session 7 (chpt 5 & 6)(3)

FIN60003 Quantitative Analysis

Probability and probability distributions

Business Statistics: A Decision-Making Approach

Chapters 5 and 6Discrete Distributions and

Continuous Probability Distributions

(Normal Distribution)

3

Chapter GoalsAfter completing this chapter, you should be able to: Distinguish between discrete and continuous probability

distributions Compute the expected value and standard deviation for a

discrete probability distribution Discuss the important properties of the normal probability

distribution Recognise when the normal distribution might apply in a

decision making process Find probabilities using a normal distribution table and

apply the normal distribution to business problems

4

Introduction to Probability Distributions

Random Variable A Variable that assigns a numerical value to each

outcome of a random experiment or trial.

Random Variables

Discrete Random Variable

ContinuousRandom VariableCh. 5 Ch. 6

5

A discrete random variable is a variable that can assume only a countable number of values

Many possible outcomes: number of complaints per day number of TV’s in a household number of rings before the phone is answered

Only two possible outcomes: gender: male or female defective: yes or no spreads peanut butter first vs. spreads jelly first

Discrete Probability Distributions

6


A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches

These can potentially take on any value, depending only on the ability to measure with sufficient precision.

7

Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twiceLet x be the number of times 4 comes up (then x could be 0, 1, or 2 times)

Toss a coin 5 times. Let x be the number of heads

(then x = 0, 1, 2, 3, 4, or 5)

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Experiment: Toss 2 Coins. Let x = Number of heads.

T

T

Discrete Probability Distribution

4 possible outcomes

T

T

H

H

H H

Probability Distributionx Value Probability

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

0 1 2 x

.50

.25

Prob

abili

ty

9

Example: Toss 2 coins, x = Number of heads

compute expected value of x:E(x) = (0 x .25) + (1 x .50) + (2 x .25)

= 1.0

Discrete Random Variable Summary Measures

Expected Value of a discrete distribution (Weighted Average)

E(x) = xi P(xi)

x P(x)

0 .25

1 .50

2 .25

Value Probability

10

Rules of probability Each probability must be at least zero

and less than 1

Sum of probabilities must be 1

10 )(xp

1)(xp

Rule 1

Rule 2

11


Example: Toss 2 coins, x = Number of heads, compute expected value of x:

Calculator into STAT mode:0 (x,y) 25 DATA1 (x,y) 50 DATA2 (x,y) 25 DATATo obtain MEAN Press [ALPHA] [x bar] => 1.0 Note: You must enter the value of x first,

then its probability x 100 (as probabilities are 2 decimal places) (0r x 1000 if probabilities are 3 decimal places, etc)

x P(x)

0 .25

1 .50

2 .25

Value ProbabilityInstructions for Sharp 735S/738/738FB

12

Standard Deviation of a discrete distribution

where:E(x) = Expected value of the random variable

x = Values of the random variableP(x) = Probability of the random variable having

the value of x


P(x)E(x)}{xσ 2x

(continued)

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Example: Toss 2 coins, x = Number of heads, compute standard deviation (recall E(x) = 1)


P(x)E(x)}{xσ 2x

.707.50(.25)1)(2(.50)1)(1(.25)1)(0σ 222x

(continued)

Possible number of heads = 0, 1, or 2

Or use [ALPHA] [σx] = 0. 707 (use POPULATION SD)

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Problem 5-4 parts a and c

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Exercise: Discrete Probability Distribution

What is the probability that the sum of the values thrown is:

1. Equal to 7?2. Greater than 9?3. Less than 3?4. Greater than 12?

The probability of two successive rolls of 6?

Let x = sum of the values on the two dice.

1. P( x = 7 ) 2. P( x > 9 )3. P( x < 3 )4. P( x >12 )

P( x = 12 )

Experiment: Tossing 2 Fair Six Sided Dice.

Next map out all possible outcomes of the throws of two dice.

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Possible Outcomes

1 2 3 4 5 6

1 1, 1 1, 2 1, 3 1, 4 1, 5 1, 62 2, 1 2, 2 2, 3 2, 4 2, 5 2, 63 3, 1 3, 2 3, 3 3, 4 3, 5 3, 64 4, 1 4, 2 4, 3 4, 4 4, 5 4, 65 5, 1 5, 2 5, 3 5, 4 5, 5 5, 66 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6 There are 36 equally likely ordered outcomes possible.

First Die O

utcome

Second Die Outcome

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Possible Outcomes

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

There are 36 equally likely ordered outcomes possible The sums of the throws range from 2-12.

First Die O

utcome

Second Die Outcome

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Possible Outcomes

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 83 4 5 6 7 8 9

4 5 6 7 8 9 105 6 7 8 9 10 11

6 7 8 9 10 11 12 There are 36 equally likely ordered outcomes possible The sums of the throws range from 2 to12. The values of x (sum of the values on the two dice) are

not equally likely.

First Die O

utcome

Second Die Outcome

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Discrete Probability Distributionx 2 3 4 5 6 7 8 9 10 11 12

P1/36.0277

2/36.055

3/36.0833

4/36.111

5/36.1388

6/36.1666

5/36.1388

4/36.1111

3/36.0833

2/36.055

1/36.0277

Prob

abili

ty

1/36 2/36 3/36 4/36 5/36 6/36

2 3 4 5 6 7 8 9 10 11 12 x

ExerciseFind: i) P( x = 9 ) ii) P( x > 9 ) iii) P( x < 3 ) iv) P( x >12 )

iv) The probability of two successive experiments resulting in a 12.

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Approaches to Forming Probability Distributions

The probability distributions above have been formed using the Classical Approach to probability assignment.

Applications of this approach are fairly limited. OK for game situations like tossing coins or throwing dice. Can be used in some business application, like the Real

Computer Co example on p150.

How do we form distributions for real life situations like: The weights of students studying at Swinburne The incomes of students studying at Swinburne The number of Jams occurring on a manufacturing machine in

a fixed period of operation?

21

Probability Distributionsin Real Life

The Relative Frequency Of Occurrence approach to probability assignment can be used to form the probability distribution if we have enough data.

Observation of similar items (man made or natural) under similar conditions tend to vary.

The pattern of values forms a distribution.

Applicable to a broad number of situation, however collecting sufficient data is time consuming and costly!

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Probability Distributionsin Real Life

A great variety of distributions occur in practice. Many may be approximated by a few theoretical models

called Reference Distributions. The models are expressed as a mathematical function,

permitting calculation of the proportion of values lying in various regions along the scale of measurement.

Values of the distribution are available in tables. The models can then be used as a Reference

Distribution for solving problems.

!xe)t()x(P

tx

22 2/)(2

1)(

xexf

Poisson Normal

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Reference Distributions

The information required to build these “Theoretical” Reference Distributions is:

The appropriate mathematical model. Estimates of the appropriate model’s generating

parameters like: Mean Standard Deviation Lambda Proportion p

These parameters can be estimated from historical data, provided that the process producing the output is statistically stable.

Estimating parameters requires less data than forming the distribution from Relative Frequencies Of Occurrence.

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Many Probability Distributions

Continuous Probability

Distributions

Binomial

Hypergeometric

Poisson

Probability Distributions

Discrete Probability

Distributions

Normal

Uniform

Exponential

Ch. 5 Ch. 6

25

Probability DistributionsAlthough there are many probability distributions, most situations in

industry can be described by one of the following:

Probability Distributions

Discrete Probability Distributions


Data from Counts Data from Measurement

Binomial

PoissonNormal Measurements

Attributes

Events

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A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height, in inches

These can potentially take on any value, depending only on the ability to measure with sufficient precision.

27

The Normal Distribution ‘Bell Shaped’ Symmetrical Mean, Median and Mode are EqualLocation is determined by the mean, μSpread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + to

Mean = Median = Mode

x

f(x)

μ

σ

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The Normal Distribution… Many data distributions

are approximately normal – both man made and natural

For example: height, weight IQ scores and scientific

measures Most important continuous

distribution in StatisticsBy varying the parameters μ and σ, we obtain different

normal distributions

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The Normal Distribution Shape

x

f(x)

μ

σ

Changing μ shifts the distribution left or right.

Changing σ increases or decreases the spread.

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Finding Normal Probabilities

a b x

f(x) P a x b( )

Probability is measured by the area under the curve

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f(x)

xμ

Probability as Area 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 ( μ 0 . 5μ )xP (

32

Empirical Rules

μ ± 1σencloses about 68% of x’s

f(x)

xμ μσμσ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%

33

The Empirical Rule μ ± 2σ covers about 95% of x’s

xμ2σ 2σ

95.44%

(continued)

34

The Empirical Rule μ ± 2σ covers about 95% of x’s μ ± 3σ covers about 99.7% of x’s

xμ2σ 2σ

xμ3σ 3σ

95.44% 99.72%

(continued)

35

Importance of the Rule If a value is about 2 or more standard deviations away from the mean in a normal distribution, then it is far from the mean

The chance that a value that far or farther away from the mean is (highly) unlikely, given that particular mean and standard deviation

Occurrence ~ 1 in 20 opportunities

36

An example The heights of women are approx. normally distributed

with = 160 cm and = 8 cm

37

Example...

(i) What % of women are between 144cm and 176 cms?

2 s.d. either side of the mean, therefore approximately 95%

(ii) Between 152 cm and 168 cm? 1 s.d. either side of the mean, therefore

approximately 68%

= 160 cm = 8 cm

38

Example...

(iii) Greater than 168cm? = 160 cm = 8 cm

39

Example...

(iii) Greater than 168cm?

That is more than 1 s.d. bigger than the mean Therefore 16%

= 160 cm = 8 cm

40

Example...

(iv) Less than 136cm? Very few people (100-99.7)/2 = 0.15%

= 160 cm = 8 cm

41

Example...

(b) Tallest 2.5% of women? = 160 cm = 8 cm

42

Example... (b) Tallest 2.5% of women?

Over 176 cm tall

= 160 cm = 8 cm

43

The Standard Normal Distribution

Also known as the “z” distribution Mean is defined to be 0 Standard Deviation is 1

44

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 the mean have negative z-values

45

The Standard Normal

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normal distribution (z)

Need to transform x units into z units

46

Translation to the Standard Normal Distribution

Translate from x to the standard normal (the “z” distribution) by subtracting the mean of x and dividing by its standard deviation:

Z is the number of standard deviations away from the mean

σμxz

47

Example If x is distributed normally with mean of 100

and standard deviation of 50, the z value for x = 250 is

This says that x = 250 is three standard deviations (3 increments of 50 units) above the mean of 100.

3 .05 0

1 0 02 5 0σ

μxz

3 .05 0

1 0 02 5 0σ

μxz

3 .05 0

1 0 02 5 0σ

μxz

48

Comparing x and z units

z100

3.00250 x

We can express the problem in original units (x)

or in standardized units (z)

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

μ = 100

σ = 50

49

Problem 6-1

50

The Standard Normal Table

The Standard Normal table in the textbook(Appendix D) Also inside the back of the book.

gives the probability from the mean (zero) up to a desired value for z

52


The Standard Normal table in the textbook (Appendix D, page 821-9h ed/880-8th ed)

gives the probability from the mean (zero) up to a desired value for z

z0 2.00

.4772Example: P(0 < z < 2.00) = .4772

53


The value within the table gives the probability from z = 0 up to the desired z value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = .4772

The row shows the value of z to the first decimal point

The column gives the value of z to the second decimal point

2.0

.

.

.

(continued)

54

General Procedure for Finding Probabilities

1. Draw the normal curve for the problem in terms of x

2. Translate x-values to z-values

3. Use the Standard Normal Table

To find P(a < x < b) when x is distributed normally:

55

Z Table exampleSuppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 11)

P(8 < x < 11)

= P(0 < z < 0.60)

Z0.60 0x11 80

588

σμxz

0 .605

811σ

μxz

1st StepDraw normal curve forproblem in terms of x valuesCalculate z-values:

2nd Step

56

Z Table exampleSuppose x is normal with mean 8.0 and standard

deviation 5.0. Find P(8 < x < 11)

P(0 < z < 0.60)

z0.60 0x11 8

P(8 < x < 11)

= 8 = 5

= 0 = 1

(continued)

57

Z

0.60

z .00 .01

0.5 .1915 .1950 .1985

.2257 .2291

0.7 .2580 .2611 .2642

0.8 .2881 .2910 .2939

Solution: Finding P(0 < z < 0.60)

.2257.02

0.6 .2324

Standard Normal Probability Table (Portion)

0.00

= P(0 < z < 0.60)P(8 < x < 11)

3rd Step

58

Finding Normal ProbabilitiesSuppose x is normal with mean 8.0 and

standard deviation 5.0. Now Find P(x < 11)


x

118.0

59


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

Now Find P(x < 11)

(continued)

Z

0.600.00

.5000 .2257 P(x < 11)

= P(z < 0.60)

= P(z < 0) + P(0 < z < 0.60)

= .5 + .2257 = .7257

60

Upper Tail ProbabilitiesSuppose x is normal with mean 8.0 and

standard deviation 5.0. Now Find P(x > 11)

x

118.0

61

Now Find P(x > 11)…(continued)

Z

0.60 0

Z

0.60

.2257

0

.5000 .50 - .2257= .27

43

P(x > 11) = P(z > 0.60) = P(z > 0) - P(0 < z < 0.60)

= .5 - .2257 = .2743

Upper Tail Probabilities

62

Lower Tail ProbabilitiesSuppose x is normal with mean 8.0 and

standard deviation 5.0. Now Find P(5 < x < 8)

x

58.0

63

Lower Tail Probabilities

Now Find P(5 < x < 8)…

x

58.0

The Normal distribution is symmetric, so we use the same table even if z-values are negative:

P(5 < x < 8)

= P(-0.6 < z < 0)

= .2257

(continued)

.2257

64

Class Tutorial Problem Problem 6-15 (a) and (b)

66

Problems 6-11 6-13

67

Computing Normal Probabilities

Assembly of motor cycle parts is normally distributed with a mean of 75 seconds and a standard deviation of 5 seconds

Find the interquartile range of assembly times, that is the middle 50%, 25th percentile to the 75th percentile

68

Example …

For the 25th percentile, find the z value below which 25% of the values lie

Look up 0.25 in the BODY of the table

0.25

0.25

69

Z

Example …

.25

Standard Normal Probability Table (Portion)

0.00

z .06 .07

0.5 .2123 .2157 .2190

.2454 .2486

0.7 .2764 .2794 .2823

0.8 .3051 .3078 .3106

.08

0.6 .2517

Look up 0.25 in the BODY of the table

.24860.6

.07

- 0.67Gives a z value of - 0.67

70

Example …

Now we need to convert z = -0.67 to an x value

X = µ + zσ= 75 + (-0.67 x 5)=71.65

0.25

What is the X

value ?

0.25

We know the Z value is -

0.67

71

Example... For the upper quartile (75% value)

0.75What is X?

Look up 0.25 in the body of the table,

Z = 0.67 X =µ + zσ = 75 + (0.67 x 5 )=78.35

72

Thus the Interquartile range is: 71.63 to 78.37 seconds

This means that the middle 50% of assembly times is between these two values

Example...

73

Class Tutorial Problem Problem 6-15 (c)

75

Key Terms Continuous Random

Variable Discrete Random Variable Expected value Normal distribution

Standard Normal Distribution Standard Normal Table

z-Value

76

Exercises 8th EditionChapter 5: 5.12,

5.13 (omit variance), 5.15, 5.16, 5.17

Chapter 6: 6.8, 6.11, 6.13, 6.14, 6.15, 6.18, 6.19, 6.22, 6.28, 6.65

9th EditionChapter 5: 5.12,

5.13 (omit variance), 5.15, 5.16, 5.17

Chapter 6: 6.8, 6.11, 6.13, 6.14, 6.15, 6.16, 6.19, 6.22, 6.28, 6.69

session 7 (chpt 5 & 6)(3)

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