doane chapter 07

8/22/2019 Doane Chapter 07

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Continuous Distributions

Chapter

7

Continuous Variables

Describing a Continuous Distribution

Uniform Continuous Distribution

Normal Distribution

Standard Normal Distribution

Normal Approximation to the Binomial (Optional)

Normal Approximation to the Poisson (Optional)

Exponential Distribution

Triangular Distribution


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Continuous Variables

Discrete Variable each value ofXhas its own

probability P(X).

Continuous Variable events are intervals andprobabilities are areas underneath smooth

curves. A single point has no probability.

McGraw-Hill/Irwin 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Events as In tervals


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Probability Density Function (PDF)

For a continuous

random variable,the PDF is an

equation that shows

the height of the

curve f(x) at eachpossible value ofX

over the range ofX.


PDFs and CDFs

Normal PDF


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Continuous PDFs:

Denoted f(x)

Must be nonnegative Total area under

curve = 1

Mean, variance and

shape depend onthe PDFparameters

Reveals the shapeof the distribution


PDFs and CDFs

Normal PDF


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Continuous CDFs:

Denoted F(x)

Shows P(X


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Continuous probability functions are smooth curves.

Unlike discrete

distributions, thearea at any

single point = 0.

The entire area under

any PDF must be 1. Mean is the balance

point of the distribution.


Probabi l i t ies as Areas


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Expected Value and Variance


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Character ist ics of the Uni form Distr ibu t ion

IfXis a random variable that is uniformly

distributed between a and b, its PDF has

constant height. Denoted U(a,b)

Area =

base x height =

(b-a) x 1/(b-a) = 1


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The CDF increases linearly to 1.

CDF formula is

(x-a)/(b-a)


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Example: Anesthesia Effect iveness

An oral surgeon injects a painkiller prior to

extracting a tooth. Given the varying

characteristics of patients, the dentist views the

time for anesthesia effectiveness as a uniform

random variable that takes between 15 minutes

and 30 minutes.

Xis U(15, 30) a = 15, b = 30, find the mean and standard

deviation.


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m =a + b

2=

15 + 30

2= 22.5 minutes

s =(ba)2

12= 4.33 minutes(30 15)

2

12=

Find the probability that the anesthetic takes between

20 and 25 minutes.P(c


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P(20


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Normal Distribution


Character ist ics of the Normal Distr ibu t ion

Normal or Gaussian distribution was named for

German mathematician Karl Gauss (1777

1855).

Defined by two parameters, m and s

Denoted N(m, s)

Domain is


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Normal Distribution




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Normal Distribution



Normal PDF f(x) reaches a maximum at m and

has points of inflection at m + s

Bell-shaped curve


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Normal Distribution



Normal CDF


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Normal Distribution



All normal distributions have the same shape but

differ in the axis scales.

Diameters of golf balls

m = 42.70mms = 0.01mm

CPA Exam Scores

m = 70s = 10


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Normal Distribution


What is Normal?

A normal random variable should:

Be measured on a continuous scale.

Possess clear central tendency. Have only one peak (unimodal).

Exhibit tapering tails.

Be symmetric about the mean (equal tails).


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Character ist ics o f the Standard Normal

Since for every value ofm and s, there is a

different normal distribution, we transform a

normal random variable to a standard normal

distribution with m = 0 and s = 1 using the

formula:

z=xm

s Denoted N(0,1)


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Character ist ics o f the Standard Normal


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Character ist ics o f the Standard Normal Standard normal PDF f(x) reaches a maximum at

0 and has points of inflection at +1.

Shape is

unaffected by the

transformation.

It is still a bell-

shaped curve.


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Character ist ics o f the Standard Normal Standard normal CDF


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Character ist ics o f the Standard Normal A common scale from -3 to +3 is used.

Entire area under the curve is unity.

The probability of an event P(z1 < Z< z2) is adefinite integral off(z).

However, standard normal tables or Excel

functions can be used to find the desired

probabilities.


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Normal A reas from Append ix C-1 Appendix C-1 allows you to find the area under

the curve from 0 to z.

For example, findP(0 < Z< 1.96):


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Normal A reas from Append ix C-1

.5000

.5000 - .4750 = .0250

Now find P(Z< 1.96):


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McGraw-Hill/Irwin 2007 The McGraw-Hill Companies, Inc All rights reserved

Normal A reas from Append ix C-1 Now find P(-1.96 < Z< 1.96).

Due to symmetry, P(-1.96 < Z) is the same as

P(Z< 1.96).

So, P(-1.96 < Z< 1.96) = .4750 + .4750 = .9500

or 95% of the area under the curve.

.9500


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McGraw-Hill/Irwin

2007 The McGraw-Hill Companies Inc All rights reserved

Basis fo r the Emp ir ical Rule Approximately 95% of the area under the curve

is between + 2s

Approximately 99.7% of the area under the curve

is between + 3s


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McGraw-Hill/Irwin

2007 The McGraw-Hill Companies Inc All rights reserved

Normal A reas from Append ix C-2 Appendix C-2 allows you to find the area under

the curve from the left ofz(similar to Excel).

For example,

.9500

P(Z< -1.96)P(Z< 1.96) P(-1.96 < Z< 1.96)


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Normal Areas from Append ices C-1 or C-2 Appendices C-1 and C-2 yield identical results.

Use whichever table is easiest.

Findingz for a Given Area Appendices C-1 and C-2 be used to find the

z-value corresponding to a given probability.

For example, what z-value defines the top 1% of

a normal distribution?

This implies that 49% of the area lies between 0

and z.


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Look for an

area of .4900

in Appendix

C-1:

Findingz for a Given Area

Without

interpolation,

the closest wecan get is

z = 2.33


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Some important Normal areas:

Findingz for a Given Area


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Finding Normal A reas w ith Excel


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d d l b


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S d d l b


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S d d N l Di ib i


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Find ing A reas by us ing Standardized Variables

Suppose John took an economics exam and

scored 86 points. The class mean was 75 with a

standard deviation of 7. What percentile is John

in (i.e., find P(X< 86)?

zJohn =xm

s=

86 75

7= 11/7 = 1.57

So Johns score is 1.57 standard deviations about

the mean.

S d d N l Di ib i


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Find ing A reas by us ing Standardized Variables

P(X< 86) = P(Z< 1.57) = .9418

(from Appendix C-2)

So, John is approximately in the 94th percentile.

St d d N l Di t ib ti


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Inverse Normal You can manipulate the transformation formula to

find the normal percentile values (e.g., 5th, 10th,

25th, etc.): x= m + zs Here are some common percentiles

St d d N l Di t ib ti


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Using Excel Without Standard izing Excels NORMDIST and NORMINV function allow

you to evaluate areas without standardizing.

For example, let m = 2.040 cm and s = .001 cm,

what is the probability that a given steel bearingwill have a diameter between 2.039 and 2.042cm?

In other words, P(2.039


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Using Excel Without Standardizing

P(X < 2.042) = .9773 P(X < 2.039) = .1587

P(2.039


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Normal Approximation to the Binomial


When is App rox imat ion Needed? Binomial probabilities are difficult to calculate

when n is large.

Use a normal approximation to the binomial.

As n becomes large, the binomial bars become

more continuous and smooth.

N l A i ti t th Bi i l


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When is App rox imat ion Needed?

Rule of thumb: when np > 10 and n(1-p) > 10,

then it is appropriate to use the normal

approximation to the binomial. In this case, the binomial mean and standard

deviation will be equal to the normal m and s,

respectively.

m = nps = np(1-p)



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Example Coin Flips If we were to flip a coin n = 32 times and p = .50,

are the requirements for a normal approximation

to the binomial met?

Are np > 5 and n(1-p) > 10? np = 32 x .50 = 16

n(1-p) = 32 x (1 - .50) = 16

So, a normal approximation can be used.

When translating a discrete scale into a

continuous scale, care must be taken about

individual points.



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Example Coin Flips For example, find the probability of more than 17

heads in 32 flips of a fair coin.

This can be written

as P(X> 18). However, more

than 17 actually

falls between 17

and 18 on a discretescale.



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Example Coin Flips Since the cutoff point for more than 17 is halfway

between 17 and 18, we add 0.5 to the lower limit

and find P(X> 17.5).

This addition toXis called the Continuity

Correction.

At this point, the problem can be completed as

any normal distribution problem.



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Cont inu i ty Correct ion The table below shows some events and their

cutoff point for the normal approximation.

Normal Approximation to the Poisson


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When is App rox imat ion Needed? The normal approximation to the Poisson works

best when lis large (e.g., when l exceeds thevalues in Appendix B).

Set the normal m and s equal to the Poisson mean

and standard deviation.

m = ls = l



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Example Util i ty B i l ls On Wednesday between 10A.M. and noon

customer billing inquiries arrive at a mean rate of

42 inquiries per hour at Consumers Energy. What

is the probability of receiving more than 50 calls?

l = 42 which is too big to use the Poisson table. Use the normal approximation with

m = l = 42s = l = 42 = 6.48074



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Example Util i ty B i l ls

To find P(X> 50) calls, use the continuity-

corrected cutoff point halfway between 50 and 51

(i.e.,X= 50.5).

At this point, the problem can be completed as

any normal distribution problem.



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Character ist ics of the Exponent ia l Distr ibut ion If events per unit of time follow a Poisson

distribution, the waiting time until the next event

follows the Exponential distribution.

Waiting time until the next event is a continuousvariable.



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Character ist ics of the Exponent ia l Distr ibut ion



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Character ist ics of the Exponent ia l Distr ibut ion

Probability of waiting more thanx Probability of waiting less thanx



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Example Cus tomer Wait ing Time Between 2P.M. and 4P.M. on Wednesday, patient

insurance inquiries arrive at Blue Choice

insurance at a mean rate of 2.2 calls per minute.

What is the probability of waiting more than 30seconds (i.e., 0.50 minutes) for the next call?

Set l = 2.2 events/min andx= 0.50 min P(X> 0.50) = e

lx= e

(2.2)(0.5)

= .3329or 33.29% chance of waiting more than 30

seconds for the next call.



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Example Cus tomer Wait ing Time

P(X> 0.50) P(X< 0.50)



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Inverse Exponential If the mean arrival rate is 2.2 calls per minute, we

want the 90th percentile for waiting time (the top

10% of waiting time).

Find thex-valuethat defines the

upper 10%.



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Inverse Exponential P(Xx) = .10

So, elx = .10 -lx= ln(.10)

= -2.302585

x= 2.302585/l= 2.302585/2.2

= 1.0466 min. 90% of the calls will arrive within 1.0466 minutes

(62.8 seconds).



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Inverse Exponential Quartiles for Exponential with l = 2.2



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p


Mean Time Between Events Exponential waiting times are described as

Mean time between events (MTBE) = 1/l 1/MTBE = l = mean events per unit of time In a hospital, if an event is patient arrivals in an

ER, and the MTBE is 20 minutes, thenl = 1/20 = 0.05 arrivals per minute (or 3/hour).



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p


Using Excel In Excel, use =EXPONDIST(x,l,1) to return the

left-tail area P(X


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g


Character ist ics of the Triangular Distr ibut ion A simple distribution that can be symmetric or

skewed.

Ranges from a to band has a mode or peak at c

Denoted T(a,b,c)



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g


Character ist ics of the Triangular Distr ibu t ion



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g


Special Cases: Symmetric Triangu lar A symmetric triangular distribution is centered at 0

Lower limit is identical to the upper limit except for

the sign, with mode 0.

Mean m = 0, standard deviation s = b/ 6 =2.45

This distribution closely

resembles a standard

normal distribution N(0,1) Generate random triangular

data in Excel by summing

RAND()+RAND()


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Applied Statistics inBusiness and Economics

End of Chapter 7

doane chapter 07

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