Post on 20-Jan-2016
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
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xf
1
Probability Distribution for a Uniform Random Variable x
Probability density function:
Mean: Standard Deviation:
dxc
2
dc
12
cd
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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
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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
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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
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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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