2005 for the Institute of Statistics,
CAS, UP Los Banos
PROBABILITY
AND PROBABILITY
DISTRIBUTIONS

Session 2.2
TEACHING BASIC STATISTICS …
Motivation for Studying Chance
Sample Statistic Estimates Population Parameter
e.g. Sample Mean X = 50 estimates Population Mean µ
Questions:
1. How do we assess the reliability of our estimate?
2. What is an adequate sample size? [ We would expect a
large sample to give better estimates. Large samples
more costly.]

Session 2.3
An Approach to Solve the Questions
If sample was chosen through
chance processes, we have to
understand the notion of
probability and sampling
distribution.

Session 2.4
To introduce probability….
 Random experiment
 Sample space
 Event as subset of sample
space
 Likelihood of an event to occur
- probability of an event

Session 2.5
Features of a Random Experiment
 All outcomes are known in
advance.
 The outcome of any one trial
cannot be predicted with
certainty.
 Trials can be repeated under
identical conditions.

Session 2.6
EXAMPLES
 Rolling a die and
observing the
number of dots on
the upturned face
 Tossing a one-peso
coin and observing
the upturned face
 Measuring the
height of a student
enrolled this term

Session 2.7
SAMPLE SPACE
 It is a set such that each element
denotes an outcome of a random
experiment.
 Any performance of the
experiment results in an outcome
that corresponds to exactly one
and only one element.
 It is usually denoted by S or Ω.

Session 2.8
ILLUSTRATION
Rolling a die and observing
the number of dots on the
upturned face
S={ , , , , , }
S={1, 2, 3, 4, 5, 6}

Session 2.9
EVENT
 A subset of the sample space whose
probability is defined
 Usually denoted by capital letters like E, A
or B
 Observance of the elements of the subset
implies the occurrence of the event, i.e.,
an event occurred if the outcome of the
experiment belongs in the event
 Can either be classified as simple or
compound event

Session 2.10
ILLUSTRATION
S = {1, 2, 3, 4, 5, 6}
An event of
observing odd-
number of dots
in a roll of a die
E1 = { 1, 3, 5}
An event of
observing even-
number of dots
in a roll of a die
E2 = { 2, 4, 6}

Session 2.11
Visualizing Events
 Contingency Tables
 Tree Diagrams
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck
of Cards
Red Cards
Black Cards
Not an Ace
Ace
Ace
Not an Ace

Session 2.12
Mutually Exclusive Events
Two events are mutually exclusive if
the two events cannot occur
simultaneously.
Example:
Coin toss: either a head or a tail, but not
both. The events head and tail are
mutually exclusive.

Session 2.13
 The numerical measure of
the likelihood that an event
will occur
 Between 0 and 1
Note: Sum of the
probabilities of all mutually
exclusive and
collective exhaustive events
is 1
Certain
Impossible
0.5
1
0
PROBABILITY

Session 2.14
Assigning Probabilities
 Subjective
confident student views chances of passing
a course to be near 100 %
 Logical
symmetry/equally likely: coin, dice, cards etc.
(A PRIORI assignment)
 Empirical
chances of rain 75 % since it rained 15 out of
past 20 days (A POSTERIORI)

Session 2.15
If all possible outcomes can be listed and
are equally likely to occur, we can compute
the Probability of an Event E:
OutcomesTotal
OutcomesEventofNumber
EP =)(
Example: If we select a card at random from a well-
shuffled deck of cards then,
P(ace in a deck of cards) = 4/52
since there are 4 aces in a deck of (52) cards.
Computing Probability: A Priori

Session 2.16
Computing Joint Probability
If the sample space contains equiprobable outcomes then the
probability of a joint event, A and B:
( and ) = ( )
number of outcomes from both A and B
total number of possible outcomes in sample space
P A B P A B∩
=
E.g. (Red Card and Ace)
2 Red Aces 1
52 Total Number of Cards 26
P
= =

Session 2.17
A POSTERIORI APPROACH
 The random experiment has to be
performed (under uniform condition for
a large number of times) and the event
of interest is observed.
 The probability of the event is the
(limiting value) of the relative frequency
of the occurrence of such event if the
experiment is endlessly repeated.

Session 2.18
ILLUSTRATION
 Suppose the experiment was done for
100 times and it was observed that an
odd-number of dots occurred 60 times
and even-number of dots occurred 40
times.
 The (empirical) probability of an event
of observing odd-number of dots in a
roll of a die is the relative frequency of
the event or P[E1] = 60/100 = 0.6

Session 2.19
Rules on Probability
 Property 1. The probability of an event E
is any number between 0 and 1 inclusive.
and P(Ω)=1 while P(∅)=0.
 Property 2. The sum of the probabilities
of a set of mutually exclusive and
exhaustive events is 1. (n events are
mutually exclusive if no pair of events
among the n can occur simultaneously)

Session 2.20
Rules on Probability
 Property 3. Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
A
B

Session 2.21
Computing Probability
 P(King or Spade)
= P(King) + P(Spade) - P(King and Spade)
=
 P(King or Queen) = P(King) + P(Queen)
=
4 13 1 16 4
52 52 52 52 13
+ − = =
4 4 8 2
52 52 52 13
+ = =
since King and Queen are mutually exclusive then P(King and Queen)=0

Session 2.23
Definition of Conditional Probability
The conditional probability of event B given
that event A has already occurred, denoted
by P(B|A) is defined as:
if P(A)>0. Otherwise, it is undefined.
( )
( | )
( )
P A B
P B A
P A
∩
=

Session 2.24
Conditional Probability
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
(Ace and Red) 2/52 2
(Ace | Red)
(Red) 26/52 26
P
P
P
= = =
A Deck of 52 Cards

Session 2.25
Definition of Independent Events
The events A and B are independent if and
only if:
This condition is equivalent to saying that
P(A|B)=P(A), or that P(B|A)=P(B).
( ) ( ) ( )P A B P A P B∩ =

Session 2.26
Examples:*
1. Consider the following events in the toss of a single
die:
A: Observe an odd number
B: Observe an even number
Are A and B independent events?
2. The probability that Robert will correctly answer the
toughest question in an exam is ¼. The probability
that Ana will correctly answer the same question is
4/5. Find the probability that both will answer the
question correctly, assuming that they do not copy
from each other.
*from Stat101 Manual, UP Stat, Diliman

Session 2.27
UNEQUALLY LIKELY OUTCOME
ASSUMPTION
 The outcomes have different
likelihood to occur.
 The probability of an event E is
then computed as the sum of the
probabilities of the outcomes
found in the event E, that is,
P[E] = sum of p{e}
where e is an element of event E.

Session 2.28
ILLUSTRATION
S = {1, 2, 3, 4, 5, 6}
 Assuming that the probability of each of the
outcomes 1,2, and 3 is 1/12 while each of the
outcomes 4, 5 and 6 has likelihood to occur equal
to 1/4.
 The probability of an event of observing odd-
number of dots in a roll of a die is P[E1] = sum of
p{1}, p{3} and p{5} = 1/12 + 1/12 + 1/4 = 5/12.

Session 2.29
Random Variable
 Defined on a random experiment
 A rule or a function that maps each element
of the sample space to one and only one
real number
 The mapping produces mutually exclusive
partitioning on the set of real numbers.
 A random variable defined on a sample
space that is countable is a discrete random
variable.

Session 2.30
ILLUSTRATION
Rolling two dice and observing the
number of dots on the upturned faces.
S={ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Session 2.31
ILLUSTRATION
We define a random variable X as the total number of dots on the
upturned faces.
Sample Points x
2
3
4
5
6
7
8
9
10
11
12
(1,1),
(1,2), (2,1),
(1,3), (2,2), (3,1),
(1,4), (2,3), (3,2), (4,1),
(1,5), (2,4), (3,3), (4,2), (5,1),
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1),
(2,6), (3,5), (4,4), (5,3), (6,2),
(3,6), (4,5), (5,4), (6,3),
(4,6), (5,5), (6,4),
(5,6), (6,5),
(6,6)

Session 2.32
ILLUSTRATION
 The random variable takes on the values 2, 3, 4,
5, 6, 7, 8, 9, 10, 11 and 12.
 Some of the values had more corresponding
elements in the sample space. For example, 2
corresponds to only one outcome while 3
corresponds to 2 outcomes.
 The probability that the (discrete) random
variable will take a value is equal to the sum of
the probabilities of the corresponding outcomes
in the sample space.

Session 2.33
ILLUSTRATION
 The probability that the random variable will take
the value 4 is equal to the sum of the
probabilities of the corresponding outcomes. The
probability that the total number of dots on the
upturned faces of the dice is 4 is then equal to
the sum of the probabilities of the outcomes
(1,3), (2,2), and (3,1).
 Each outcome in the sample space has
probability of 1/36 if the dice are fair. Thus, the
probability that the total number of dots is 4 is
equal to 3/36 or 1/12.

Session 2.34
PROBABILITY DISTRIBUTION
 The discrete probability distribution of a
discrete random variable is a table or a
function that presents the possible values of
the random variable and its corresponding
probabilities.
 The probability density function of a
continuous random variable is a curve or a
function f such that P(a ≤ X ≤ b) is the area
bounded by the curve f(x), the x-axis and
the lines x=a and x=b.

Session 2.35
ILLUSTRATION
The probability distribution of the random variable, X defined
as the total number of dots on the upturned faces in a roll of
two dice, is presented as a table below:
X 2 3 4 5 6 7 8 9 10 11 12
P[X=x] 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
0.00
0.05
0.10
0.15
0.20
2 3 4 5 6 7 8 9 10 11 12
X = Total Number of Dots on the Upturned faces

Session 2.36
Types of Probability Distributions
 Discrete Probability Distributions:
e.g. Bernoulli, Binomial,
Geometric, Hypergeometric,
Negative Binomial,
 Continuous Probability
Distributions:
e.g. Normal, Exponential,
Gamma, Beta, Uniform,

Session 2.37
MEAN OF A DISCRETE RANDOM
VARIABLE
 If X is a discrete random variable and its discrete probability
distribution is as follows:
x x1 x2 … xn
P(X=x) P(X=x1) P(X=x2) … P(X=xn)
Then the expected value of X, also referred to as the mean of X is:
E(X) = µ = x1P(X=x1) + x2P(X=x2) + … + xnP(X=xn)
 In interpreting the mean of X, the collection of data points that we
are summarizing is now an infinite collection containing all of the
realized values of X if we are to repeat the random experiment over
and over again. Thus, the mean of X can be interpreted as the
average value generated by continually repeating the random
experiment.

Session 2.38
VARIANCE OF A DISCRETE
RANDOM VARIABLE
 If X is a discrete random variable and its discrete probability
distribution is as follows:
x x1 x2 … xn
P(X=x) P(X=x1) P(X=x2) … P(X=xn)
then the variance of X is:
Var(X)= E(X- µ)2
=(x1- µ)2
P(X=x1)+(x2- µ)2
P(X=x2)+ …+(xn- µ)2
P(X=xn)
 The variance of X is a measure of dispersion. It is the average
squared deviation between the realized value of X and µ. It tends
to have a larger value if the values of X are likely to be far from the
mean (the center of the distribution), than if the values are
concentrated about the mean. If there is no variation in the values
generated by X then Var(X) will be 0.

Session 2.39
Bernoulli Probability Distribution
 Named after Bernoulli
 Discrete random variable with
only two possible values; 0 and 1
 The value 1 represents success
while the value 0 represents
failure
 The parameter p is the probability
of success.

Session 2.40
Bernoulli Probability Distribution
 Its probability
distribution function
is given by:
 Graphically, the
distribution is illustrated
as follows:
1
( ) 1 0
0
p x
P X x p x
otherwise
=

= = − =


0 1
p
1-p

Session 2.41
Binomial Probability Distribution
 Composed of n independent
Bernoulli trials
 The parameter p is the probability of
success remains constant from one
trial to another
 Discrete random variable defined as
the number of success out of n trials
 Possible values; 0, 1, 2, .., n

Session 2.42
Binomial Probability Distribution
 Its probability
distribution function is
given by:
 Graphically, the
distribution is illustrated
as follows:
( )( ) 1 , 0, 1, 2,
n xxn
P X x p p x n
x
− 
= = − = ÷
 
K
0 1 2 …. n
and the function is 0
elsewhere.

Session 2.43
Illustration: Binomial Distribution*
The probability that a patient recovers from a rare
blood disease is 0.4. If 15 people are known to have
contracted this disease, what is the probability that
exactly 5 survive.
Solution: Let X be the number of people that survive.
P(X=5) = b(5;15, 0.4) = = 0.1859
*from Walpole, Introduction to Statistics
105
)4.01(4.0
5
15
−






Session 2.44
• ‘Bell-Shaped’
• Symmetric
• Range of possible values
is infinite on both
directions. Mean
Median
Mode
X
f(X)
µ
Normal Probability Distribution

Session 2.45
The Mathematical Model
( )
( )
( )
21
2
2
1
2
: density of random variable
3.14159; 2.71828
: population mean
: population standard deviation
x
f x e
f x X
e
µ
σ
πσ
π
µ
σ
2
− −
=
≈ ≈

Session 2.46
THE NORMAL CURVE
0.00
0.05
0.10
0.15
0.20
0.25
-15 -10 -5 0 5 10 15 20
Two normal distributions with the same mean but
different variances.
N(5,4)
N(5,9)

Session 2.47
Two normal distributions with the different means
but equal variances
0.00
0.05
0.10
0.15
0.20
0.25
-5 0 5 10 15 20
N(5,4)
N(10,4)
THE NORMAL CURVE

Session 2.48
By varying the parameters σ and µ, we obtain
different normal distributions
There are an infinite number of normal curves
Many Normal Distributions

Session 2.49
Normal Distribution Properties
For a normal curve, the area within:
a) one standard deviation from the
mean is about 68%,
b) two standard deviations from the
mean is about 95%; and
c) three standard deviations from
the mean is about 99.7%.

Session 2.50
Probability is the area
under the curve!
c d X
f(X)
P c X d( ) ?≤ ≤ =
Areas Normal Distributions

Session 2.51
Infinitely Many Normal Distributions imply
Infinitely Many Tables to Look Up!
Each distribution
has its own table?
Which Table???

Session 2.52
Standard Normal Distribution
Since there are many normal curves,
often it is important to standardize,
and refer to a STANDARD NORMAL
DISTRIBUTION (or curve) where the
mean µ = 0 and the σ =1

Session 2.53
THE Z-TABLE
P[Z ≤ z]
Examples:
1. P[Z ≤ 0] = 0.5
2. P[Z ≤ 1.25] = 0.8944
3. P[Z ≤ 1.96] = 0.9750
0 z
This table summarizes the cumulative probability
distribution for Z (i.e. P[Z ≤ z])

Session 2.54
Standardizing Example
6.2 5
0.12
10
X
Z
µ
σ
− −
= = =
Shaded Area Exaggerated
Normal Distribution
10σ =
5µ =
6.2 X
Standard Normal Distribution
Z
0Zµ =
0.12
1Zσ =

Session 2.55
Solution: The Cumulative
Standardized Normal Curve
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.5478
.02
0.1 .5478
Cumulative Standard Normal Distribution Table (Portion)
Probabilities
Shaded
Area
Exaggerated
Only One Table is Needed
0 1Z Zµ σ= =
Z = 0.12
0

Session 2.56
Normal Distribution Standardized Normal Curve
10σ = 1Zσ =
5µ =
7.1 X Z
0Zµ =
0.21
2.9 5 7.1 5
.21 .21
10 10
X X
Z Z
µ µ
σ σ
− − − −
= = = − = = =
2.9 0.21−
.0832
( )2.9 7.1 .1664P X≤ ≤ =
.0832
Example:

Session 2.57
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.5832
.02
0.1 .5478
Cumulative Standard Normal
Distribution Table (Portion)
Shaded
Area
Exaggerated
0 1Z Zµ σ= =
Z = 0.21
(continued)
0
( )2.9 7.1 .1664P X≤ ≤ =Example:

Session 2.58
Z .00 .01
-03 .3821 .3783 .3745
.4207 .4168
-0.1.4602 .4562 .4522
0.0 .5000 .4960 .4920
.4168
.02
-02 .4129
Cumulative Standard
Normal Distribution Table
(Portion)
Shaded
Area
Exaggerated
0 1Z Zµ σ= =
Z = -0.21
( )2.9 7.1 .1664P X≤ ≤ =
(continued)
0
Example:

Session 2.59
( )8 .3821P X ≥ =
Normal Distribution Standard Normal
Distribution
10σ =
1Zσ =
5µ =
8 X Z0Zµ =
0.30
8 5
.30
10
X
Z
µ
σ
− −
= = =
.3821
Example:

Session 2.60
(continued)
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.6179
.02
0.1 .5478
Shaded Area
Exaggerated
0 1Z Zµ σ= =
Z = 0.30
0
( )8 .3821P X ≥ =
Example:

Session 2.61
.1217
Finding Z Values for Known Probabilities
Z .00 0.2
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
.6179 .6255
.01
0.3
What is Z Given area between
0 and Z is 0.1217 ?
Shaded
Area
.6217
0 1Z Zµ σ= =
.31Z =
0

Session 2.62
Example
Suppose that women’s heights can be modeled by a
normal curve with a mean of 1620 mm and a
standard deviation of 50 mm
Solution: The 10th percentile of the height distribution
may be obtained by firstly getting the 10th percentile
of the standard normal curve, which can be read off
as -1.282. This means that the 10th percentile of the
height distribution is 1.282 standard deviations below
the mean. This height is
–1.282(50)+1620 =1555.9

Session 2.63
RULES IN COMPUTING PROBABILITIES
P[Z = a] = 0
P[Z ≤ a] can be obtained directly
from the Z-table
P[Z ≥ a] = 1 – P[Z ≤ a]
P[Z ≥ -a] = P[Z ≤ +a]
P[Z ≤ -a] = P[Z ≥ +a]
P[a1 ≤ Z ≤ a2] = P[Z ≤ a2] – P[Z ≤ a1]

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