BEE31702 ENGINEERING MATHEMATICS V
CHAPTER 1: RANDOM VARIABLES
LECTURERS:
Dr. Chew Chang Choon (Coordinator)
PM Dr. Muhammad Ramlee Bin Kamarudin
Dr. Nor Surayahani Binti Suriani
Dr. Nordiana Azlin Binti Othman
Dr. Suhaila Sari Dr. Suhaila binti Sari
Ts. Shamsul Bin Mohamad
1
CONTENTS
1. Introduction
2. Random variables (RV)
3. Discrete random variables:
i. Probability density function (PDF)
ii. Cumulative distribution function (CDF)
iii. Mean, variance and standard variation
4. Continuous random variables
i. Probability density function (PDF)
ii. Cumulative distribution function (CDF)
iii. Mean, variance and standard variation
2
1. INTRODUCTION
ROLE OF STATISTICS AND PROBABILITY
• What do engineers do?
– An engineer is someone who solves problems of interest to
society with the efficient application of scientific principles
by:
• Refining existing products
• Designing new products or processes
Statistics & Probability
4
WHAT IS PROBABILITY?
• Why we learn probability?
– Nothing in life is certain. In everything we do, we gauge the
chances of successful outcomes, from business to
engineering to medicine to the weather
• In situation in which one of a number of possible outcomes
may occur, the theory of probability produces methods for
quantifying the chances or the likelihood associated with
various outcomes.
• It provides a bridge between
descriptive and inferential
statistics.
5
WHAT IS STATISTICS?
• To guess is cheap. To guess wrongly is expensive - Chinese
proverb
• Statistics is the scientific application of mathematical principles
to the collection, analysis, and presentation of data:
– at the foundation of all of statistics is data
6
PROBABILISTIC VS STATISTICAL REASONING
• Suppose I know exactly the proportions of car makes in
Malaysia. Then I can find the probability that the first car I see
in the street is a Proton. This is probabilistic reasoning as I
know the population and predict the sample.
• Now suppose that I do not know the proportions of car makes
in Malaysia, but would like to estimate them. I observe a
random sample of cars in the street and then I have an
estimate of the proportions of the population. This is statistical
reasoning
7
2. RANDOM VARIABLES (RV)
TERMINOLOGY
• Random variables (RV):
– Assumes any of several different values as a result of some
random event or experiment
– Denoted by a capital letter such as X,Y and Z
• Population:
– A group of individuals of items that share one or more
characteristics from which data can be gathered and
analyzed
• Sample:
– A subset of the population
– Elements are selected intentionally as a representation of
the population being studied
9
TERMINOLOGY
• Sample space:
– The set of all possible outcome or events of an experiment
– Denoted by S
• Sample size:
– The number of items in a sample
• Random sample:
– The sample selected in a way that allows every member of
the population to have the same chance of being chosen
10
BASIC CONCEPTS
• Use graphs and numerical measures to describe data sets
which were usually samples
• We measured “how often” using:
Relative frequency = f/n
• As n gets larger,
Sample
And “How often”
= Relative frequency
Population
Probability
11
BASIC CONCEPTS
• An experiment is the process by which an observation (or
measurement) is obtained
• An event is an outcome of an experiment, usually denoted by
a capital letter:
– The basic element to which probability is applied
– When an experiment is performed, a particular event either
happens, or it doesn’t!
• Experiment: Record an age
– Event X: person is 30 years old
– Event Y: person is older than 65
• Experiment: Toss a die
– Event A: observe an odd number
– Event B: observe a number greater than 2
12
BASIC CONCEPTS
• Two events are mutually exclusive if, when one event
occurs, the other cannot, and vice versa
• Experiment: Toss a die
Not Mutually
Exclusive
– A: observe an odd number
– B: observe a number greater than 2
– C: observe a 6
B and C?
Mutually
B and D?
– D: observe a 3
Exclusive
13
BASIC CONCEPTS
• Simple event, Ei:
– An event that cannot be decomposed
– Denoted by E with a subscript
– Each simple event will be assigned a probability,
measuring “how often” it occurs
• Sample space, S:
– The set of all simple events of an experiment
14
SIMPLE EVENT, SAMPLE SPACE
• Experiment – tossing a die
• Simple events:
• Sample space:
15
EVENT, SIMPLE EVENT
• An event is a collection of one or more simple events
• The die toss:
– A: an odd number
– B: a number > 2
16
PROBABILITY OF AN EVENT
• The probability of an event A measures “how often” A
will occur. We write P(A)
• Suppose that an experiment is performed n times.
The relative frequency for an event A is:
=
• If we let n get infinitely large:
= lim
→
17
PROBABILITY OF AN EVENT
• P(A) must be between 0 and 1:
– If event A can never occur, P(A) = 0.
– If event A always occurs when the experiment is performed,
P(A) = 1
• The sum of the probabilities for all simple events in S equals 1
• The probability of an event A is found by adding the
probabilities of all the simple events contained in A
18
FINDING PROBABILITIES
• Probabilities can be found using:
– Estimates from empirical studies
– Common sense estimates based on equally likely events.
• Examples:
– Toss a fair coin
P(Head) = 1/2
– Suppose that 90% of the Malaysia population has black
hair. Then for a person selected at random,
P(Black hair) = .90
19
USING SIMPLE EVENTS
• The probability of an event A is equal to the sum of
the probabilities of the simple events contained in A
•
• If the simple events in an experiment are equally
likely, you can calculate
=
=
20
EXAMPLE 1
• Toss a fair coin twice. What is the probability of
observing at least one head?
21
SOLUTION 1
1st Coin
2nd Coin
Ei
P(Ei)
H
HH
1/4
P(at least 1 head)
T
HT
1/4
= P(E1) + P(E2) + P(E3)
H
= 1/4 + 1/4 + ¼
H
TH
1/4
T
TT
1/4
T
= 3/4
22
EXAMPLE 2
• The sample space of throwing a pair of dice is
23
SOLUTION 2
Event
Simple events
Probability
Dice add to 3
(1,2), (2,1)
2/36
Dice add to 6
(1,5), (2,4), (3,3), (4,2), (5,1)
5/36
Red die show 1
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
6/36
Green die show 1
(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)
6/36
24
EXAMPLE 3
• An experiment of tossing a fair coin three times. Let
X will be a random variable of getting head. Find the
probability of getting head by using probability
distribution function (PDF) table, graph and equation.
25
SOLUTION 3
PDF table
HHH
HHT
HTH
THH
HTT
1/8
1/8
1/8
1/8
1/8
x
3
2
2
2
1
THT
1/8
1
TTH
1/8
1
TTT
1/8
0
P(x = 0) =
P(x = 1) =
P(x = 2) =
P(x = 3) =
1/8
3/8
3/8
1/8
x
p(x)
0
1/8
1
3/8
2
3
3/8
1/8
PDF graph
Probability
Histogram for x
PDF equation
26
RANDOM VARIABLES TYPES
• Discrete random variables have a countable number
of outcomes
– Examples: Dead/alive, treatment/placebo, dice,
counts
• Continuous random variables have an infinite
continuum of possible values
– Examples: blood pressure, weight, the speed of a
car, the real numbers from 1 to 6
27
3. DISCRETE RANDOM
VARIABLES
PROBABILITY DISTRIBUTION FOR A DISCRETE RV
• A probability distribution for a discrete random
variable is a complete set of an possible outcomes
and their possibilities of occurring
X, representing
Possible values of X
Number of dots appear when tossing a die
1, 2, 3, 4, 5, 6
Number of books in the bag
0, 1, 2, 3, 4, 5
Number of male students in a class
15, 16, 17, 22
29
PROBABILITY DISTRIBUTION FUNCTION (PDF)
FOR A DISCRETE RV
• Theory 1
– The Probability Distribution Function (PDF) of a
discrete random variable X is described as the
function P(X=x) = P(x) which is satisfied:
0≤ ( )≤1
=1
= ( =
)
30
EXAMPLE 4
• Consider the table as follows:
• From the table of probability
distribution function, proof that the
distribution is a probability distribution
function(PDF) of discrete variable
then find the probability of:
– P(X<2)
– P(X≤2)
– P(X≥1)
x
P(x)
0
1/8
1
3/8
2
3/8
3
1/8
31
SOLUTION 4
• For P(X<2):
꞊ P(X=0) + P(X=1)
꞊ 1/8 + 3/8
꞊ 4/8
꞊ 1/2
x
P(x)
0
1/8
1
3/8
2
3/8
3
1/8
• For P(X≤2):
꞊ P(X=0) + P(X=1) + P(X=2)
꞊ 1/8 + 3/8 + 3/8
꞊ 7/8
32
SOLUTION 4
• For P(X≥1):
꞊ P(X=1) + P(X=2) + P(X=3)
꞊ 3/8 + 3/8 + 1/8
꞊ 7/8
x
P(x)
0
1/8
1
3/8
2
3/8
3
1/8
33
CUMULATIVE DISTRIBUTION FUNCTION (CDF) OF DISCRETE RV
• Theory 2
– The Cumulative Distribution Function (CDF) of a
discrete RV X is described as the function:
=
≤
=
( = )
which satisfies:
• F(x) is an increase function:
= 1 , which is the maximum value is 1
– lim
→
= 0 , which is the minimum value is 0
– lim
→
34
CUMULATIVE DISTRIBUTION FUNCTION (CDF)
OF DISCRETE RV
• Theory 3
– The CDF of a discrete RV X can be calculated by
using the following formulas:
≤
=
>
<
=1−
=
≤
−1 =
−1
=
≤
=
≤
− ( − 1)
=
−
+ ( )
≤
<
<
<
=
=
−
−
<
≤
=
− ( )
+
− ( )
− ( )
*f(r) = f(r)- f(r-1)
*f(s) = f(s)- f(s-1)
35
EXAMPLE 5
• Customers purchase a particular make of automobile with a
variety of options. The probability distribution function of the
number of options selected is:
X
7
P(X = x)
0.11
8
9
10
11
12
13
0.10 0.22 0.23 0.12 0.13 0.09
• Find the CDF of X, then by using the CDF, find the probability:
– Exactly ten number of options was selected
– More than nine number of options was selected
– Between eight and twelve number of options was selected
36
SOLUTION 5
• Find the CDF of X:
= ( ≤ )
X
7
8
9
10
11
12
13
P(X = x)
0.11
0.10
0.22
0.23
0.12
0.13
0.09
7 =
≤7 =
= 7 = 0.11
8 =
≤8 =
=7 +
= 8 = 0.11 + 0.10 = 0.21
9 =
≤9 =
=7 +
=8 +
= 9 = 0.11 + 0.10 + 0.22 = 0.43
10 = 0.66
11 = 0.78
12 = 0.91
13 = 1
X
7
8
9
10
11
12
13
P(X = x)
0.11
0.10
0.22
0.23
0.12
0.13
0.09
F(x)
0.11
0.21
0.43
0.66
0.78
0.91
1
37
SOLUTION 5
• Probability of exactly ten number of options was
selected:
X
7
8
9
10
11
12
=
=
= 10 =
− ( − 1)
P(X = x)
0.11
0.10
0.22
0.23
0.12
13
0.13
0.09
10 − (9)
= 10 = 0.23
X
7
8
9
10
11
12
13
P(X = x)
0.11
0.10
0.22
0.23
0.12
0.13
0.09
F(x)
0.11
0.21
0.43
0.66
0.78
0.91
1
38
SOLUTION 5
• Probability of more than nine number of options was
selected:
X
7
8
9
10
11
12
13
>
=1− ( )
P(X = x)
0.11
0.10
0.22
0.23
0.12
0.13
0.09
> 9 = 1 − (9)
> 9 = 1 − 0.43 = 0.57
• Probability of between eight and twelve number of
options was selected:
<
<
=
−
12 −
− ( )
8<
< 12 =
8 − (12)
8<
< 12 = 0.91 − 0.21 − 0.13 = 0.57
39
EXAMPLE 6
• Suppose X denotes the number of telephone receivers in a
single family residential house. From an examination of the
phone subscription records of 1000 residence in a city, the
following probability function of X is obtained:
0.17,
0.23,
=
0.2,
0,
= 0, 1
= 2, 3
=4
ℎ
• Find the CDF of X, then by using the CDF, calculate the value
of
–
–
–
( < 3)
(1 ≤
< 4)
(0 <
≤ 4)
40
SOLUTION 6
• Find the CDF of X:
= ( ≤ )
0 =
≤0 =
= 0 = 0.17
1 =
≤1 =
=0 +
0.17,
0.23,
=
0.2,
0,
= 0, 1
= 2, 3
=4
ℎ
= 1 = 0.17 + 017 = 0.34
2 = 0.57
3 = 0.8
4 =1
41
SOLUTION 6
• Calculate the value of
<3 =
≤3−1 =
3−1 =
2 = 0.57
0.17,
0.23,
=
0.2,
0,
= 0, 1
= 2, 3
=4
ℎ
• Calculate the value of
1≤
<4 =
4 −
1 +
1 −
4 = 1 − 0.34 + 0.17 − 0.2 = 0.63
• Calculate the value of
<
0<
≤
=
≤4 =
− ( )
4 −
0 = 1 − 0.17 = 0.83
42
EXPECTED VALUE, VARIANCE AND
STANDARD DEVIATION
• All probability distributions are characterized by an expected
value (mean), a variance (standard deviation squared) and
standard deviation
• The expected value of a discrete RV is defined as its weighted
average over all possible outcomes, with the weight for each
outcome being the relative frequency or probability associated
with the outcome
• Theory 4
– Expected value (mean), µ @ E(X) of discrete RV:
=
ℎ
ℎ
+
,
,
=
+
( )
43
EXPECTED VALUE, VARIANCE AND
STANDARD DEVIATION
• The variance of a discrete RV is defined as the
weighted average of the squared differences
between each possible outcome and the average
value of the outcomes, with the weights being the
probabilities associated with each of the outcomes
• Theory 5
– Variance,
of discrete RV:
,
Var aX + b =
+
44
EXPECTED VALUE, VARIANCE AND
STANDARD DEVIATION
• Theory 6
– The standard deviation, of the probability
distribution of a discrete RV is the square root of
the variance:
45
EXAMPLE 7
• The discrete RV X has range space {1,2,3,4,5} and
its CDF takes the value as follows:
– Tabulate the probability distribution function
– Compute the mean and variance
46
SOLUTION 7
• Tabulate the probability distribution function:
≤
=
=
= CDF
=
− ( − 1)
1 +5∗1
=1 = 1 − 0 =
− 0 = 0.12
50
= 2 = 0.28 − 0.12 = 0.16
= 3 = 0.48 − 0.28 = 0.2
= 4 = 0.72 − 0.48 = 0.24
= 5 = 1 − 0.72 = 0.28
47
SOLUTION 7
• Compute the mean and variance:
=
=
=
= 1 ∗ 0.12 + 2 ∗ 0.16 + 3 ∗ 0.2 + 4 ∗ 0.24 + 5 ∗ 0.28 = 3.4
= 1 ∗ 0.12 + 2 ∗ 0.16 + 3 ∗ 0.2 + 4 ∗ 0.24 + 5 ∗ 0.28 = 13.4
= 13.4 − 3.4 = 1.84
48
EXAMPLE 8
• A random variable Y has the following probability
distribution:
y
-2
0
2
P(Y = y)
k
1-2k
k
– Show that Var(Y)=8k
– If given k=1/3, find mean and variance
49
SOLUTION 8
• Show that Var(Y)=8k:
=
=
= (−2) ∗
= (−2) ∗
y
-2
0
2
P(Y = y)
k
1-2k
k
+0 ∗ 1 − 2
+0 ∗ 1 − 2
+2 ∗
+2∗
=8
=0
=8 −0 =8
50
SOLUTION 8
• If given k=1/3, find mean and variance:
=
=
=
y
-2
0
2
P(Y = y)
k=1/3
1-2k=1/3
k=1/3
1
1
1
= (−2) ∗ + 0 ∗ + 2 ∗ = 0
3
3
3
1
1
1 8
= (−2) ∗ + 0 ∗ + 2 ∗ = = 2.66
3
3
3 3
= 2.66 − 0 = 2.66
51
EXAMPLE 9
• The following table lists the probability distribution of
the number of student taken course per semester in
science centre:
x
3
4
5
6
7
P(x)
0.37
0.26
0.18
0.11
0.08
• Calculate the mean and standard deviation for this
probability distribution
52
SOLUTION 9
• Calculate the mean for this probability distribution:
=
=
x
3
4
5
6
7
P(x)
0.37
0.26
0.18
0.11
0.08
= 3 ∗ 0.37 + 4 ∗ 0.26 + 5 ∗ 0.18 + 6 ∗ 0.11 + 7 ∗ 0.08 = 4.27
53
SOLUTION 9
• Calculate the standard deviation for this probability
distribution:
=
x
3
4
5
6
7
P(x)
0.37
0.26
0.18
0.11
0.08
= 3 ∗ 0.37 + 4 ∗ 0.26 + 5 ∗ 0.18 + 6 ∗ 0.11 + 7 ∗ 0.08 = 19.87
= 19.87 − 4.27 = 1.64
= 1.64 = 1.28
54
EXAMPLE 10
• Let the probability distribution function of X be
defined by:
– Find
– Calculate mean and variance
– Find the value of E(2X-9) and Var(3X+10)
55
SOLUTION 10
• Find
≤3 =
3 =P X=1 +P X=2 +P X=3
2
2
2
=
+
+
= 0.875
64
64
64
56
SOLUTION 10
• Calculate mean and variance
=
1∗
=
2
2
2
2
2
2
1
+2∗
+3∗
+4∗
+5∗
+6∗
+7∗
= 1.985
64
64
64
64
64
64
64
=
2
2
2
2
2
2
1
1 ∗
+2 ∗
+3 ∗
+4 ∗
+5 ∗
+6 ∗
+7 ∗
= 5.728
64
64
64
64
64
64
64
= 5.728 − 1.985 = 1.78
57
SOLUTION 10
• Find the value of E(2X-9)
2 −9 =2
− 9 = 2 1.985 − 9 = −5.03
58
SOLUTION 10
• Find the value of Var(3X+10)
3 + 10 = 3
+
10 = 9 1.78 + 0 = 16.02
59
4. CONTINUOUS RANDOM
VARIABLES (RV)
CONTINUOUS RV
• A random variable that can take any numeric value
within a range of value. The range may be infinite or
bounded at either both ends.
X, representing
Possible values of X
The height of the students
150cm <X<170cm
The weight of the students
45kg<X<85kg
The time to complete a quiz
5 minutes to 10 minutes
61
PROBABILITY DENSITY FUNCITON (PDF) FOR CONTINUOUS RV
• Theory 7
– Properties of the Probability Density Function (PDF) of
Continuous RV
62
EXAMPLE 11
• Let X be a continuous RV of X with probability density
function (PDF)
• Show that the function is the PDF of X
63
SOLUTION 11
• Show that the function is the PDF of X
( )
=
2
=
2
2
1
=1
0
64
EXAMPLE 12
• The continuous RV of X has the probability
distribution function:
• Find
– The value of k
– P(0.5<X<1)
– P(X>0.25)
65
SOLUTION 12
• Find the value of k:
( )
=
2 ( −
)
=
2
2
−
2
3
1
=1
0
=3
66
SOLUTION 12
• Find P(0.5<X<1), k=3
6( −
)
=
.
6
6
−
2
3
1
= 0.5
0.5
• Find P(X>0.25), k=3
6( −
.
)
6
6
=
−
2
3
1
= 0.84
0.25
67
CUMULATIVE DISTRIBUTION FUNCTION OF CONTINUOUS RV
• Theory 8
– The CDF of a continuous RV X is described as the
function
=
≤
=
( )
−∞<
<∞
• F(x) is an increase function:
=1
– lim
→
=0
– lim
→
68
CUMULATIVE DISTRIBUTION FUNCTION OF CONTINUOUS RV
• The CDF of a continuous RV X can be calculated by
using the following formulas:
≤
= ( )
–
>
=1−
≤
=1− ( )
–
< ≤
=
≤ ≤
=
≤ <
=
< <
=
− ( )
–
69
EXAMPLE 13
• The continuous RV of X has the probability
distribution function
• Find the CDF of X. By using it, calculate
– P(X<0.5)
– P(X>=0.8)
– P(0.5<=X<0.8)
70
SOLUTION 13
• Find the CDF of X, k=3:
=
≤
=
< 0,
0<
( )
=
( )
−∞<
<∞
=0
< 1,
=
( )
+
( )
=0+
( )
+
( )
+
(6 − 6
)
=3
−2
(6 − 6
)
> 1,
=
( )
=0+
+
0
=1
= 3
0, < 0
− 2 ,0 ≤
1, ≥ 1
<1
71
SOLUTION 13
• P(X<0.5):
≤
= ( )
= 3
0.5 = 3(0.5) −2 0.5
= 0.5
0, < 0
− 2 ,0 ≤
1, ≥ 1
<1
• P(X>=0.8):
≤
= ( )
≥ 0.8 = 1 −
≤ 0.8 = 1 −
0.8 = 1 − 3 0.8
− 2 0.8
= 0.104
• P(0.5<=X<0.8):
<
≤
=
0.5 ≤ < 0.8 =
= 3 0.8 − 2 0.8
≤
≤
=
≤
<
=
<
<
=
− ( )
−
= 0.8 − 0.5
− 3 0.5 − 2 0.5
= 0.396
72
EXPECTED VALUE, VARIANCE, AND STANDARD VARIATION
• Theory 9
– Expected value, µ @ E(X) of continuous RV
• Theory 10
– Variance,
of continuous RV
• Theory 11
– The standard deviation, of the probability
distribution of a continuous RV is the square root
of the variance
73
EXAMPLE 14
• Let X be a continuous RV of X with probability density
function (p.d.f)
• Find the mean and variance of X
74
SOLUTION 14
• Mean and variance of X:
=
=
=
1
2
= −
2
3
.
.
=
=
.2
2
=
3
.2
=
2
4
1 2
= = 0.66
0 3
1 1
= = 0.5
0 2
1
=
18
75
CHAPTER 1
FINISHED