The Uniform Probability
Distribution
1
A rectangle
The height
of the
rectangle
The base of the rectangle
Do you remember how to calculate the area of a rectangle
from your earlier schooling? No problem, it is the base
times the height!
What if the base equals the height? We would have b times
h or b times b (since b = h) or b2, where we say b squared.
What do we call that special type of rectangle where the
base equals the height? A circle, right?
2
I will come back to the rectangle, it will be important for what
we will do.
Let’s talk about a problem. Say a plane flying to the Big Apple
from the old Second City takes anywhere from 120 to 140
minutes (from take off to touch down.)
Say the distribution of flying time is uniform over the interval
120 to 140. This basically means the probability of each 1
minute interval between 120 and 140 has the same probability
of occurring. IN other words, the plane has the same chance of
landing in the 120 to 121 interval as the 121 to 122 interval and
so on.
3
f(x)
1/20
0
120
140
a
b
x = minutes of
flight time.
1/(b – a)
4
On the previous screen you see a rectangle. It is on the
interval 120 to 140 because this is the number of minutes a
plane would take when flying to New York from Chicago.
In general we say the interval is from a to b. Then the base of
the rectangle is b – a = 140 –120 = 20 in our example.
Now, here is an important point based on rules of probability.
Remember that the sum of all the probabilities of the
outcomes must sum to 1. Since we want the area to represent
probability the area of the rectangle must equal =1. We do
this by making the height 1/(b – a) = 1/(140 – 20) = 1/20.
Base time height is then (20)(1/20) = 1
Now we can answer more questions dealing with the
probability of an event when we have a uniform distribution.
5
What is the probability the plan will land between 122 and 124
minutes? Picture is on the next screen.
We just want to find the area inside the rectangle between 122
and 124. The base is 124 – 122 = 2 and the height is 1/20, so
we have a probability 2(1/20) = .1.
You might see the same problem asked in the following way:
P(122 <= x <= 124), or what is the probability x is greater than
or equal to 122, but less than or equal to 124?
Special case
P(x = 125)? What is the probability the plane lands in 125
minutes? Well 125 – 125 = 0 and 0(1/20) = 0. Since
probabilities are areas for continuous variable, the probability
for any one value is zero. We can only do probabilities over an
6
interval.
f(x)
1/20
0
120 122 124
a
140
x = minutes of
flight time.
b
1/(b – a)
7
In our example the mathematical expression for the probability
density function is
1/20
for 120 <= x <= 140
0
elsewhere,
f(x) =
1/(b – a)
and in general is
for a <= x <= b
f(x) =
0
elsewhere
All this really means is that the height of the rectangle is
just the inverse of the length of the rectangle.
8
The expected value for a uniform variable is (I do not have a
proof here so hopefully you can accept this.)
E(x) = (a + b)/2 and in our example equals
(120 + 140)/2 = 130.
The variance is
Var(x) = (b – a)2/12 and in our example equals
(140 – 120)2/12 = 400/12 = 33.33
9
Let’s do problem 6 page 246
The uniform distribution suggested by the problem looks like
8
11.975
12.1
.125
10
Part a is probability that we are between 12 and 12.05
We want this area and it is
(12.05 – 12)8 = .05(8) = .4
8
12
12.05
11.975
12.1
.125
11
Part b is probability that we are above 12.02
We want this area and it is
(12.1 – 12.02)8 = .08(8) = .64
8
12.02
11.975
12.1
.125
12
Part c is probability that we are above 12.02 or below 11.98
We want these areas and they
add up to
8
(12.1 – 12.02)8 + (11.98 –
11.975)8 = .08(8) + .005(8) =
.64 + .04 = .68
11.98 12.02
11.975
12.1
This is also equal to
1 – [(12.02 – 11.98)8] =
.125
1 - .32 = .68
13