12.SPECIAL PROBABILITY DISTRIBUTIONS
CONTINUOUS
DISCRETE
1. UNIFORM
2. BINOMIAL
3. POISSON
1.
UNIFORM
CONTINUOUS
2. NORMAL
12.1 Discrete Probability Distributions
(Uniform Distribution)
LEARNING OUTCOMES
At the end of the lesson students are able to
(a) Understand the discrete uniform
distribution
(b) Find the mean and variance for
discrete uniform distributions.
12.1 Discrete Uniform Distributions.
A discrete random variable X is said to have
a uniform distribution , if its probability
function has a constant value, p and
defined as
P(X=x)  p
for x  x1 , x2 , x3 , ..., xn
Example 1
Suppose a fair die is rolled. The
number obtained is a random variable
X with a uniform distributions as
shown in the following probability
distribution table.
Solution
x
1
2
3
4
5
6
P(X=x)
1
6
1
6
1
6
1
6
1
6
1
6
From the table , we find that the
1
probability for all variables are
.
6
The uniform distribution can be written as
1
P(X=x) 
for x  1, 2, 3,4,5,6
6
If X is a discrete random variable
with a uniform distribution , then
1
i ) P ( X  xi ) 
n
for i  1, 2, 3,..., n
mean
n
ii ) E ( x )   xP ( X  x )
i
variance

iii ) V ( x )    x 2 P ( X  x )    xP ( X  x )
n
i
n
i
= E(x2) - [E(x)]2

2
Example 2
An unbiased spinner , numbered 1,2,3,4 is
spun. The random variables X is the number
obtained. Draw a probability distribution
table and find
a)the probability distribution function and
sketch the graph.
b)The mean of the probability distribution .
c)The variance of the probability
distribution .
Solution:
1
a) P( X  x) 
4
for x  1, 2, 3,4
PROBABILITY DISTRIBUTION TABLE IS
x
1
2
3
4
P(X=x)
1
4
1
4
1
4
1
4
f(x)
1
probability distribution function
4
1 2 3 4
b ) mean ,   xP ( X  x )
x
4

1
1
1
1
1
 1( )  2( )  3( )  4( )
4
4
4
4
10

4
 2.5
c) variance,   Var ( x )
2
= E(x2) - μ2
= 7.5 - (2.5)2
=1.25
1
1
1
1
E( X )  1( )  4( )  9( )  16( )
4
4
4
4
 7 .5
2
Example 3
For the following uniform distribution
function
1
P( X  x) 
n
for x   3, 2, 1,0,1, 2, 3
a)Show that n=7.
b)Calculate the mean and variance
c)Find P(|x-1|≥1)
Solution:
A) PROBABILITY DISTRIBUTION TABLE IS
x
P(X=x)
3
 P( X
3
-3
1
n
-2
-1
1 1
n n
 x)  1
1
7
n
0
1
2
3
1 1 1 1
n n n n

 1

n=7
b) mean ,   E ( x )   x P ( X  x )
3
3
1
1
1
1
1
1
1
 3( )  2( )  1( )  0( )  1( )  2( )  3( )
7
7
7
7
7
7
7
=0
variance,   Var ( x ) = E(x2) - μ2
2
= 4-0
=4
1
1
1
1
1
1
1
E(X )  9( )  4( )  1( )  0( )  1( )  4( )  9( )
7
7
7
7
7
7
7
4
2
c) P(|x-1| ≥1)
P(|x-1|≥1)
|x-1|≥1
= P(x≤ 0) + P(x≥2)
= 1 - P(x=1)
x-1≤ -1
x-1≥1
x≥2
x≤ 0
|
0
|
2
1
=17
6

7
Example 4
A DISCRETE RANDOM VARIABLE X HAS THE
FOLLOWING PROBABILITY DISTRIBUTION .
x
2
4
6
8
10
P(X=x)
0.2
0.2
0.2
0.2
0.2
Find
a ) P (2  X  10)
b ) P ( X  6)
c ) mean, E ( x )
d) Standard deviation
Solution:
a ) P (2  X  10)
= P(X=4) + P(X=6) + P(X=8)
= 0.2 + 0.2 + 0.2
= 0.6
b ) P ( X  6)
= P(X=6) + P(X=8) + P(X=10)
= 0.2 + 0.2 + 0.2
= 0.6
b) mean ,   E ( x )   x P ( X  x )
 2(0.2)  4(0.2)  6(0.2)  8(0.2)  10(0.2)
=6
variance,   Var ( x ) = E(x2) - μ2
2
=44 – 36
=8
  8  2.828
E(X 2 )  4(0.2)  16(0.2)  36(0.2)  64(0.2)  100(0.2)
 44
CONCLUSION
11.1 Discrete Uniform Distributions.
f ( x )  p for x  x1 , x2 , x3 , ..., xn
x
1
2
3
…
n
P(X=x)
p
p
p
…
p