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“Teach A Level Maths”
Statistics 1
The Discrete Uniform
Distribution
© Christine Crisp
The Discrete Uniform Distribution
Statistics 1
Edexcel
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The Discrete Uniform Distribution
We may have a situation where the probabilities
of each event are the same.
For example, if we roll a fair die, we assume that the
1
probability of obtaining each number is .
6
If X is the random variable (r.v.) “ the number
showing” , the probability distribution table is
x
1
2
3
4
5
6
P(X = x)
1
6
1
6
1
6
1
6
1
6
1
6
The probability distribution function (p.d.f.) is
1
P( X  x)  ,
6
x  1, 2, 3, 4, 5, 6
The distribution with equal probabilities is called
“uniform”
The Discrete Uniform Distribution
It is also possible to have a continuous uniform
distribution where the r.v. can be any number in a
given interval.
The continuous uniform distribution is also called the
rectangular distribution. You will not be studying it in
this module.
The Discrete Uniform Distribution
A diagram for the distribution
1
P( X  x)  ,
6
x  1, 2, 3, 4, 5, 6
looks like this:
p
1
6
1
2
3

4
5
6 x
The mean value of X is given by the average of the
1st and last values of x, so,
1 6

 35
2
The Discrete Uniform Distribution
However, we could also use the formula for the mean
of any discrete distribution of a random variable:
   xP( X  x )
1
For P ( X  x )  ,
6
x  1, 2, 3, 4, 5, 6
we would get
1
1
1
  1  2  . . .  6
6
6
6
1
 21
6
 35
The Discrete Uniform Distribution
e.g. 1 The r.v. X has p.d.f. given by
P( X  x)  k ,
x  1, 2, 3, 4
where k is constant.
(a) Find the value of k.
(b) Find the mean of the distribution.
Solution:
(a) The sum of the probabilities,  P ( X  x )  1
1
 k  Be
k  careful
k  k  1here!
 kThe
 probabilities
4
the same
(b) The mean, ,are
is given
by for all the values of x.
Either
1 4

 25
2
Or
   xP( X  x )
   1k  2k  3k  4k
   10k  10( 14 )
   25
The Discrete Uniform Distribution
The Variance of the Uniform Distribution
We can find the variance for any discrete random
variable X using
Var ( X )   2   x 2 P ( X  x )   2
e.g. The random variable X has p.d.f. given by
1
P( X  x)  ,
x  1, 2, 3, 4, 5, 6
6
1
1
1
2
2
2
2
So, Var ( X )  1 
 2  ... 6   
6
6
6
We found earlier that   3·5, so
91
Var ( X ) 
 3  5 2  2  92 ( 3 s. f . )
6
The Discrete Uniform Distribution
The formulae for the mean and variance of the
discrete uniform distribution are not in the formulae
booklet.
However, the mean can either be seen directly from
the x-values or by using the general formula for the
mean of a discrete distribution.
A formula for the variance is given by
2
n
1
2
Var ( X )   
12
where n is the number of values of X.
However, if you forget the formula, you can use the
general formula for variance as I just did in the
example.
The Discrete Uniform Distribution
SUMMARY
 If a random variable X has p.d.f. given by
P( X  x)  k
where k is constant and the values of x are
discrete, then X has a discrete uniform distribution.
 The value of k is found by using  P ( X  x )  1
 The mean of the distribution can be found by
averaging the 1st and last x-values
The variance is given by 2
n 1
2
Var ( X )   
12
where n is the number of values of X.
The mean and variance can both be found using the
formulae for general discrete distributions.
The Discrete Uniform Distribution
Exercise
1. The r.v. X has p.d.f. given by
P( X  x)  k ,
x  4, 5, 6, 7, 8
where k is constant.
(a) Find the value of k.
(b) Find the mean and variance of the distribution.
Solution:
(a)  P ( X  x )  1
 k  k  k  k  k 1
1
 k
5
(b)   6 or E ( X)     xP( X  x )
We can get the mean directly from
 4k  5k  6k  7k  8k
the x values or by using the formula.
1
 30k  30   6
 5
The Discrete Uniform Distribution
1. The r.v. X has p.d.f. given by
P( X  x)  k ,
x  4, 5, 6, 7, 8
where k is constant.
(a) Find the value of k.
(b) Find the mean and variance of the distribution.
1
k , 6
5
Either:
n2  1 52  1
Var ( X ) 

2
12
12
use
Or: If youVar
( Xthis
)  formula,
x 2 P ( X bex )careful.
 2
n is the number2 of values
of x, which
in this
2
2

4
k

5
k

.
.
.

8
k

36
example is 5, not the final value, 8.

 190k  36
 1
 190   36  2
 5
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
The Discrete Uniform Distribution
SUMMARY
 If a random variable X has p.d.f. given by
P( X  x)  k
where k is constant and the values of x are
discrete, then X has a discrete uniform distribution.
 The value of k is found by using  P ( X  x )  1
 The mean of the distribution can be found by
averaging the 1st and last x-values
The variance is given by 2
n 1
2
Var ( X )   
12
where n is the number of values of X.
The mean and variance can both be found using the
formulae for general discrete distributions.
The Discrete Uniform Distribution
e.g. 1 The r.v. X has p.d.f. given by
P( X  x)  k ,
x  1, 2, 3, 4
where k is constant.
(a) Find the value of k.
(b) Find the mean and variance of the distribution.
Solution:
(a) The sum of the probabilities,  P ( X  x )  1
Be careful here! The probabilities
are the same for all the values of x.
1
 k  k  k  k 1  k 
4
(b) The mean, , is given by
Either
1 4

 25
2
The Discrete Uniform Distribution
Or
   xP( X  x )
   1k  2k  3k  4k
   10k    2  5
The variance is given by
2
2
n

1
4
1
Either Var ( X )   2 

 1  25
12
12
Or
 2   x 2 P( X  x)   2
 12 k  2 2 k  3 2 k  4 2 k   2
 30k   2
1
k
4
and   2·5, so
 1
Var ( X )  30   2  5 2  1  25
 4