[email protected] MTH16_Lec-19_sec_10-3_Taylor_Series.pptx 1 Bruce Mayer, PE Chabot College...

[email protected] • MTH16_Lec-19_sec_10-3_Taylor_Series.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics§11.1

Probability& Random-

Vars



Review §

Any QUESTIONS About• §10.3 Power & Taylor Series

Any QUESTIONS About HomeWork• §10.3

→ HW-19

10.3



§11.1 Learning Goals

Define outcome, sample space, random variable, and other basic concepts of probability

Study histograms, expected value, and variance of discrete random variables

Examine and use geometric distributions



Random Experiment

A Random Experiment is a PROCESS• repetitive in nature• the outcome of any trial is uncertain • well-defined set of possible outcomes• each outcome has an associated probability

Examples• Tossing Dice• Flipping Coins• Measuring Speeds of Cars On Hesperian



6.5

Random Experiment…

A Random Experiment is an action or process that leads to one of several possible outcomes.

Some examples:

Experiment OutComes

Flip a coin Heads, Tails Exam Scores Numbers: 0, 1, 2, ..., 100 Assembly Time t > 0 seconds Course Grades F, D, C, B, A, A+



OutComes, Events, SampleSpace

OutCome → is a particular result of a Random Experiment.

Event → is the collection of one or more outcomes of a Random Experiment.

Sample Space → is the collection or set of all possible outcomes of a random experiment.

http://drkatesview.files.wordpress.com/2010/10/rolling-dice.jpg



Example OutComes, etc.

Roll one fair die twice and record the sum of the results.

The Sample Space is all 36 combinations of two die rolls

1st Roll 2nd Roll Total OutComes1 1 2 3 4 5 6 62 1 2 3 4 5 6 63 1 2 3 4 5 6 64 1 2 3 4 5 6 65 1 2 3 4 5 6 66 1 2 3 4 5 6 6

GRAND Total OutComes 36

https://larspsyll.files.wordpress.com/2012/10/dice011.jpg



Example OutComes, etc.

One outcome: 1st Roll = a five,2nd Roll = a two → which canbe represented by theordered pair (5,2)

One Event (or Specified Set of OutComes) is that the sum is greater than nine (9), which consists of the (permutation) outcomes

(6,4), (6,5), (6,6), (4,6), and (5,6)

https://larspsyll.files.wordpress.com/2012/10/dice011.jpg



Random Variable

A Random Variable is a function X that assigns a numerical value to each outcome of a random experiment.

A DISCRETE Random Variable takes on values from a finite set of numbers or an infinite succession of numbers such as the positive integers

A CONTINUOUS Random Variable takes on values from an entire interval of real numbers.



Probability

Probability is a Quotient of the form

Example: Consider 2 rolls of a Fair Die• Probability of (3, 4)

• Probability that the Sum > 9

Probabilityof an Event

=

Total Number of SPECIFIED OutComes

Total Number of POSSIBLE OutComes

Probability =ONE OutCome = 1

=2.78%36 OutComes 36

Probability =FIVE OutComes = 5

= 13.89%36 OutComes 36



Probability OR (U) vs AND (∩)

The Sum>9 is an example of the OR Condition.• The OR Probability is the SUM of the

INDIVIDUAL Probabilities

• The AND Probability is the MULTIPLICATION of the INDIVIDUAL Probabilities

36

5

36

1

36

1

36

1

36

1

36

19

6,46,54,65,66,69

P

PPPPPP



AND Probability

Probability The LIKELYHOOD that a Specified OutCome Will be Realized• The “Odds” Run from 0% to 100%

Class Question: What are the Odds of winning the California MEGA-MILLIONS Lottery?



258 890 085 ... EXACTLY???!!!

To Win the MegaMillions Lottery• Pick five numbers from 1 to 75 • Pick a “MEGA” number from 1 to 15

The Odds for the 1st ping-pong Ball = 5 out of 75

The Odds for the 2nd ping-pong Ball = 4 out of 75, and so On

The Odds for the MEGA are 1 out of 15



258 890 085... Calculated

Calc the OverAll Odds as the PRODUCT of each of the Individual OutComes (AND situation)

15

1

!75

!70!5

15

1

71

1

72

2

73

3

74

4

75

5

Odds

• This is Technically a COMBINATION

085,890,258

1

000,902,066,31

120



258 890 085... is a DEAL!

The ORDER in Which the Ping-Pong Balls are Drawn Does NOT affect the Winning Odds

If we Had to Match the Pull-Order:

Current theX120000,902,066,31

1!7115

!70

15

1

71

1

72

1

73

1

74

1

75

1

Odds

• This is a PERMUTATION



Probability Distribution Function

A probability assignment has been made for the Sample Space, S, of a Particular Random Experiment, and now let X be a Discrete Random Variable Defined on S. Then the Function p such that:

for each value x assumed by X is called a Probability Distribution Function

xXPxp



Probability Distribution Function

A Probability Distribution Function (PDF) maps the possible values of x against their respective probabilities of occurrence, p(x)

p(x) is a number from 0 to 1.0, or alternatively, from 0% to 100%.

The area under a probability distribution function Curve or BarChart is always 1 (or 100%).



Discrete Example: Roll The Die

1/6

1 4 5 62 3

x

xp all

1

x

x p(x)

1 p(x=1)=1/6

2 p(x=2)=1/6

3 p(x=3)=1/6

4 p(x=4)=1/6

5 p(x=5)=1/6

6 p(x=6)=1/6

xp



Example B-School Admission

A business school’s application process awards two points to applications for each grade of A, one point for each grade of B or C, and zero points for lower grades

If Each category of grades is equally likely, what is the probability that a given student meets the admission requirement of five total points from grades from 3 different courses?




SOLUTION: The sample space is the set of 27

outcomes (using “A” to represent a grade of A, “B” to represent a B or C, and “N” to represent a lower grade)

The Entire Sample Space Listed:




The event “student meets admission requirement of five points” consists of any outcomes that total at least five points according to the given scale. i.e. the outcomes

This acceptance Criteria Thus has a Probability

%8.14148.0



Expected Value

The EXPECTED VALUE (or mean) of a discrete Random Variable, X, with PDF p(x) gives the value that we would expect to observe on average in a large number of repetitions of the experiment

That is, the Expected Value, E(X) is a Probability-Weighted Average, µX k

n

kkk

n

kkX xpxxXPxXE

11

)(



Example Coin-Tossing µX

A “friend” offers to play a game with you: You flip a fair coin three times and she pays you $5 if you get all tails, whereas you pay her $1 otherwise

Find this Game’s Expected Value SOLUTION: The sample space for the experiment of

flipping a coin three times:




The expected value is the sum of the product of each probability with its “value” to you in the game:

Since Each outcome is equally likely, All the Probabilities are 1/8=0.125=12.5%:

kkX xPxEV

8

151111111 kk xPxEV




Calculating the Probability Weighted Sum Find

Thus, in the long run of playing this game with your friend, you can expect to LOSE 25¢ per 8-Trial Game

4

1

8

2

8

175

kkX xPxEV



Discrete Random Var Spread

The Expected Value is the “Central Location” or Center of a symmetrical Probability Distribution Function

The VARIANCE is a measure of how the values of X “Spread Out” from the mean value E(X) = µX

The Variance Calculation

2

1

2

1

2Var

n

kkXk

n

kkk xpµxxpXExX



Discrete Random Var Spread

The Square Root of the Variance is called the STANDARD DEVIATION

Quick Example → The standard deviation of the random variable in the coin-flipping game

n

kkXk xpµxX

1

22 Var

046.2

125.0)25.05(

125.0)25.01(...125.0)25.1(

2

times7

22

X



Geometric Random Variable

Consider Again Coin Tossing Take a fair coin and toss as many times

as needed to Produce the 1st Heads. Let X ≡ number of tosses needed for

FIRST Heads. Sample points={H, TH, TTH, TTTH, …} The Probability Distribution of X

X 1 (H) 2 (TH) 3 (TTH) 4 (TTTH) N (TTT…TH)

P(X) 1/2 1/4 1/8 1/16 1/2N




Consider now an UNfair Coin Tossing Flip until the 1st head a biased coin with

70% of getting a tail and 30% of seeing a head,

Let X ≡ number of tosses needed to get the first head.

The Probability Distribution of X

X 1 (H) 2 (TH) 3 (TTH) 4 (TTTH) N (TTT…TH) P(X) 0.3 0.7·0.3 0.7·0.7·0.3 0.7·0.7·0.7·0.3 0.7·0.7·…·0.7·0.3




In these two example cases the OutCome value can be interpreted as “the probability of achieving the first success directly after n-1 failures.” Let:• p ≡ Probability of SUCCESS• Then (1-p) = Probability of FAILURE

Then the OverAll Probability of 1st Success

pppppppnXP n 111111

n−1 Failures Success



Example Exam Pass Rate

The Electrical Engineering Version of the Professional Engineer’s Exam has a Pass (Success) Rate of about 63%

Find the probability of Passing on the • SECOND Try• FOURTH Try

Assuming GeoMetric Behavior

%3.23233.063.063.012 12 XP

%2.3032.063.063.014 14 XP




After Some Algebraic Analysis Find for a GeoMetric Random Variable• Expected Value

• Standard Deviation

p

µXEV X

1

p

pX

1



WhiteBoard PPT Work

Problems From §11.1• P31 → HighWay Safety Stats

TelsaModel S



All Done for Today

RolltheDice



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2



HiWay Safety Stats The Data Probability of any

Given No. of Accidents Per Day

No. Accidentsper Day, X

No. Days of Observation

TotalAccidents

0 6 01 8 82 7 143 2 64 3 125 0 06 2 127 0 08 1 89 1 9

Σtotals = 30 69

No. Accidentsper Day, X P(X) P(X)·X

0 6/30 = 0.2 01 8/30 = 0.2667 0.26672 7/30 = 0.2333 0.46673 2/30 = 0.0667 0.24 3/30 = 0.1 0.45 0/30 = 0 06 2/30 = 0.0667 0.47 0/30 = 0 08 1/30 = 0.0333 0.26679 1/30 = 0.0333 0.3

Σtotals = 1 2.3



HiWay Safety Stats The Expected Value

µX = 2.3 Accidents/Day

EV(X) Interpretation• The Expected Value

of 2.3 Accidents per Day is, on Average, the No. of Accidents likely to occur on any random day of observations

No. Accidentsper Day, X P(X) P(X)·X

0 6/30 = 0.2 01 8/30 = 0.2667 0.26672 7/30 = 0.2333 0.46673 2/30 = 0.0667 0.24 3/30 = 0.1 0.45 0/30 = 0 06 2/30 = 0.0667 0.47 0/30 = 0 08 1/30 = 0.0333 0.26679 1/30 = 0.0333 0.3

Σtotals = 1 2.3



HiWay Safety Stats HistoGram

0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

7

8

No. of Accidents/Day

No

. of D

ays

MTH16 • P11.1-31 HiWay Safety

0 1 2 3 4 5 6 7 8 90

5

10

15

20

25

30

No. of Accidents/Day

% o

f No

. of D

ays

(N

= 3

0 d

ays

)

MTH16 • P11.1-31 HiWay Safety



HiWay Safety Stats

The σ2 Calc →

Then the StdDeviation fromthe Variance

No. Accidentsper Day, xk

(xk − µX)2 P(xk) (xk − µX)2·P(xk)

0 5.29 0.2000 1.05801 1.69 0.2667 0.45072 0.09 0.2333 0.02103 0.49 0.0667 0.03274 2.89 0.1000 0.28905 7.29 0.0000 0.00006 13.69 0.0667 0.91277 22.09 0.0000 0.00008 32.49 0.0333 1.08309 44.89 0.0333 1.4963

Σtotals = 1.000 5.3433

Day

Accidents331.2

Day

Accidents 433.5

2

2



Exam 1st Timers RepeatsChemical 67% 40%Civil 64% 29%Electrical and Computer 63% 28%Environmental 63% 35%Mechanical 72% 41%Structural Engineering (SE) Vertical Component 50% 34%Structural Engineering (SE) Lateral Component 38% 43%

PE Exam Pass RatesGroup 1 PE Exams, October 2013 Pass Rates



Exam1st

TimersRepeats

Agriculture (October 2013) 69% 50%

Architectural 74% 43%

Control Systems (October 2013) 76% 53%

Fire Protection (October 2013) 69% 37%

Industrial 72% 50%

Metallurgical and Materials (October 2013) 62% 0%

Mining and Mineral Processing (October 2013) 71% 37%

Naval Architecture and Marine Engineering 58% 46%

Nuclear (October 2013) 54% 44%

Petroleum (October 2013) 75% 53%

Software 50% NA

Group 2 PE Exams, October 2013 and April 2013 Pass RatesIn most states, and for most exams, the Group 2 exams are given only in October, as indicated in parentheses. The following table shows pass rates of Group 2 examinees.

[email protected] MTH16_Lec-19_sec_10-3_Taylor_Series.pptx 1 Bruce Mayer, PE Chabot College...

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