


Can take on any value in a given interval (or a
union of intervals)
 Examples: height, weight, length, mass,
temperature, concentration
They are continuous in theory but in practice
we measure their values to the nearest inch,
gram, milligram, etc.
For continuous rv’s, there is no positive
probability at a point.

In the study of the ecology of a lake, depth
measurements are taken at randomly chosen
locations. Then the depth X at a location is a
continuous rv.

Let X be the amount of time that a random
customer spends waiting for a haircut. Then
X is neither discrete not continuous, since
there is a positive probability that X=0.
3

If we “discretize” lake depth and draw a
histogram, so that the area of the rectangle
above any possible integer k is the proportion
of the lake whose depth is (to the nearest
meter) k, then the total area under the
rectangles is one.
4

If we measure depth more accurately, we
obtain a histogram with a finer mesh.

In the limit (as the mesh size gets smaller) we
obtain a smooth curve, the probability
density function.
The area under the curve is one.

5




pdf defines the distribution of a continuous rv
pdf f(x) has to be nonnegative: f  x   0
f(x) can be greater than 1 but the integral
(probability) is between 0 and 1
Total area under the curve y = f(x) is one, and area
under the curve for a specific interval is the
probability for that interval.

b
P (a  X  b)   f ( x)dx
a
 f ( x)dx  1

a
P( X  a)  P(a  X  a)   f ( x)dx  0
a

Thus, we don’t need to worry about < or 
when we talk about probabilities for a
continuous random variable.

f(x)=2x, 0  x  1
=0 otherwise
 f(x) is nonnegative for 0<=x<=1

 integrates to one


1
f ( x)dx   2 xdx  x
2 1
0
0
• Calculate P(1/4<=X<=2/3)
2/3
P(1/ 4  X  2 / 3) 
 2 xdx  4 / 9  1/16  55 /144
1/ 4
1

A continuous rv X is said to have a uniform
distribution on the interval [A,B] if the pdf of
X is
 1

f  x; A, B    B  A
 0
A x B
otherwise
9

Consider a reference line from the stem of a
tire to an imperfection on a tire, and let X be
the angle (in degrees) measured clockwise to
the location of an imperfection. Then
45
1
P  0  X  45   
dx  1/ 8
360
0
10