Lecture 7: Continuous Probability Distributions - Biostatistics, Study notes of Biostatistics

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BiostatisticsBiostatistics Lecture 7Lecture

BIL 311 Lecturer: Dr. Patricia Buendia

Lecture 7 OutlineLecture

7 Outline

„

Chapter 5 – Continuous ProbabilityDistributionsDistributions

What is

Φ

(^

0 62

)?

What

is

Φ

( -0.

)?

1

Remember: Table 3, Column A, in the Appendix of the bookpresents

Φ

( x )=Pr(X

≤

x ) for various positive values of x!

presents

Φ

( x )=Pr(X

≤

x ) for various positive values of x!

T

he Inverse Normal

Function „^

The (100 * u)th percentile of a standarddistribution is denoted by z

It is

distribution is denoted by z

. It isu^

defined by Pr(X<z

)=u, where X~N(0,1).u^

C

„

Example: Compute z

„

From table 3 we know

~ means: isdistributed

as

)= Pr(X

)=0.95, therefore

z^ 0 9

Use

z^ 0.

Use NORMSINV in

Excel

Book pages:133 & 134

Standardization: From an N(μ,

Distribution to an N(0,1) Distribution „^

Probabilities of

any

normally distributed

random variable can be evaluated byrandom variable can be evaluated by standardizing

the normal random

variables

as described before and by

variables

as

described before and by

using the tables. „

Example: The probability of being a mild „

Example: The probability of being a mildhypertensive among 35-to-44-year oldmen can now be calculated men can now be calculated. Book page:

Standardization: From an N(μ,

Distribution to an N(0,1) Distribution

F

th

DBP

l^

d

„

F

or the DBP example: μ=80,

σ

=12 and

we are looking for Pr(90<X<100) = …

Book page:136 & 137

Tree diameters are normally distributedwith μ=9 in. and

σ

=2 in.

μ

What is Pr(X>13)?

1

P (Z >(

9)/2)

1-Pr(Z >(13-9)/2)

1-Pr(X<13) =1-Pr(Z < (13-9)/2)P (Z

Pr(Z > 13)

Pr((13-9)/2 <Z < 1)

Standardization ExampleStandardization

Example

We have X~N(9,4) and want to know

Pr(X>13)=1-Pr(X<13)=1-Pr(Z < (13-9)/2) =1-Pr(Z<2.0)=1-0.977=0. Thus 2 3% of trees from this area haveThus, 2.3% of trees from this area have

an unusually large diameter.

Use 1

-NORMDIST(13 9 2 TRUE) Use

1 NORMDIST(13,9,2,TRUE)

In Excel

Book page:

Normal Approximation to the

pp

Binomial Distribution „^

Special Cases: „

For Pr(X=0) approximate to the left of ½ „

For Pr(X=n) approximate to the right of^ n- ½ Book page:

Normal Approximation to the

pp

Binomial Distribution

p too large forPoisson approx.

„

Example: We want to compute theprobability that between 50 to 75 of 100probability that between 50 to 75 of 100white blood cells will be neutrophils,with p=0 6 for neutrophilswith p=0.6 for neutrophils. „

50 to 75 is range of neutrophils in^ normal people and we wish to predictwhat proportion of people will be in that^ range.Book page:147&

How do we obtain the probability that between 50to 75 of 100 white blood cells will be neutrophils (given that p=0.6 for neutrophils) by using

the

normal approximation?^ 1.

Find Pr(

≤Y

≤75) for an N(60,24)

distributed random variable Y 2

Find Pr(49 5

≤Y

≤75 5) for an

2.^

Find Pr(49.

≤Y

≤75.5) for an

N(0,1) distributed random variableY 3

Find Pr(49 5

≤Y

≤75 5) for an

3.^

Find Pr(49.

≤Y

≤75.5) for an

N(60,24) distributed randomvariable Y

We have Pr(49.

≤

Y≤

75.5) with Y~N(60,24), but

we cannot easily find that probability unless we…^ 1.

Use the Binomial formula

2.^

Use the Poisson approximation 3

St

d^

di^

Y

3.^

Standardize Y

Normal Approximation to the

pp

Binomial Distribution „^

Pr(49.

≤

Y

≤75.5)=

⎞

⎛ ⎞

⎛^

60

49 5

60

75 5 Φ

(3 164)

Φ

( 2 143)

⎞ ⎟ ⎠

⎛^ ⎜ ⎝

−

⎞−⎟ ⎠

⎛^ ⎜ ⎝

−

=^

24

60

Φ

24

60

Φ =Φ

(3.164)-

Φ

(-2.143)

=Φ

(3.164)+

Φ

(2.143)-

=0.9992+0.9840-1=0.983^ Therefore 98 3% of people will be normalTherefore 98.3% of people will be normalBook page:

Normal Approximation to the

pp

Binomial Distribution „^

Restriction: The normal distribution withmean np and variance npq can be usedmean np and variance npq can be usedto approximate a binomial distributionwith parameters n and p when npq

with parameters n and p when npq

Book page: