CHAPTER 4

• 4.1 - Discrete Models General distributions Classical: Binomial, Poisson, etc.

• 4.2 - Continuous Models General distributions Classical: Normal, etc.

X

2

~ The Normal Distribution ~(a.k.a. “The Bell Curve”)

Johann Carl Friedrich Gauss 1777-1855

μ

σ

mean

standard deviation

X ~ N(μ, σ)

• Symmetric, unimodal• Models many (but not

all) natural systems• Mathematical

properties make it useful to work with

Standard Normal Distribution

Z ~ N(0, 1)

Z

2

21

( )2

z

z e

density function

The cumulative distribution function (cdf) is denoted by (z). It is tabulated, and computable in R via the command pnorm.

SPECIA

L CASE

Total Area = 11

Z

1

Standard Normal Distribution

Z ~ N(0, 1)

Example

Find P(Z 1.2).

1.2

“z-score”

Total Area = 1

Z

Standard Normal Distribution

Z ~ N(0, 1)

Example

Find P(Z 1.2).

1

1.2

Use the included table.

“z-score”

Total Area = 1

6

Lecture Notes Appendix…

7

Z

Standard Normal Distribution

Z ~ N(0, 1)

Example

Find P(Z 1.2).

1

1.2

Use the included table.

0.88493

Use R:

> pnorm(1.2) [1] 0.8849303

“z-score”

P(Z > 1.2)

0.11507

Total Area = 1

Note: Because this is a continuous distribution, P(Z = 1.2) = 0, so there is no difference between P(Z > 1.2) and P(Z 1.2), etc.

Standard Normal Distribution

Z ~ N(0, 1)

Z

XZ

μσ

X ~ N(μ, σ)

1

Any normal distribution can be transformed to the standard normal distribution via a simple change of variable.

Why be concerned about this, when most “bell curves” don’t have mean = 0, and standard deviation = 1?

Year 2010X ~ N(25.4, 1.5)

μ = 25.4

σ = 1.5

10

Example

Random Variable X = Age at first birth

POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?

P(X < 27.2) = ?

27.2

11

Example

Random Variable X = Age at first birth

POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?

P(X < 27.2) = ?

Year 2010X ~ N(25.4, 1.5) σ = 1.5

μ = 33

The x-score = 27.2 must first be transformed to a corresponding z-score.

μ = 25.4μ = 25.4 27.2

12

Example

Random Variable X = Age at first birth

POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?

P(X < 27.2) = ?

XZ

27.2 25.4

1.5Z

1.2Z

σ = 1.5

μ = 33

P(Z < 1.2) = 0.88493

Using R:

> pnorm(27.2, 25.4, 1.5) [1] 0.8849303

Year 2010X ~ N(25.4, 1.5)

μ = 25.4 27.2

Z

What symmetric interval about the mean 0 contains 95% of the population values?

That is…

1

Standard Normal Distribution

Z ~ N(0, 1)

Z

0.95

0.025 0.025

+z.025 = ?-z.025 = ?

What symmetric interval about the mean 0 contains 95% of the population values?

That is…

Standard Normal Distribution

Z ~ N(0, 1)

Use the included table.

15

Lecture Notes Appendix…

16

Use the included table.

+z.025 = ?+z.025 = +1.96-z.025 = ?

Standard Normal DistributionZ ~ N(0, 1)

Z

0.95

0.025 0.025

What symmetric interval about the mean 0 contains 95% of the population values?

-z.025 = -1.96 “.025 critical values”

Use R:

> qnorm(.025) [1] -1.959964

> qnorm(.975) [1] 1.959964

+z.025 = ?+z.025 = +1.96-z.025 = ?

Standard Normal DistributionZ ~ N(0, 1)

Z

0.95

0.025 0.025

What symmetric interval about the mean 0 contains 95% of the population values?

-z.025 = -1.96 “.025 critical values”

25.41.96

1.5

X

XZ

25.4 (1.96)(1.5)X

What symmetric interval about the mean age of 25.4 contains 95% of the population values?

25.4 2.94X

22.46 X 28.34 yrs

X ~ N(μ, σ) X ~ N(25.4, 1.5)

> areas = c(.025, .975)> qnorm(areas, 25.4, 1.5) [1] 22.46005 28.33995

Use the included table.

Standard Normal DistributionZ ~ N(0, 1)

Z

0.90

0.05 0.05

+z.05 = ?-z.05 = ?

What symmetric interval about the mean 0 contains 90% of the population values?

Similarly…

20

…so average 1.64 and 1.65 0.95 average of 0.94950 and 0.95053…

Use the included table.

-z.05 = ?-z.05 = -1.645

Standard Normal DistributionZ ~ N(0, 1)

Z

0.90

0.05 0.05

+z.05 = ?

What symmetric interval about the mean 0 contains 90% of the population values?

Similarly…+z.05 = +1.645“.05 critical values”

Use R:

> qnorm(.05) [1] -1.644854

> qnorm(.95) [1] 1.644854

-z.05 = ?-z.05 = -1.645

Standard Normal DistributionZ ~ N(0, 1)

Z

0.90

0.05 0.05

+z.05 = ?

What symmetric interval about the mean 0 contains 100(1 – )% of the population values?

Similarly…+z.05 = +1.645“.05 critical values”

In general….

1 –

/ 2 / 2

-z / 2 +z / 2“ / 2 critical values”

23

Normal Approximation to the Binomial Distribution

continuous discrete

Suppose a certain outcome exists in a population, with constant probability .

P(Success) = P(Failure) = 1 –

We will randomly select a random sample of n individuals, so that the binary “Success vs. Failure” outcome of any individual is independent of the binary outcome of any other individual, i.e., n Bernoulli trials (e.g., coin tosses).

Discrete random variableX = # Successes in sample(0, 1, 2, 3, …,, n)

Discrete random variableX = # Successes in sample(0, 1, 2, 3, …,, n)

Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability function”

f(x) = , x = 0, 1, 2, …, n.

x n xnx

(1 )

24

> dbinom(10, 100, .2) [1] 0.00336282

Area

25

> pbinom(10, 100, .2) [1] 0.005696381

Area

26

27

28

29

30

Therefore, if…

X ~ Bin(n, ) with n 15 and n (1 – ) 15,

then… , (1 .X N n n

That is…

(1 )ˆ ,X

Nn n

“Sampling Distribution” of ̂

31

Classical Continuous Probability Distributions

● Normal distribution

● Log-Normal ~ X is not normally distributed (e.g., skewed), but Y = “logarithm of X” is normally distributed

● Student’s t-distribution ~ Similar to normal distr, more flexible

● F-distribution ~ Used when comparing multiple group means

● Chi-squared distribution ~ Used extensively in categorical data analysis

● Others for specialized applications ~ Gamma, Beta, Weibull…