Probability and Statistics
Lukmanul Hakim Jurusan Teknik Elektro Universitas Lampung
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Course Overview
Reference : Probability and Statistics for Engineers 4th Edition, Irwin Miller – John E. Freund – Richard A.
Johnson, Prentice – Hall International Editions
Introduction
Treatment of Data
Probability
Probability Distributions
Probability Densities
Introduction
Everything dealing with the collection, processing, analysis, and interpretation of numerical data belongs to the
domain of statistics.
In electrical engineering, this includes such tasks as calculating the average length of the downtimes of a computer, calculating the reliability indices of a
power system, failure rate of a power system component, determining
tolerance of a resistor, or error rate in a digital communication.
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A Brief History
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Games of Chance Political Science
Error Treatment
Probability Descriptive Statistics
Data Summary
Statistical Inference Origin
Mid-Eighteenth
Recent Times
Treatment of Data
Pareto and Dot Diagrams
Frequency Distributions and Graphs
Stem-and-Leaf Displays
Descriptive Measures
Quartiles and Other Percentiles
Calculation of x and s
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Pareto and Dot Diagrams
6 Dot Diagram of Cutting Speed Deviations
-2 0 2 4 6 8
Unstable Error Power Tool Other
A Pareto Diagram of Failures
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13
6
2 5
Frequency Distributions (1)
Frequency Distribution is a table that divides a set of data into a suitable number of classes (categories),
showing also the number of items belonging to each class.
Class of Number Numerical Distribution
Class of Quality or Attribute Categorical Distribution
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Frequency Distributions (2) Construction
Deciding how many classes to use
Choosing the limits for each class
Tally Observation
Determining Class Frequencies Notes :
• Number of classes depends on the number of observation and data range.
• Classes do not overlap,
accommodate all data, and are all of the same size.
• Determination of class limits should avoid ambiguities where one of data included in two classes.
• This can be achieved by using more decimal places than the raw data results in CLASS BOUNDARIES
• Representation of each class by its mid-point is called CLASS MARK.
Frequency Distributions (3) Example
See page 8 of the book.
80 determinations of the daily emission (in tons) of sulfur oxides from an industrial plant.
Class marks are (5.0+8.9)/2=6.95 ; (9.0+12.9)/2=10.95 ; 14.95 ; …
Class Interval = (10.95 – 6.95) = 4
Class Boundaries = 4.95 – 8.95 ; 8.95 – 12.95 ; … and so forth
Class Limits
Class Limits TallyTally FrequencyFrequency
5.0 – 8.9
5.0 – 8.9 IIIIII 33
9.0 – 12.9
9.0 – 12.9 IIII IIIIIIII IIII 1010 13.0 – 16.9
13.0 – 16.9 IIII IIII IIIIIIII IIII IIII 1414 17.0 – 20.9
17.0 – 20.9 IIII IIII IIII IIII IIIIIIII IIII IIII IIII IIII 2525 21.0 – 24.9
21.0 – 24.9 IIII IIII IIII IIIIII IIII IIII II 1717 25.0 – 28.9
25.0 – 28.9 IIII IIIIIIII IIII 99 29.0 – 32.9
29.0 – 32.9 IIII 22
Total
Total 8080
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Frequency Distributions (4)
Tons of Sulfur Oxides
Cumulative Frequency
< 4.95 0
< 8.95 3
< 12.95 13
< 16.95 27
< 20.95 52
< 24.95 69
< 28.95 78
< 32.95 80
Tons of Sulfur Oxides
Cumulative Percentage
< 4.95 0
< 8.95 3.75 %
< 12.95 16.25 %
< 16.95 33.75 %
< 20.95 65.00 %
< 24.95 86.25 %
< 28.95 97.50 %
< 32.95 100.00 %
Cumulative Frequency Distributions
Cumulative Percentage Distributions
Graphs of Frequency Distributions (5)
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6.95
10.9514.95
18.9522.95
26.9530.95 3
10
14
25
17
9
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A Histogram of Frequency Distributions of Sulfur Oxides Emission in Tons
4.95
8.95 12.95
16.9520.95
24.9528.95
32.95
Dot Diagrams of Cumulative Frequency Distributions of Sulfur Oxides Emission in
Tons
Stem-and-Leaf Displays (1)
10 -19 2 7 5
20 - 29 9 1 5 3 4 7 1 8 30 - 39 4 9 2 4 7
40 - 49 4 8 2 50 - 59 3
11 II
50 – 59 50 – 59
33 IIIIII
40 – 49 40 – 49
55 IIIIIIII
30 – 39 30 – 39
88 IIII III
IIII III 20 – 29
20 – 29
33 IIIIII
10 – 19 10 – 19
Frequency Frequency Tally
Tally Humidity Readings
Humidity Readings
1. Group data 2. Replace tally with the last digit of data Data is humidity readings rounded to the nearest percent:
29; 44; 12; 53; 21; 34; 39; 25; 48; 23;
17; 24; 27; 32; 34; 15; 42; 21; 28; 37
Stem-and-Leaf Displays (2)
1* 2 7 5
2* 9 1 5 3 4 7 1 8 3* 4 9 2 4 7
4* 4 8 2
5* 3
1 2 7 5
2 9 1 5 3 4 7 1 8 3 4 9 2 4 7
4 4 8 2
5 3
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3. Rewrite 4. * is a placeholder for the last digit data
Each line in the table is a STEM and each digit on a stem to the right of the vertical line is a LEAF. To the left of the vertical line are the
STEM LABELS.
Descriptive Measures (1)
The median of n observations x1, x2, …, xn can be defined loosely as the “middlemost” value once data are arranged according to size.
Median is the value of the observation
numbered
[ (n + 1) / 2 ] ; if n is an odd number If n is an even number, then median is defined as the mean or average of the observations numbered [ n / 2 ] and [ (n + 2) / 2 ]
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n x x
n i
i
1 This is arithmetic mean, also referred as sample meanDescriptive Measures (2) Examples
1. Find mean and median of 15, 14, 2, 27, 13 Mean,
Median, the third largest value is 14
2. Find mean and median of 11, 9, 17, 19, 4, 15 Mean,
Median, the average value of 3rd and 4th largest value which are 15 and 11, result in 26/2 = 13
2 . 5 14
13 27
2 14
15
x
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Descriptive Measures (3)
VARIANCE
STANDARD DEVIATION
Descriptive Measures (4)
1. Examples : Delay times for cutting six parts on an engine lathe are 0.6, 1.2, 0.9, 1.0, 0.6, and 0.8 minutes. Calculate s.
2. Coefficient of Variation
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Quartiles and Other Percentiles
When an ordered data set is divided into quarters, the resulting division points are called the sample QUARTILE.
The sample 100 pth percentile is a value such that at least 100 p % of the observations are at or below this value and at least 100 (1 – p) % are at or above this value.
The quartiles are the 25th, 50th, and 75th percentiles.
Probability
Sample Spaces and Events
Counting
Probability
The Axioms of Probability
Some Elementary Theorems
Conditional Probability
Bayes’ Theorem
Mathematical Expectation and Decision Making
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Sample Spaces and Events (1)
Sample Space a set of all possible outcomes of an experiment and usually denoted by the letter S
Experiment may consist of the simple
process of noting whether a switch is turned on or off; determining the time it takes a car to accelerate to 30 miles per hour; or the
complicated process of finding the mass of electron
Outcome of an experiment may be a simple choice between two alternatives; the result of a direct measurement or count; or an answer obtained after extensive measurements and calculations.
Sample Spaces and Events (2)
Discrete Sample Space
It has
finitely many or countable infinity of elements e.g. various ways in which a president and a vice-president can be selected from among the 25 members of an organization
Continuous Sample Space
Elements of sample space constitute a continuum e.g. a person is interested in the nitrogen oxide emission of a
given car in grams per mile
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Sample Spaces and Events (3)
EVENT is any subset of a sample space
Mutually Exclusive Events have no elements in common
Example : A government agency must decide where to locate two new computer research facilities and that (for a certain purpose) it is of interest to indicate how many of them will be located in Texas and how many in
California.
Sample Spaces and Events (4)
The sample space can be written as :
S={ (0,0), (1,0), (0,1), (2,0), (1,1), (0,2) }
1. C = { (1,0), (0,1) } is the event that, between them, Texas and California will get one of the two research facilities
2. D = { (0,0), (0,1), (0,2) } is the event that Texas will not get either of the two research facilities
3. E = { (0,0), (1,1) } is the event that Texas and California will get equally many of the facility
4. Events C and E have no elements in common mutually exclusive events
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Sample Spaces and Events (5)
With reference to the above example,
list the outcomes comprising each of
the following events and also express
the events in words:
Sample Spaces and Events (6)
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is the event that neither Texas or California will get both of the new research facilities
is the event that Texas will not get either of the two new facilities and California will get one
is the event that Texas will get at least one of the new computer research facilities
Counting (1)
Theorem 3.1. “ If sets A1, A2,…, Akcontain respectively, n1, n2,…, nk elements, there are n1∙n2∙∙∙nk ways of choosing first an
element of A1, then an element of A2,…, and finally an element of Ak.
Example : In how many different ways can an organization with membership of 25 choose a vice president and a president?
Solution : Vice president can be chosen in 25 ways, Solution : Vice president can be chosen in 25 ways,
subsequently the president in 24 ways and altogether are subsequently the president in 24 ways and altogether are 25∙25∙24 = 600 ways in which the whole choice can be made.24 = 600 ways in which the whole choice can be made.
Counting (2)
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Theorem 3.2. The number of permutations of r objects selected from a set of n distinct objects is:
or, in factorial notation
Counting (3)
Example : In how many ways can one make a first, second, third, and fourth choice among 12 firms
leasing construction equipment ?
Solution : For n=12 and r=4, the first formula yields
and the second formula yields
Counting (4)
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Theorem 3.3. The number of ways in which r objects can be selected from a set of n distinct objects is
or, in factorial notation,
Counting (5)
Example : In how many different ways can the
director of a research laboratory choose two chemist from among seven applicants and three physicists from among nine applicants?
Solution : Two chemists can be chosen in three physicists can be chosen in
the whole selection can be made in
Probability (1)
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Classical Probability Concept
If there are n equally likely possibilities, of which one must occur and s are regarded as favorable, or as a “success”, then the probability of a “success” is given by s/n
Example : If 3 of 20 tires are defective and 4 of them are randomly chosen for inspection (that is, each tire has the same chance of being
selected), what is the probability that only one of
the defective tires will be included?
Probability (2)
Solution : There are equally likely ways of choosing 4 of 20 tires, so n = 4,845. The number of favorable outcomes is the number of ways in which one of the defective tires and three of the non-defective tires can be selected, or
The probability is
Probability (3)
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The Frequency Interpretation of Probability
The probability of an event (happening or outcome) is the proportion of times the event would occur in a long run of repeated experiments
Example : If records show that 294 of 300 ceramic
insulators tested were able to withstand a certain thermal shock, what is the probability that any one such insulator will be able to withstand the thermal shock?
Solution : Among the insulators tested, 294/300 = 0.98 were able to withstand the thermal shock, and we use this figure as an estimate of the probability
The Axioms of Probability (1)
The first axiom states that probabilities are real numbers on the interval from 0 to 1.
This axiom states that the sample space as a whole is
assigned a probability of 1 and this expresses the idea the probability of a certain event which must happen is equal to 1
The third axiom states that probability functions must be additive.
The Axioms of Probability (2) Example
If an experiment has the three possible and mutually
exclusive outcomes A, B, and C, check in each case whether the assignment of probabilities is permissible:
(a) P(A)=1/3, P(B)=1/3, and P(C)=1/3
(b) P(A)=0.64, P(B)=0.38, and P(C)= - 0.02 (c) P(A)=0.35, P(B)=0.52, and P(C)=0.26 (d) P(A)=0.57, P(B)=0.24, and P(C)=0.19
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Some Elementary Theorems (1)
Theorem 3.4. If A1, A2,…,An are mutually exclusive events in a sample space S, then
Example : The probability that a customer testing service will rate a new antipollution device for cars very poor, poor, fair, good, very good, or excellent are 0.07, 0.12, 0.17, 0.32,0.21, and 0.11. What are the probabilities that it will rate the device:
(a) Very poor, poor, fair, or good;
(b) Good, very good, or excellent?
Solution : (a) 0.07+0.12+0.17+0.32=0.68 (b) 0.32+0.21+0.11=0.64
Some Elementary Theorems (2)
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Theorem 3.5. If A is an event in the finite sample space S, then P(A) equals the sum of the
probabilities of the individual outcomes comprising A
Figure 3.8, page 60 of the book.
Find P(E1), P(P1), P(C1), P(E1 ∩ P1), and P(E1 ∩ C1) Solution:
P(E1) = 0.07 + 0.13 + 0.06 + 0.05 + 0.07 + 0.02 = 0.40 P(P1) = 0.07 + 0.13 + 0.06 + 0.07 + 0.14 + 0.07 + 0.02
+ 0.03 + 0.01 = 0.60
P(C1) = 0.07 + 0.05 + 0.07 + 0.08 + 0.02 + 0.01 = 0.30 P(E1 ∩ P1) = 0.07 + 0.13 + 0.06 = 0.26
and
P(E1 ∩ C1) = 0.07 + 0.05 = 0.12
Some Elementary Theorems (3)
Theorem 3.6. If A and B are any events in S, then P(A
UB) = P(A) + P(B) – P(A
∩B)
Example : If the probabilities are 0.87, 0.36, and 0.29 that a family, randomly chosen as part of a sample survey in a large metropolitan area, owns a color television set, a
black-and-white set, or both, what is the probability that a family in this area will own one or the other or both kinds of sets?
Solution : P(C)=0.87, P(BW)=0.36, and P(C ∩ BW)=0.29 P(C U BW) = 0.87 + 0.36 – 0.29 = 0.94
Some Elementary Theorems (4)
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Theorem 3.7. If A is any event in S, then P(A’) = 1 – P(A)
Example : Referring again to the lawn-mower-rating example and the results on page 61, find
(a) the probability that a lawn mower will not be rated easy to operate
(b) the probability that a lawn mower will be rated as either not being easy to operate or not having a high average repair cost.
Solution :
(a) P(E1’) = 1 – P(E1) = 1 – 0.40 = 0.60
(b) P(E1’ U C1’) = P[(E1 ∩ C1)’] = 1 – P(E1 ∩ C1) = 1 – 0.12 = 0.88
Conditional Probability (1)
Example : If the probability that a communication system will have high fidelity is 0.81 and the probability that it will have high fidelity and high selectivity is 0.18, what is the
probability that a system with high fidelity will also have high selectivity?
If A and B are any events in S and P(B) ≠ 0, the conditional probability of A given B is
Solution : If A is the event that a communication system has high selectivity and B is the event that it has high
fidelity, we have P(B)=0.81 and P(A ∩ B)=0.18, and substitution into the formula yields
Example : The supervisor of a group of 20 construction workers wants to get the opinion of 2 of them (to be
selected at random) about certain new safety regulations. If 12 of them favor the new regulations and the other 8 are against it, what is the probability that both of the workers chosen by the supervisor will be against the new safety regulations?
Conditional Probability (2)
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Theorem 3.8. If A and B are any events in S, then
Conditional Probability (3)
Solution :
Assuming equal probabilities for each
selection (which is what we mean by the selections being random), the probability that the first worker selected will be
against the new safety regulations is (8/20), and the probability that the
second worker selected will be against
the new safety regulations given that the
first one is against them is (7/19). Thus
the desired probability is (8/20)∙(7/19) =
(14/95)
Conditional Probability (4)
Theorem 3.9. If A and B are independent events, then
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Example : Two cards are drawn at random from an ordinary deck of 52 playing cards. What is the probability of getting two aces if (a) the first card is replaced before the second card is drawn; (b) the first card is not replaced before the second card is drawn?
Solution : (a) Since there are four aces among 52 cards, we get (4/52)∙(4/52) = (1/169) INDEPENDENT EVENT (b) Since there are only three aces among the 51 cards that remain after one ace has been removed from the deck, we get (4/52)∙(3/51)=(1/221)
(1/221) ≠ (4/52)∙(4/52) NOT INDEPENDENT EVENT
Conditional Probability (5)
Example : If P(C)=0.65, P(D)=0.40, and
P(C∩D)=0.24, are the events C and D independent?
Solution : Since P(C)∙P(D)=(0.65)∙(0.40)=0.26 and not 0.24, the two events are not independent.
P(A∩B) is the probability of drawing the two cards of ∩B) is the probability of drawing the two cards of being aces
being aces
P(A) is the probability of the drawing the first card of being P(A) is the probability of the drawing the first card of being aceace
P(B) is the probability of the drawing the second card of P(B) is the probability of the drawing the second card of being ace given that the card of the first draw is replaced.
being ace given that the card of the first draw is replaced.
Since
Since P(A∩B) is not equal to P(A)∩B) is not equal to P(A)∙P(B) then the two events ∙P(B) then the two events are NOT INDEPENDENT EVENTS.
are NOT INDEPENDENT EVENTS.
Bayes’ Theorem (1)
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Theorem 3.10. If B1, B2, …, Bn are mutually exclusive events of which one must occur, then
Theorem 3.11. If B1, B2, …, Bn are mutually exclusive events of which one must occur, then
Bayes’ Theorem (2)
0.60
0.10 0.30
B1
B2
B3
0.95
0.80
0.65
A
A
A P(B1) = 0.60, P(B2) = 0.30,
P(B3) = 0.10
P(A | B1) = 0.95, P(A | B2) = 0.80, P(A | B3) = 0.65
The four attendants of a gasoline service station are supposed to wash the windshield of each customer’s car. Janet, who services 20% of all cars, fails to wash the windshield one time in 20; Tom, who services 60% of all cars, fails to wash the windshield one time in 10; Georgia, who services 15% of all cars, fails to wash the windshield one time in 10; and Peter, who services 5% of all cars, fails to wash the windshield one time in 20. If a customer complains later that her windshield was not washed, what is the probability that her car was serviced by Janet?
Bayes’ Theorem (3)
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Mathematical Expectation and Decision Making (1)
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If the probabilities of obtaining the amounts a
1, a
2,
…, or a
kare p
1, p
2, …, and p
k, then the mathematical expectation is
Example : An engineering firm is faced with the task of preparing a proposal for a research contract. The cost of preparing the proposal is $5,000 and the probabilities for potential gross profits of $50,000; $30,000; $10,000; or $0 are 0.20, 0.50, 0.20 and 0.10, provided that proposal is accepted. If the probability is 0.30 that the firm’s proposal will be accepted, what is its expected net profit?
Mathematical Expectation and Decision Making (2)
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Solution : The probability that the firm will make a net profit of
$45,000 ($50,000 minus the cost of the proposal) is (0.30)
(0.20) = 0.06. Similarly, the probabilities that the firm will make net profits of $25,000 or $5,000 are (0.30)(0.50) = 0.15 and (0.30)(0.20) = 0.06, whereas the probability of a $5,000 loss is (0.30)(0.10) + 0.70 = 0.73, allowing for the possibility that the proposal will not be accepted. Thus, the expected net profit is 45,000(0.06) + 25,000(0.15) + 5,000(0.06) – 5,000(0.73) =
$3,100
Risking $5,000 with 0.73 chance of losing it to make an expected profit of $3,100.
Tutorial 1
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The probability that a man will be alive in 10 years is 0.8 and the probability that his wife will be alive in 10 years is 0.9. Find the probability that in 10
years, (a) both will be alive, (b) only the man will be alive, (c) only the wife will be alive, (d) at least one will be alive, (e) none of them will be alive.
(a) 0.8 x 0.9 = 0.72 (b) 0.8 x 0.1 = 0.08 (c) 0.2 x 0.9 = 0.18
(d) 0.8 + 0.9 – (0.8 x 0.9) = 0.98 or 0.72 + 0.08 + 0.18 = 0.98
(e) 0.2 x 0.1 = 0.02
Tutorial 2
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take one take one take one
what is the probability of this ball being white ? Event A P(white-white-white) = (1/3)(3/4)(3/5) = 9/60
Event B P(red-white-white) = (2/3)(2/4)(3/5) = 12/60 Event C P(red-red-white) = (2/3)(2/4)(2/5) = 8/60 Event D P(white-red-white) = (1/3)(1/4)(2/5) = 2/60
P(A) + P(B) + P(C) + P(D) = (9/60) + (12/60) + (8/60) + (2/60)
= (31/60)
Tutorial 3
An electronic controlling mechanism requires five identical memory chips. In how many ways can this mechanism be assembled using 5 given chips?
Tutorial 4
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If the probability that a research project will be well planned is 0.80 and the probability that it will be
well planned and well executed is 0.72, what is the probability that a research project that is well
planned will also be well executed?
Substitution into the formula for a conditional
probability yields (0.72/0.80) = 0.90
