continuous random variables lecture 22 section 7.5.4 mon, feb 25, 2008
TRANSCRIPT
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Continuous Random Variables
Lecture 22
Section 7.5.4
Mon, Feb 25, 2008
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Random Variables
Random variable Discrete random variable Continuous random variable
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Continuous Probability Distribution Functions Continuous Probability Distribution
Function (pdf) – For a random variable X, it is a function with the property that the area below the graph of the function between any two points a and b equals the probability that a ≤ X ≤ b.
Remember, AREA = PROPORTION = PROBABILITY
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Example
The TI-83 will return a random number between 0 and 1 if we enter rand and press ENTER.
These numbers have a uniform distribution from 0 to 1.
Let X be the random number returned by the TI-83.
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Example
The graph of the pdf of X.
x
f(x)
0 1
1
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Example
What is the probability that the random number is at least 0.3?
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Example
What is the probability that the random number is at least 0.3?
x
f(x)
0 1
1
0.3
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Example
What is the probability that the random number is at least 0.3?
x
f(x)
0 1
1
0.3
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Example
What is the probability that the random number is at least 0.3?
x
f(x)
0 1
1
0.3
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Area = 0.7
Example
Probability = 70%.
x
f(x)
0 1
1
0.3
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0.25
Example
What is the probability that the random number is between 0.25 and 0.75?
x
f(x)
0 1
1
0.75
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Example
What is the probability that the random number is between 0.25 and 0.75?
x
f(x)
0 1
1
0.25 0.75
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Example
What is the probability that the random number is between 0.25 and 0.75?
x
f(x)
0 1
1
0.25 0.75
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Example
Probability = 50%.
x
f(x)
0 1
1
0.25 0.75
Area = 0.5
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Uniform Distributions
The uniform distribution from a to b is denoted U(a, b).
a b
1/(b – a)
x
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A Non-Uniform Distribution
Consider this distribution.
5 10x
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A Non-Uniform Distribution
What is the height?
5 10
?
x
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A Non-Uniform Distribution
The height is 0.4.
5 10
0.4
x
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A Non-Uniform Distribution
What is the probability that 6 X 8?
5 10
0.4
x6 8
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A Non-Uniform Distribution
It is the same as the area between 6 and 8.
5 10
0.4
x6 8
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Uniform Distributions
The uniform distribution from a to b is denoted U(a, b).
a b
1/(b – a)
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Hypothesis Testing (n = 1)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
One value of X is sampled (n = 1).
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Hypothesis Testing (n = 1)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
One value of X is sampled (n = 1). If X is more than 0.75, then H0 will be
rejected.
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Hypothesis Testing (n = 1)
Distribution of X under H0:
Distribution of X under H1:
0 0.5 1 1.5
1
0 0.5 1 1.5
1
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Hypothesis Testing (n = 1)
What are and ?
0 0.5 1 1.5
1
0 0.5 1 1.5
1
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Hypothesis Testing (n = 1)
What are and ?
0.75
0.75
0 0.5 1 1.5
1
0 0.5 1 1.5
1
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Hypothesis Testing (n = 1)
What are and ?
0.75
0.75
0 0.5 1 1.5
1
0 0.5 1 1.5
1
Acceptance Region Rejection Region
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Hypothesis Testing (n = 1)
What are and ?
0.75
0.75
0 0.5 1 1.5
1
0 0.5 1 1.5
1
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Hypothesis Testing (n = 1)
What are and ?
0.75
0.75
0 0.5 1 1.5
1
0 0.5 1 1.5
1
= ¼ = 0.25
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Hypothesis Testing (n = 1)
What are and ?
0.75
0 0.5 1 1.5
1
0 0.5 1 1.5
1
= ¼ = 0.25
= ¼ = 0.25
0.75
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Example
Now suppose we use the TI-83 to get two random numbers from 0 to 1, and then add them together.
Let X2 = the average of the two random numbers.
What is the pdf of X2?
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Example
The graph of the pdf of X2.
y
f(y)
0 0.5 1
?
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Example
The graph of the pdf of X2.
y
f(y)
0 0.5 1
2
Area = 1
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Example
What is the probability that X2 is between 0.25 and 0.75?
y
f(y)
0 0.5 10.25 0.75
2
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Example
What is the probability that X2 is between 0.25 and 0.75?
y
f(y)
0 0.5 10.25 0.75
2
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Example
The probability equals the area under the graph from 0.25 to 0.75.
y
f(y)
0 0.5
2
10.25 0.75
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Example
Cut it into two simple shapes, with areas 0.25 and 0.5.
y
f(y)
0 0.5 10.25 0.75
2
Area = 0.5
Area = 0.250.5
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Example
The total area is 0.75.
y
f(y)
0 0.5 10.25 0.75
2
Area = 0.75
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Hypothesis Testing (n = 2)
An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5). H0: X is U(0, 1).
H1: X is U(0.5, 1.5).
Two values of X are sampled (n = 2). Let X2 be the average.
If X2 is more than 0.75, then H0 will be rejected.
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Hypothesis Testing (n = 2)
Distribution of X2 under H0:
Distribution of X2 under H1:0 0.5 1 1.5
2
0 0.5 1 1.5
2
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Hypothesis Testing (n = 2)
What are and ?
0 0.5 1 1.5
2
0 0.5 1 1.5
2
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Hypothesis Testing (n = 2)
What are and ?
0 0.5 1 1.5
2
0 0.5 1 1.5
2
0.75
0.75
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Hypothesis Testing (n = 2)
What are and ?
0 0.5 1 1.5
2
0 0.5 1 1.5
2
0.75
0.75
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Hypothesis Testing (n = 2)
What are and ?
0 0.5 1 1.5
2
0 0.5 1 1.5
2
0.75
0.75
= 1/8 = 0.125
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Hypothesis Testing (n = 2)
What are and ?
0 0.5 1 1.5
2
0 0.5 1 1.5
2
0.75
0.75
= 1/8 = 0.125
= 1/8 = 0.125
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Conclusion
By increasing the sample size, we can lower both and simultaneously.