Download Continuous Random Variables

Continuous Random Variables
and the Normal Distribution
What are Continuous Random
Variables?
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FRE C408
Fall 1999
Dr. Ilvento
When dealing with a PDF
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It is not particularly useful to think of a
probability when our variable takes on a
particular value
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P(a # x # b) = some proportion of the
curve
For every distribution with a mean (:)
and a standard deviation (F) there is a
different normal curve
Thus, there are an infinite number of
normal curves
If x is distributed as a normal variable
then it is designated as:
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x~N
PDF
f(x)
Normal Distribution
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P(x=a) = 0
Normal Distribution
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One bell shaped, symmetrical
distribution is the normal distribution
It is defined by two parameters
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But, we can think of areas under the
curve as reflecting a probability
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Unlike Discrete Random Variables,
Continuous Random Variables take on any
point in the interval
Thus the probability function is continuous
It is referred to as a Probability Density
Function
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:
F
f ( x) =
2
1
e − (1/ 2)[( x − µ ) / σ ]
σ 2π
Properties of the Normal
Distribution
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Symmetrical, Bell-shaped curve
Defined by the mean and standard deviation
Mean = Median = Mode
Since its properties are defined by a formula,
we can a priori define probabilities associated
with the curve
If we convert our variable to a z-score, we
make it possible to use one table for all
normal pdf
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mean = 0
std dev = 1
1
Standard Normal Distribution
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:=0
F=1
If x ~ N, then
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Finding Areas under the Curve
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Use the Table on page
779
Steps
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z = (x - :)/F ~ N
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Draw the curve and the
area we are interested in
Convert the values to zscores
Read the proportions in
the table
Do any calculations
necessary
Look at the table on page 779
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Only ½ of the curve since the distribution is
symmetrical
Allows for two decimal places
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Vertical axis is the ones and first decimal place
Horizontal axis is the second decimal place
The probabilities in the table represent the
probability up to that half of the curve
Thus, a z-score of
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3.00 corresponds to .4987 of one half of the curve
3.09 corresponds to .4990 of one half of the curve
Problem
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Find the area under the standard normal
curve for a z-score between 0 and 2.00.
Answer:
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A z-score of zero is at the mean, with a probability
of zero
A z-score of 2 is two standard deviations above
the mean, which corresponds to a probability of
.4772
We want the area from the mean to 2 standard
deviations from the mean
Equal to .4772
Problems in Class
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Suppose a variable is distributed normally
with a mean = 300 and a standard deviation
of 30
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X~N
: = 300 F = 30
What is the probability that x is more
than 2 standard deviations from the
mean?
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Draw it out
Calculate z-score
Check the table
Do any final calculations
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Problem
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X~N
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Problem
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X~N
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Z=2
In table when z = 2.00 we have a probability
up to that point on one side of the curve of
.4772
.5 - .4772 = .0228
one side of curve
2 x .0228 = .0456
both sides of curve
Problem
: = 300 F = 30
More than 3 std deviations
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: = 300 F = 30
More than 2 std deviations
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Z = 3.00
In table when z = 3.00 we have a probability
up to that point on one side of the curve of
.4987
.5 - .4987 = .0013
one side of curve
2 x .0013 = .0026
both sides of curve
X~N
: = 300 F = 30
Probability that x is between 260
and 360?
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Draw it
Calculate z-scores
Look up in the table
Do any calculations
Problem
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X~N
: = 300 F = 30
Probability that x is between 260
and 360?
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X =300 z = (260 – 300)/30 = -1.33
X = 340 z = (360 – 300)/30 = 2.00
Z for 1.33 = .4082
Z for 2.00 = .4772
.4082 + .4772 = .8854
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What is the value at the 80th
percentile?
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Given X ~ N
: = 300 F = 30
What am I looking for?
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The 80th percentile reflects everything up to the
mean (50th percentile)
Plus .30 more
Look in the table for .30
It is between .84 (p=.2995) and .85 (p=.3023)
I could extrapolate, but I know it is a lot closer to
.84
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The normal distribution can be used as an
approximation for the Binomial distribution of
x successes in n trials
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.841 = (x – 300)/30
30 A .841 = x – 300
300 + (30 A .841) = x
325.23 = x
The 80th percentile is at .841
Example of test calculation
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Yes, because
: = np = .06(200) = 12
F = (200 A.06 A.94).5 = (11.28).5 = 3.36
and 12.0 " 3(3.36) = 1.92 to 22.08
To solve the problem
Example on page 214
Quality control for pocket calculators
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Calculate
: "3F = np " 3 (npq).5
If this interval lies in the range of 0 to n, we can
go forward
And we use a correction for continuity of .5
Can we use Binomial Approximation?
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This save a lot of calculations
Calculator problem
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Solve for x
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Provided we can assume the following
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.841 is a good approximation
Normal Distribution as an approximation
of the Binomial Distribution
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Next -
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Defective rate is .06 or 6%
Sample is 200 (I.e., n= 200 trials)
What is the probability of 20 or more
defects are observed?
Calculator Problem
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Next we look at what we want to solve
for
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Probability that 20 or more defects are
observed
Or we can look at the complement, which
is 1- p(19 or less)
We use the correction for continuity of 19
+ .5 = 19.5
And solve for the probability of 19.5 or
less.
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Calculator Problem
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Z = (19.5 – 12)/3.36
Z = 2.23
Look it up in the table
= .4871
The probability of less than 19.5 =
.5 + .4871 = .9871
The probability of the complement is
1 - .9871 = .0129
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