Continuous Probability Distributions

•Continuous Random Variable: Values from Interval of Numbers

Absence of Gaps

•Continuous Probability Distribution:Distribution of a Continuous Variable

•Most Important Continuous Probability Distribution: the Normal Distribution

The Normal Distribution

• ‘Bell Shaped’

• Symmetrical

• Mean, Median and

• Mode are Equal

•Random Variable has• Infinite Range

Mean Median Mode

X

f(X)

The Mathematical Model

f(X) = frequency of random variable X

= 3.14159; e = 2.71828

= population standard deviation

X = value of random variable (- < X < )

= population mean

f(X) =1 e

(-1/2)((X- )2

2

Many Normal Distributions

Varying the Parameters and , we obtain Different Normal Distributions.

There are an Infinite Number

Which Table?

Infinitely Many Normal Distributions Means Infinitely Many Tables to Look Up!

Each distribution has its own table?

Z Z

Z = 0.12

Z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .0179 .0217 .0255

The Standardized Normal Distribution

.0478.02

0.1 .0478

Standardized Normal Probability Table (Portion) = 0 and = 1

ProbabilitiesShaded Area Exaggerated

Z = 0

Z = 1

.12

Standardizing Example

Normal Distribution

Standardized Normal Distribution

X = 5

= 10

6.2

12.010

52.6

XZ

Shaded Area Exaggerated

0

= 1

-.21 Z.21

Example:P(2.9 < X < 7.1) = .1664

Normal Distribution

.1664

.0832.0832

Standardized Normal Distribution

Shaded Area Exaggerated

5

= 10

2.9 7.1 X

2110

592.

.xz

2110

517.

.xz

Z = 0

= 1

.30

Example: P(X 8) = .3821

Normal Distribution

Standardized Normal Distribution

.1179

.5000

.3821

Shaded Area Exaggerated

.

X = 5

= 10

8

3010

58.

xz

Z .00 0.2

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

Z = 0

= 1

.31

Finding Z Values for Known Probabilities

.1217.01

0.3

Standardized Normal Probability Table (Portion)

What Is Z Given P(Z) = 0.1217?

Shaded Area Exaggerated

.1217

Z = 0

= 1

.31X = 5

= 10

?

Finding X Values for Known Probabilities

Normal Distribution Standardized Normal Distribution

.1217 .1217

Shaded Area Exaggerated

X 8.1 Z= 5 + (0.31)(10) =

Assessing Normality

• Compare Data Characteristics

• to Properties of Normal • Distribution

• Put Data into Ordered Array

• Find Corresponding Standard

• Normal Quantile Values

• Plot Pairs of Points

• Assess by Line Shape

Normal Probability Plot for Normal Distribution

Look for Straight Line!

30

60

90

-2 -1 0 1 2

Z

X

Normal Probability Plots

Left-Skewed Right-Skewed

Rectangular U-Shaped

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X

30

60

90

-2 -1 0 1 2

Z

X

Estimation

•Sample Statistic Estimates Population Parameter

• e.g. X = 50 estimates Population Mean,

•Problems: Many samples provide many estimates of the

Population Parameter.

• Determining adequate sample size: large sample give better

estimates. Large samples more costly.

• How good is the estimate?

•Approach to Solution: Theoretical Basis is Sampling

Distribution.

_

• Population Mean Equal to

• Sampling Mean

• The Standard Error (standard deviation) of the Sampling distribution is Less than Population Standard Deviation

• Formula (sampling with replacement):

Properties of Summary Measures

x

As n increase, decrease. x =

xn__

XX

Central Limit Theorem

As Sample Size Gets Large Enough

Sampling Distribution

BecomesAlmost Normal

regardless of shape of

population

• Categorical variable (e.g., gender)

• % population having a characteristic

• If two outcomes, binomial distribution

– Possess or don’t possess characteristic

• Sample proportion (ps)

sizesample

successes of number

n

XPs

Population Proportions

• Approximated by normal distribution– n·p 5– n·(1 - p) 5

• Mean

• Standard error

pP

n

ppP

1

Sampling Distribution of Proportion

p = population proportion

Sampling Distribution

P(ps)

.3

.2

.1 0

0 . 2 .4 .6 8 1ps

Standardizing Sampling Distribution of Proportion

Sampling Distribution

StandardizedNormal Distribution

Z pp s s - pp

=p -

n

)p(p 1

psZ = 0

p

p

= 1