Download Continuous Random Variables

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 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
Probability Distribution for a Uniform Random Variable x
1
cxd
Probability density function: f  x  
d c
cd
d c
Mean:  
Standard Deviation:  
2
12
Pa  x  b  b  a /d  c, c  a  b  d
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
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
1
1 2  x     2
f x  
e
 2
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
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
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
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0
.1
.2
.3
.0000
.0398
.0793
.1179
.0040
.0438
.0832
.1217
.0080
.0478
.0871
.1255
.0120
.0517
.0910
.1293
.0160
.0557
.0948
.1331
.0199
.0596
.0987
.1368
.0239
.0636
.1026
.1406
.0279
.0675
.1064
.1443
.0319
.0714
.1103
.1480
.0359
.0753
.1141
.1517
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
x
z

P(8<x<12) = P(-1.33<z<1.33) = 2(.4082) = .8164
The Normal Distribution
Steps for Finding a Probability Corresponding to a
Normal Random Variable
• Sketch the distribution, locate mean, shade area
of interest
x
• Convert to standard z values using z 

• Add z values to the sketch
• Use tables to calculate probabilities, making use
of symmetry property where necessary
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 100,000 and
standard deviation of 10,000, what
value of x identifies the top 10%
of the distribution?
x0   
x0  100,000 


Px  x0   P z 
  .90
  P z 
 
10,000 


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
Descriptive Methods for Assessing
Normality
• Evaluate the shape from a histogram or stemand-leaf display
• Compute intervals about mean x  s, x  2s, x  3s
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
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   3  np  3 npq
• 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