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Chapter 6.1:
Continuous Distributions and Continuous Random Variables
Chris Morgan, MATH G160
[email protected]
February 22, 2012
Lecture 17
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Continuous Random Variables
So far we have only discussed random variables that can take on
only discrete values.
–Remember discrete values are values that are countable.
But these distributions do not describe all situations that we
encounter in the real world. So there is another important class of
distributions that describe variables that can be measured but are
not countable.
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Continuous Random Variables
Examples:
- Measurements like height, weight, or diameter of any object or
person.
- Time
- Distance
These types of random variables take on values in an interval so
they are called continuous random variables.
We can have an inclusive interval – [ , ] – or an exclusive interval
denoted by – ( , ).
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Continuous Random Variables
A random variable X is called continuous if there is a function f(x),
called the probability density function (PDF) of X, such that:
a) f(x) ≥ 0 or all real numbers x
b)



f ( x)dx  1
i. Because of this, we cannot find P(X=x) because the integral
would cancel out and give 0.
ii. P(X=x) = 0
c) The probability that X takes on a value between a and b (a< b) is
given by:
b
P(a  X  b)   f ( x)dx
a
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Probability Density Function
• Abbreviated PDF
• Sometimes called Probability Distribution Function
• A function that describes the density of probability at an interval.
• The probability is found by taking an integral of the density
function over the given interval.
• f(x) denotes we are using a PDF
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Probability Density Function
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Cumulative Density Function
• Abbreviated CDF
• The cumulative distribution function is the integral of the PDF:

CDF  F ( x)   f ( x)dx

• This is a simple way of finding probabilities
–If you already have the CDF, then you do not need to
integrate to find the probability.
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Cumulative Density Function
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PDF/CDF Example
Let X be a continuous RV with PDF:
cx 0  x  4
f ( x)  
 0 otherwise
a) Find the value of the constant c:
2

cx
4 
0 (cx)dx  1  2 |0  c 2   1  c * 8  1
4
2
4
So: c = 1/8
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PDF/CDF Example
b. Draw a sketch of the PDF:
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PDF/CDF Example
Find the CDF:
2
x
x
F ( X )   dx 
8
16
 0
x0
1 2
f ( x)   x 0  x  4
16
x

4
1

12
PDF/CDF Example
d. Draw a sketch of the CDF:
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PDF/CDF Example
e. Compute P(2 < X ≤ 5) using both the PDF and the CDF:
PDF:
P(2  X  5)  
4
2
2
2
2
x
x 4 4 2
3
dx  |2   
8
16
16 16 4
CDF:
42 22 3
P(2  X  5)  F (4)  F (2)   
16 16 4
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Expected Value and Variance

E( X )   x * f ( x)dx


E ( X )   x 2 * f ( x)dx
2

Var( X )  E ( X 2 )  [ E ( X )]2
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Expectation/Variance Example
Suppose we are given the following PDF:
1
 x 0 x4
f ( x)   8
otherwise
 0
We have already found the CDF:
 0
x0
1 2
f ( x)   x 0  x  4
16
x

4
1

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Expectation/Variance Example
Find the expected value of X:
E( X )  
4
0
x
x 3 4 43
x dx  | 
 2.6667
8
24 0 24
Find the variance of X:
E( X 2 )  
4
0
4
4
4
x
x
4
x 2 dx  | 
8
8
24 0 24
2
8
8
Var ( X )  E ( X )  [ E ( X )]  8    
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3
2
2
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