chapter 5 · the uniform distribution uniform probability distribution –distribution resulting...

Chapter 5

Continuous Random Variables

Continuous Probability Distributions

Continuous Probability Distribution – areas under

curve correspond to probabilities for x

Area A corresponds to the probability that x lies

between a and b

Do you see the similarity in shape between the continuous and discrete

probability distributions?

The Uniform Distribution

Uniform Probability

Distribution – distribution

resulting when a continuous

random variable is evenly

distributed over a particular

interval

cd

xf

1

Probabillity Distribution for a Uniform Random Variable x

Probability density function:

Mean: Standard Deviation:

dxc

2

dc

12

cd

dbaccdabbxaP ,/

The Uniform Distribution

The Normal Distribution

A normal random variable has a probability

distribution called a normal distribution

The Normal DistributionBell-shaped curve

Symmetrical about its mean μ

Spread determined by the value

of it’s standard deviation σ


The mean and standard deviation affect the

flatness and center of the curve, but not the

basic shape


The function that generates a normal curve is of the form

where

= Mean of the normal random variable x

= Standard deviation

= 3.1416…

e = 2.71828…

P(x<a) is obtained from a table of normal probabilities

221

2

1

xexf


Probabilities associated with values or ranges of a random

variable correspond to areas under the normal curve

Calculating probabilities can be simplified by working with a

Standard Normal Distribution

A Standard Normal Distribution is a Normal distribution with

=0 and =1

The standard normal

random variable is

denoted by the

symbol z


Table for Standard Normal Distribution contains probability

for the area between 0 and z

Partial table below shows components of table

Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

Value of z a

combination of

column and

row

Probability

associated with a

particular z value, in

this case z=.13,

p(0<z<.13) = .0517


What is P(-1.33 < z < 1.33)?

Table gives us area A1

Symmetry about the mean

tell us that A2 = A1

P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z < 1.33)=

A2 + A1 = .4082 + .4082 = .8164


What is P(z > 1.64)?



tell us that A2 + A1 = .5

P(z > 1.64) = A1 = .5 – A2=.5 - .4495 = .0505


What is P(z < .67)?



tell us that A2 = .5

P(z < .67) = A1 + A2 = .2486 + .5 = .7486


What is P(|z| > 1.96)?

Table gives us area .5 - A2

=.4750, so A2 = .0250


tell us that A2 = A1

P(|z| > 1.96) = A1 + A2 = .0250 + .0250 =.05


What if values of interest were

not normalized? We want to know

P (8<x<12), with μ=10 and σ=1.5

Convert to standard normal using

P(8<x<12) = P(-1.33<z<1.33) = 2(.4082) = .8164

xz


Steps for Finding a Probability Corresponding to a

Normal Random Variable

•Sketch the distribution, locate mean, shade area

of interest

•Convert to standard z values using

•Add z values to the sketch

•Use tables to calculate probabilities, making use

of symmetry property where necessary

xz


Making an InferenceHow likely is an observation

in area A, given an assumed normal

distribution with mean of 27 and

standard deviation of 3?

Z value for x=20 is -2.33

P(x<20) = P(z<-2.33) = .5 - .4901 = .0099

You could reasonably conclude that this is a rare event


You can also use the table

in reverse to find a z-value

that corresponds to a

particular probability

What is the value of z that will be exceeded only 10% of

the time?

Look in the body of the table for the value closest to .4, and

read the corresponding z value

Z = 1.28


Which values of z enclose the

middle 95% of the standard

normal z values?

Using the symmetry property,

z0 must correspond with a

probability of .475

From the table, we find that z0 and –z0 are 1.96 and -1.96

respectively.


Given a normally distributed

variable x with mean 100000 and

standard deviation of 10000, what

value of x identifies the top 10%

of the distribution?

The z value corresponding with .40 is 1.28. Solving for x0

x0 = 100,000 +1.28(10,000) = 100,000 +12,800 = 112,800

90.000,10

000,10000

0

xzP

xzPxxP

Descriptive Methods for Assessing

Normality

•Evaluate the shape from a histogram or

stem-and-leaf display

•Compute intervals about mean

and corresponding percentages

•Compute IQR and divide by standard

deviation. Result is roughly 1.3 if normal

•Use statistical package to evaluate a

normal probability plot for the data

sxsxsx 3,2,

Approximating a Binomial Distribution with a

Normal Distribution

You can use a Normal Distribution as an approximation of a Binomial Distribution for large values of n

Often needed given limitation of binomial tables

Need to add a correction for continuity, because of the discrete nature of the binomial distribution

Correction is to add .5 to x when converting to standard z values

Rule of thumb: interval +3 should be within range of binomial random variable (0-n) for normal distribution to be adequate approximation

Approximating a Binomial Distribution with a

Normal Distribution

Steps

Determine n and p for the binomial distribution

Calculate the interval

Express binomial probability in the form P(x<a) or

P(x<b)–P(x<a)

Calculate z value for each a, applying continuity

correction

Sketch normal distribution, locate a’s and use table

to solve

npqnp 33

The Exponential Distribution

Used to describe the amount of time between

occurrences of random events

Probability Distribution, for an Exponential Random Variable x

Probability Density function:

Mean:

Standard Deviation:

)0(,

xexfx

1

1


Shape of the distribution

is determined by the

value of

Mean is equal to

Standard deviation


To find the area A to the right of a,

A can be calculated

using a calculator or

with tables

a

eaxPA


If = .5, what is the p(a>5)?

From tables, A = .082085

Probability that A > 5 is

.082085

5.255. eeeA

a


a) If = .16, what are the and ?

b) What is the p(0<a<5)?

c) What is the p(+2<a< +2)

a) = = 1/ = 1/.16 = 6.25


b) P(x>a) = e-a

P(x>5) = e-(.16)5 = e-.8

= .449329

P(x<5) = 1-P(x>5)

= 1-.449329 = .550671


c) What is the p(+2<a< +2)?

Find the complement of the

area above +2

P = 1-P(x>18.75)

= 1- e-(18.75) = 1- e-.16(18.75)

= 1- e-3 = 1- .049787

= .950213

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