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Prepared by:

Renna Magdalena

Chapter 6Introduction to Continuous

Probability Distributions


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Chapter Goals

After completing this chapter, you should be able to:

Convert values from any normal distribution to a

standardized z-score Find probabilities using a normal distribution table

Apply the normal distribution to business problems

Recognize when to apply the uniform and exponentialdistributions


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Continuous Probability Distributions

A continuous random variable is a variable that canassume any value on a defined continuum (canassume an uncountable number of values) see

Chapter 5 thickness of an item

time required to complete a task

These can potentially take on any value, depending

only on the ability to measure accurately.


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4

Types of Continuous Distributions

Three types Normal

Uniform

Exponential

A B

Involves determining the probability for a RANGEof values rather than 1 particular incident oroutcome


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5

The Normal Distribution

Bell Shaped

Symmetrical

Mean=Median=Mode

Location is determined by themean,

Spread is determined by thestandard deviation,

The random variable has aninfinite theoretical range:+ to

MeanMedianMode

x

f(x)


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6

By varying the parameters and , we obtain different normaldistributions

Many Normal Distributions


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The Normal Distribution Shape

x

f(x)

Changing shifts the

distribution left or right.

Changing increases or

decreases the spread.


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8

Finding Normal Probabilities

a b x

f(x)P a x b( )

Probability is measured by the areaunder the curve


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f(x)

x

Probability asArea Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so

half is above the mean, half is below

1.0)xP(

0.5)xP( 0.5)xP(


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The Standard Normal Distribution

Also known as the z distribution

Mean is defined to be 0

Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values Values below themean have negative z-values


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The Standard Normal

Any normal distribution (with any mean andstandard deviation combination) can betransformed into the standard normal distribution

(z)

Need to transform x units into z units Where x is any point of interest

Can use the z value to determine probabilities


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Translation to the StandardNormal Distribution

Translate from x to the standard normal (the zdistribution) by subtracting the mean of x anddividing by its standard deviation:

z is the number of standard deviations units that

x is away from the population mean

xz


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Example

If x is distributed normally with mean of100 and standard deviation of 50, the zvalue for x = 250 is

This says that x = 250 is three standarddeviations (3 increments of 50 units) abovethe mean of 100.

3.050

100250

xz


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Comparing x and z units

z

100

3.00

250 x

Note that the distribution is the same, only the scale has changed.

We can express the problem in original units (x) or in standardized

units (z)

= 100

= 50


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The Standard Normal Table

The Standard Normal table in the textbook(Appendix D)

Gives the probability from the mean (zero)

up to a desired value for z

z0 2.00

0.4772Example:

P(0 < z < 2.00) = 0.4772


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The Standard Normal Table gives the probabilitybetween the mean and a certain z value

The z value ALWAYS refers to the area betweensome value (-z or +z) and the mean

Since the distribution is symmetrical, the Standard

Normal Table only displays probabilities for of thefull distribution

The Standard Normal Table(continued)


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The Standard Normal Table

The value within the

table gives theprobability from z =0 up to the desiredz value

z 0.00 0.01 0.02

0.1

0.2

.4772

2.0P(0 < z < 2.00) = 0.4772

The row shows thevalue of z to thefirst decimal point

The column gives the value of z to thesecond decimal point

2.0

.

.

.

(continued)


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General Procedure for FindingProbabilities

1. Determine m and s

2. Define the event of interest

e.g., P(x > x1)

3. Convert to standard normal

4. Use the table to find the probability

xz


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z Table Example

Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.120x8.68

05

88

xz

0.125

88.6

xz

Calculate z-values:


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z Table Example

Suppose x is normal with mean 8.0 andstandard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.120x8.68

P(8 < x < 8.6)

= 8 = 5

= 0 = 1

(continued)


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Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

0.0478

.02

0.1 .0478

Standard Normal ProbabilityTable (Portion)

0.00

= P(0 < z < 0.12)

P(8 < x < 8.6)


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Suppose x is normal with mean 8.0and standard deviation 5.0.

Now Find P(x < 8.6) The probability of obtaining a value less than 8.6

Z

8.6

8.0

P = 0.5


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Now Find P(x < 8.6)

(continued)

Z

0.12

0.0478

0.00

0.5000P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= 0.5000 + 0.0478 = 0.5478


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Upper Tail Probabilities


Now Find P(x > 8.6)

Z

8.6

8.0


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P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= 0.5000 - 0.0478 = 0.4522

Now Find P(x > 8.6)(continued)

Z

0.12

0Z

0.12

0.0478

0

0.5000 0.4522

Upper Tail Probabilities


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Lower Tail Probabilities


Now Find P(7.4 < x < 8)

Z

7.48.0


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Lower Tail Probabilities

Now Find P(7.4 < x < 8)the probabilitybetween 7.4 and the mean of 8

Z

7.48.0

The Normal distribution is symmetric,

so we use the same table even if z-values are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= 0.0478

(continued)

0.0478


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Normal Probabilities in PHStat

We can use Excel and PHStat to quickly

generate probabilities for any normal

distribution

We will find P(8 < x < 8.6) when x is

normally distributed with mean 8 and

standard deviation 5


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PHStat Dialogue Box

Select desired options and

enter values


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PHStat Output


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Empirical Rules

1 covers about 68% of

xs

f(x)

x

1

1

What can we say about the distribution of valuesaround the mean if the distribution is normal?

68.26%

Recall

Tchebyshev

from Chpt. 3


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The Empirical Rule

2covers about 95% of xs

3covers about 99.7% of xs

x

2 2

x

3 3

95.44% 99.72%

(continued)


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Importance of the Rule

If a value is about 2 or more standarddeviations away from the mean in a normaldistribution, then it is far from the mean

The chance that a value that far or fartheraway from the mean is highly unlikely, giventhat particular mean and standard deviation


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The Uniform Distribution

The uniform distribution is a probability distribution thathas equal probabilities for all possible outcomes of the

random variable

Referred to as the distribution of little information

Probability is the same for ANY interval of the samewidth

Useful when you have limited information about howthe data behaves (e.g., is it skewed left?)


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The Continuous Uniform Distribution:

otherwise0

bxaif

ab

1

wheref(x) = value of the density function at any x value

a = lower limit of the interval of interest

b = upper limit of the interval of interest

The Uniform Distribution(continued)

f(x) =


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The mean (expected value) is:

2

baE(x) +

where

a = lower limit of the interval from a to b

b = upper limit of the interval from a to b

The Mean and Standard Deviationfor the Uniform Distribution

The standard deviation is

12

a)(b

2


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Steps for Using theUniform Distribution

1. Define the density function

2. Define the event of interest

3. Calculate the required probability

x

f(x)


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Uniform Distribution

Example: Uniform Probability DistributionOver the range 2 x 6:

2 6

.25

f(x) = = .25 for 2 x 66 - 21

x

f(x)


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Uniform Distribution

Example: Uniform Probability DistributionOver the range 2 x 6:

42

62E(x) +

1.154712

2)(612

a)(b22


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The Exponential Distribution

Used to measure the time that elapses betweentwo occurrences of an event (the time betweenarrivals)

Examples: Time between trucks arriving at a dock

Time between transactions at an ATM Machine

Time between phone calls to the main operator

Recall l = mean for Poisson (see Chpt. 5)


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The Exponential Distribution

ae1a)xP(0

The probability that an arrival time is equal toor less than some specified time a is

where 1/l is the mean time between events and e = 2.7183

NOTE: If the number of occurrences per time period isPoisson with mean l, then the time between occurrences isexponential with mean time 1/ l and the standard deviationalso is 1/l


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Exponential Distribution

Shape of the exponential distribution

(continued)

f(x)

x

l = 1.0(mean = 1.0)

l= 0.5

(mean = 2.0)

l = 3.0(mean = .333)


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Time between arrivals is exponentially distributed with meantime between arrivals of 4 minutes (15 per 60 minutes, onaverage)

1/l = 4.0, so l = .25

P(x < 5) = 1 - e-la = 1 e-(.25)(5) = 0.7135

There is a 71.35% chance that the arrival time betweenconsecutive customers is less than 5 minutes

Example

Example: Customers arrive at the claims counter at the rate of15 per hour (Poisson distributed). What is the probability thatthe arrival time between consecutive customers is less than fiveminutes?


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Using PHStat


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Chapter Summary

Reviewed key continuous distributions normal uniform exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems


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Thank You

End of Chapter 6

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