[email protected] MTH16_Lec-21_sec_11-2_Continuous_PDFs.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical &

[email protected] • MTH16_Lec-21_sec_11-2_Continuous_PDFs.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics§11.2

ProbabilityDistribution

Fcns



Review §

Any QUESTIONS About• §11.1 Discrete Probability

Any QUESTIONS About HomeWork• §11.1

→ HW-20

11.1



§11.2 Learning Goals

Define and examine continuous probability density/distribution functions

Use uniform and exponential probability distributions

Study joint probability distributions

http://blogs.sas.com/content/iml/files/2012/07/mvnormalpdf.png



Probability Distribution

Consider Data on the Height of a sample group of 20 year old Men

Ht (in) No.

64 164.5 065 065.5 066 266.5 467 567.5 468 868.5 1169 1269.5 1070 970.5 871 771.5 572 472.5 473 373.5 174 174.5 075 1

We can Plot this Frequency Data using bar

y_abs=[1,0,0,0,2,4,5,4,8,11,12,10,9,8,7,5,4,4,3,1,1,0,1]xbins = [64:0.5:75];axes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenbar(xbins, y_abs, 'LineWidth', 2),grid, ... xlabel('\fontsize{14}Height (Inches)'), ylabel('\fontsize{14}Height (Inches)'),... title(['\fontsize{16}Height of 20 Yr-Old Men',])

62 64 66 68 70 72 74 760

2

4

6

8

10

12

Height (Inches)

He

igh

t (In

che

s)

Height of 20 Yr-Old Men



Probability Distribution Fcn (PDF)

Because the Area Under the Scaled Plot is 1.00, exactly, The FRACTIONAL Area under any bar, or set-of-bars gives the probability that any randomly Selected 20 yr-old man will be that height

e.g., from the Plot we Find • 67.5 in → 4%• 68 in → 8%• 68.5 in → 11%

Summing → 23 % Thus by this data-

set 23% of 20 yr-old men are 67.25-68.75 inches tall



Random variables can be Discrete or Continuous

Discrete random variables have a countable number of outcomes• Examples: Dead/Alive, Red/Black,

Heads/Tales, dice, deck of cards, etc.

Continuous random variables have an infinite continuum of possible values. • Examples: Battery Current, human weight,

Air Temperature, the speed of a car, the real numbers from 7 to 11.



Continuous Case

The probability function that accompanies a continuous random variable is a continuous mathematical function that integrates to 1.

The Probabilities associated with continuous functions are just areas under a Region of the curve (→ Definite Integrals)

Probabilities are given for a range of values, rather than a particular value • e.g., the probability of Jan RainFall in Hayward,

CA being between 6-7 inches (avg = 5.20”)



Continuous Probability Dist Fcn



Continuous Case PDF Example

Recall the negative exponential function (in probability, this is called an “exponential distribution”):

0 if0

0 if)(

x

xexf

x

This Function Integrates to 1 for limits of zero to infinity as required for all PDF’s

1100

0

xx ee



Continuous Case PDF Example

x

p(x)=e-x

1

For example, the probability of x falling within 1 to 2:

The probability that x is any exact value (e.g.: 1.9476) is 0 • we can ONLY assign

Probabilities to possible RANGES of x

x

1

1 2

p(x)=e-x

NO Area Under a

LINE

23% 23.368.135.

2)(1

12

2

1

2

1

ee

eexp xx



Example DownLoad Wait

When downloading OpenProject SoftWare, the website may put users in a queue as they attempt the download.

The time spent in line before the particular download begins is a random variable with approx. density function

minutes 01for0

min 01min 0for05.004.00045.0

minutes 0for02

x

xxx

x

xP




For this PDF then, What is the probability that a user waits at least five (5) minutes before the download?

SOLUTION: We need P(x) ≥ 5 which can be found by

integration and noting that if x is larger than 10, the probability is zero. Thus by the Probability:

5

5 dxxPxP




ContinuePDFReduction

Thus There is a 43.75% chance of a 5 minute PreDownLoad Wait Time

10

5

2 05.004.00045.0 dxxx

5

5 dxxPxP

1674375.0



Example Build a PDF

Find a value of k so that the following represents a Valid, Continuous Probability Distribution Function



Example Build a PDF

SOLUTION: The function is always NON-negative

for non-negative inputs, so simply need to verify that the definite integral equals 1 (that all probabilities together Add-Up, or Integrate, to 100%).

Thus, the correct value of k produces this functional behavior →

1

dxxf



Example Build a PDF

Because the function is identically zero everywhere outside of the interval [0, k], restrict the evaluation to that interval →

Solve by SubStitution; Let:

Then 1 10

2

kx

x

dxx

x

xdx

du2 dxdu

x

2

1

12

0

kx

x x

du

u

x

1 10

2

k

dxx

x



Example Build a PDF

Then 1 1

2

1

0

kx

x

duu

1ln2

1

1 00

2

kx

x

k

udxx

x

21ln0

2 k

x

210ln1ln 22 k

21ln 2 k

1 10

2

kx

x

dxx

x

21ln1ln 2 k



Example Build a PDF

Finally However, the 0 ≤ x ≤ k interval ends in a

non-negative value so need k-positive:

Thus the Desired PDF

577.212 ek

.otherwise , 0

10 if , 1)(

22

ex

x

xxf



Uniform Density Function

Definition

Graph

OtherWise0

if1

bxaabxf



Example Random No. Generator

A Random Number Generator (RNG) selects any number between 0 and 100 (including any number of decimal places).

Because each number is equally likely, a uniform distribution models the probability distribution.

What is the probability that the RNG selects a number between 50 and 60?




SOLUTION: The Probability Distribution Function:

Then the Probability of Generating a RN between 50 & 60

60

50

100

16050 dxxP




Evaluating the Integral

As Expected find the Probability of a 50-60 RN as 10%

60

50

100

16050 dxxP

60

50100

x

1.0100

50

100

60



Exponential Density Function

Definition

Graph

00

0if

x

xexf

x



Example SmartPhone LifeSpan

The battery of a popular SmartPhone loses about 20% of its charged capacity after 400 full charges.

Assuming one charge per day, the estimated probability density function for the length of tolerable lifespan for a phone that is t years old →

otherwise , 0

0 if , 12.1 12.1 tetf

t



Example SmartPhone LifeSpan

Find the probability that the tolerable lifespan of the SmartPhone is at least 500 days (500 charges).

SOLUTION: The probability of a tolerable lifespan being greater than or equal to 500 days (500/365 years):

365/500

12.1 12.1 dte t

Mt

Me

365/500

12.1

12.1

12.1lim

)365/500(12.112.1lim

ee M

M

%56.212156.00



Joint Probability Distribution Fcn

A joint probability density function f(x, y) has the following properties:

1. f(x, y) ≥ 0 for all points (x, y) in the Cartesian Plane

2. Double Integrates to 1:

3. The Probability that an Ordered Pair, (X, Y) Lies in Region R found by:

x

x

y

ydxdyyxf ,

R

dAyxfRYXP ,in ,



Joint Probability Distribution Fcn

Example joint probability density function Graph



Example Joint PDF

Consider the Joint PDF:

Find: 5 yxP

xyP 5

dxdyex

yx 65

0

5

0

32

5

0

5

032 2 dxe

xy

yyx

5

0

215 22 dxee xx

502152 xx ee

99986.0



WhiteBoard PPT Work

Problems From §11.2• P48 → Traffic Lite Roullette



All Done for Today

FittingPDFs to

Hists



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2







[email protected] MTH16_Lec-21_sec_11-2_Continuous_PDFs.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical &

Documents

chabot mathematics

discrete probability

probability function

continuous functions

continuous random variables

inches tall slide

height inches

yrold men