#### Transcript SSF1063: Statistics for Social Sciences

```SSF1063: Statistics for
Social Sciences
LU5: Probability Distribution of Discrete
Random Variables
18th February 2008
Definitions

Random Variables


A variable whose value is determined by
the outcome of a random experiment
Discrete Random Variables

Variables with countable values
Table 5.1 (p. 189)
Frequency and Relative Frequency Distributions of the
Number of Vehicles Owned by Families
Discrete Random Variable

Other examples includes:
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The number of cars sold at a dealership during
a given month
The number of babies being born during a
given week in a local hospital
The number of houses in a certain block
The number of fish caught on a fishing trip
The number of complaints received at the office
of an airline on a given day
Continuous Random Variable

A variable that assumes any value contained in
one or more intervals
0

250
Examples:

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The
The
The
The
height of a person
time taken to complete an examination
weight of a fish
price of a house
Probability Distribution of a Discrete
Random Variable

Lists all the possible values that the random
variable can assume and their corresponding
probabilities
Probability Distribution of a Discrete
Random Variable

Write the probability distribution of the
discrete random variable for Table 5.1
The Two Characteristics

The probability distribution of a discrete
random variable has two characteristics:
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The probability assigned to each value is
0≤P(x)≤1 for each x
The sum of all probabilities assigned to all
possible values of x is equal to 1.0; ΣP(x) = 1.0
These are the two conditions that a probability
distribution must satisfy
The Two Characteristics
Find the probability
of the following:
P(2) =
P(>2) =
P(≤1) =
P(>4) =
Graphical Presentation of Probability
Distribution
Validity of Probability Distribution
x
P(x)
x
P(x)
2
3
4
5
0.25
0.34
0.28
0.13
0
1
2
3
0.08
0.11
0.39
0.27
x
P(x)
7
8
9
0.70
0.50
-0.20
Example 1

The following table lists the probability
distribution of the number of breakdowns per
week for a machine based on past data.
Breakdowns per week
Probability
0
1
2
3
0.15
0.20
0.35
0.30
a. Present this probability distribution graphically
b. Find the probability that the number of
breakdowns for this machine during a given week
is:
i. exactly 2
ii 0 to 2
iii. more than 1 iv. at most 1
Example 1
i. exactly 2
ii. 0 to 2
iii. more than 1
iv. at most 1
Example 2

According to a survey, 60% of students suffer
from Math anxiety. Two students are randomly
selected from this university. Let x denote the
number of students in this sample who suffer
from math anxiety. Develop the probability
distribution of x.
N = student selected does not suffer from Math
anxiety
M = student selected suffer from Math anxiety
Example 2
Example 2
Example 3

Five percent of all cars manufactured
at a large auto company are lemons.
Let x denote the number of lemons
in this sample. Write the probability
distribution of x. Draw a tree
diagram for this problem.
Mean of a Discrete Random Variable

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Mean of the probability distribution (also
known as ‘expected value’)
The value that we expect to observe per
repetition, on average
If a man sells 1.6 cars per week, on
average, does is mean that he sells 1.6
cars every week?
Mean of a Discrete Random Variable

The mean of discrete random variable x is
denoted by E(x) or μ
E(x) = ΣxP(x)
Standard Deviation of a Discrete
Random Variable

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Measures the spread of its probability
distribution, denoted by σ
Higher value of σ indicates that x can assume
values over a larger range about the mean
Smaller value for σ indicates that values of x are
Formula:
σ = √ Σx2 P(x) – μ2
Example 4

In Panda Electronics, a few defective parts
will usually go into a shipment undetected.
Let x denote the number of defective parts
in a shipment of 400. The following table
gives the probability distribution of x.
x
0
1
2
3
4
5
P(x)
0.02
0.20
0.30
0.30
0.10
0.08
Example 4
σ = √ Σx2 P(x) – μ2
Factorials
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Value is obtained by multiplying all the
integers from that number to 1.
8x7x6x5x4x3x2x1
0! always equals to 1.
Evaluate 4!, 6!, 9!, 11!
Evaluate (12 – 4)!
Evaluate (5 – 5)!
Combinations

Gives the number of ways x elements can
be selected from n elements. Denoted
nCx
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x = The number of elements
selected per selection
n = The total number of elements
The formula is given:
n!
nC x =
x!(n – x)!
Example 5

An ice cream parlor has six flavors of ice
cream. Kristen wants to buy two flavors
of ice cream. If she randomly selects two
flavors out of six, how many possible
combinations are there?
Example 6

Three members of jury will be randomly
selected from five people. How many
different combinations are possible?
Example 7
What is the probability
of a player (who plays
this lottery once) wins
if he needs to select 6
number from 1 – 49 ?
Permutations

The concept is similar to combinations but
with one major difference – order of
selection of important.
P
n x

x = The number of elements
selected per selection
n = The total number of elements
The formula is given:
n!
nP x =
(n – x)!
Example 8

A club has 20 members. They are to
select three office holders – president,
secretary and treasurer – for the next
year. Three names will be drawn
randomly where the first person will be
selected becomes the president, the
second is the secretary and the third one
takes over as treasurer. Therefore, the
order is important.
Example 9
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An ice cream shop offers 25 flavors of ice
cream. How many ways are there to select
2 different flavors from these 25 flavors?
How many permutations are possible?
A ski patrol unit has nine members
available for duty and two of them are to
be sent to rescue and injured skier. In how
many ways two of these nine members be
selected? Suppose the order of selection is
important. How many arrangements are
possible in this case?
Binomial Probability Distribution
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The binomial probability distribution is
most widely used discrete probability
distribution
Applied to find the probability that an
outcome will occur x times in n
performances of experiment
Rules of the Binomial Experiment

A binomial experiment must satisfy ALL
FOUR conditions
1. n identical trials
2. Each trial has only TWO outcomes;
success and failure
3. Success = p, Failure = q, p + q =1.
Constant probability
4. Trials are independent. One trial does
not affect the outcome of the other
Rules of the Binomial Experiment
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Is this a binomial experiment?
Probability for a defective DVD player is
0.05. What is the probability of getting
exactly ONE defective DVD player in the
sample of 3 DVD player?
Check if the situation satisfy the FOUR
rules.
0.05 x 0.95 x 0.95
0.1354
0.95 x 0.05 x 0.95
0.95 x 0.95 x 0.05
The Binomial Formula
P(x) = nCx·px·qn-x
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n
p
q
x
n
= total number of trials
= probability of success
= probability of failure (1 – p)
= number of successes in n trials
– x = number of failures in n trials
The Binomial Formula
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The probability of exactly ONE DVD player
is defective:
P(x) = nCx·px·qn-x
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p
q
n
x
=
=
=
=
success = defective DVD player = 0.05
failure = non-defective DVD player = 0.95
total number of trials = 3
number of successes in n trial
P(ONE defect)
= 3C1·0.051·0.953-1
= 3 x 0.05 x 0.952
= 0.1354
Example 10
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2% of packages delivered by Donkey
Express do not reach their destinations on
time. Suppose a corporation mails 10
packages through Donkey Express. Find:
The probability exactly ONE package will
not reach its destination on time
The probability at most ONE of these
packages will not reach the receiver on
time
Example 10
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Success, p = package fail to be delivered
on time
Failure, q = package delivered on time
Thus, p = 0.02 and q = 1 – 0.02 = 0.98
n = 10
Find p(x=1)
Find p(x≤1) = p(x=0) + p(x=1)
P(x) = nCx·px·qn-x
Graphical Presentation of Probability
Distribution

64% of adults in Singapore complained
that the working hours is too long. Choose
3 adults randomly and see if they have
the same opinion. Write the probability
distribution and draw a bar graph to
illustrate it.
Graphical Presentation of Probability
Distribution
Table of Binomial Probabilities

30% adults in Malaysia say they cannot live without cell
phones. If 6 adults are selected randomly, find the
probability that exactly 3 adults hold the said opinion.
Table of Binomial Probabilities
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P(at most 2 hold
the said opinion)
P(at least 3 hold
the said opinion)
P(exactly 5 hold
the said opinion)
P(more than 4 hold
the said opinion)
Shape of the Binomial Distribution
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Distribution is skewed to the right if p < 0.50
Shape of the Binomial Distribution
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Distribution is symmetric if p = 0.50
Shape of the Binomial Distribution
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Distribution is skewed to the left if p > 0.50
Mean and Standard Deviation of the
Binomial Distribution
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Mean, μ = np
Standard deviation, σ = √npq
If an experiment has 60 trials, with
p=0.85 and q=0.15, what is the mean and
the standard deviation of the distribution?
The Poisson Probability Distribution
Useful for occurrences that are random
and with unpredictable patterns
 THREE conditions to apply the Poisson
Probability Distribution:
1. Discrete random variable
2. The occurrences are random
3. The occurrences are independent

The Poisson Probability Distribution
P(x) =
x
-λ
λ ·e
x!
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λ = mean number of occurrences in the
interval (weeks, month, 5m, two-hours)
e = 2.71828
Example 11
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On average, a household receives 9.5
telemarketing calls a week. Using the
Poisson distribution formula, find the
probability that a randomly selected
Exactly 6 telemarketing calls a week
At most ONE call a week
P(x) = λx·e-λ
x!
Table of Poisson Probabilities

On average, two new accounts are opened per day at
Donkey Bank. Find the probabilities that on a given day,
the number of new accounts opened at this bank will be:
Exactly 6
At most 3
At least 7
Mean and Standard Deviation of the
Poisson Distribution
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Mean, μ = λ
Variance, σ2 = λ
Standard deviation, σ = √ λ
Given λ is 0.9. Find the mean, variance
and standard deviation for the probability
distribution.
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