Chapter 4

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 bDo 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

Probability 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 μSpread determined by the value

of it’s standard deviation σ

The Normal Distribution

The mean and standard deviation affect the flatness and center of the curve, but not the basic shape

The Normal Distribution

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

The Normal Distribution

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 normalrandom variable is denoted by thesymbol z

The Normal Distribution

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

The Normal Distribution

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

Table gives us area A1

Symmetry about the meantell 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

The Normal Distribution

What is P(z > 1.64)?

Table gives us area A2

Symmetry about the meantell us that A2 + A1 = .5

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

The Normal Distribution

What is P(z < .67)?

Table gives us area A1

Symmetry about the meantell us that A2 = .5

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

The Normal Distribution

What is P(|z| > 1.96)?

Table gives us area .5 - A2

=.4750, so A2 = .0250

Symmetry about the meantell us that A2 = A1

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

The Normal Distribution

What if values of interest were not normalized? We want to knowP (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

x

z

The Normal Distribution

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

x

z

The Normal Distribution

Making an InferenceHow likely is an observationin 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

The Normal Distribution

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

The Normal Distribution

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.

The Normal Distribution

Given a normally distributed variable x with mean 100,000 and standard deviation of 10,000, 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

000,100000

x

zPx

zPxxP

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 nOften needed given limitation of binomial tablesNeed to add a correction for continuity, because of the discrete nature of the binomial distributionCorrection is to add .5 to x when converting to standard z valuesRule 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

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