Chapter 5 Continuous Random Variables Continuous Probability - - PowerPoint PPT Presentation

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

 

c d x f   1

Probabillity Distribution for a Uniform Random Variable x Probability density function: Mean: Standard Deviation:

d x c  

2 d c    12 c d   

     

d b a c c d a b b x a P         , /

The Uniform Distribution

The Normal Distribution

A normal random variable has a probability distribution called a normal distribution The Normal Distribution

Bell-shaped curve Symmetrical about its mean μ Spread determined by the value

• f 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

 

     

2

2 1

2 1

 

 

 



x

e x f

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 normal random variable is denoted by the symbol 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 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

The Normal Distribution

What is P(z > 1.64)? Table gives us area A2 Symmetry about the mean tell 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 mean tell 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 mean tell 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 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

    x z

The Normal Distribution

Steps for Finding a Probability Corresponding to a Normal Random Variable

• Sketch the distribution, locate mean, shade area
• f interest
• Convert to standard z values using
• Add z values to the sketch
• Use tables to calculate probabilities, making use
• f symmetry property where necessary

    x z

The Normal Distribution

Making an Inference

How 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

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 100000 and standard deviation of 10000, what value of x identifies the top 10%

• f 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 , 100                     x z P x z P x x P  

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

s x s x s x 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 npq np 3 3     

The Exponential Distribution

Used to describe the amount of time between

• ccurrences of random events

Probability Distribution, for an Exponential Random Variable x Probability Density function: Mean: Standard Deviation:

 

) ( ,  



x e x f

x 



  1    1 

The Exponential Distribution

Shape of the distribution is determined by the value of  Mean is equal to Standard deviation

The Exponential Distribution

To find the area A to the right of a, A can be calculated using a calculator or with tables

 

a

e a x P A

 

  

The Exponential Distribution

If  = .5, what is the p(a>5)? From tables, A = .082085 Probability that A > 5 is .082085

 

5 . 2 5 5 .   

   e e e A

a 

The Exponential Distribution

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

The Exponential Distribution

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

The Exponential Distribution

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