The Uniform Distribution

Probability and Binomial Distributions

One important distribution is the uniform one in which all possible outcomes have equal possibilities.

The random variable X is said to follow a uniform distribution when all its outcomes are equally likely. A very simple example is given by the random variable H, 'the number of heads seen when a single coin is tossed'.

The probability distribution is given by

h	0	1
p(h)

p(h) = for h = 0, 1

E(H) = 0 × \frac{1}{2} + 1 × \frac{1}{2} = \frac{1}{2}

and

V(H)	= 0^2 × \frac{1}{2} + 1^2 × \frac{1}{2} - \left(\frac{1}{2}\right)^2
	= \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

and the standard deviation of H is .

Another example of a uniform distribution is the random variable, X, the score obtained when rolling a single unbiased die. In this case

P(X = x) = for x = 1, 2 ..., 6

giving

E(X) = = 3

which can be written down by considering symmetry, and

V(X)	= (1 + 4 + 9 + 16 + 25 + 36) -
	=

This can be generalised to a random variable, X, having n equally likely outcomes for which the probability distribution is given by

P(X = x) = for x = 1, 2 ..., n

Worked Examples

For a 10-sided spinner, let X be the score obtained. Determine E(X) and V(X)

Show me If X is the random variable 'score on the spinner', then it has probability distribution

x	1	2	3	4	5	6	7	8	9	10
P(X = x)

and

E(X)	= \sum_{\ }^{\ } x p(x)
	= 1 × \frac{1}{10} + 2 × \frac{1}{10} + 3 × \frac{1}{10} + 4 × \frac{1}{10} + 5 × \frac{1}{10} + 6 × \frac{1}{10} + 7 × \frac{1}{10} + 8 × \frac{1}{10} + 9 × \frac{1}{10} + 10 × \frac{1}{10}
	= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) ÷ 10
	= 5\frac{1}{2} (as expected)

Also

var(X)	= E(X)^2 - (E(X))^2
	= 1 × \frac{1}{10} + 4 × \frac{1}{10} + 9 × \frac{1}{10} + 16 × \frac{1}{10} + 25 × \frac{1}{10} + 36 × \frac{1}{10} + 49 × \frac{1}{10} + 64 × \frac{1}{10} + 81 × \frac{1}{10} + 100 × \frac{1}{10} - \left(5\frac{1}{2}\right)^2
	= (1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100) ÷ 10 - (5.5)^2
	= \frac{33}{4}

You may be able to spot patterns in the results with n equal to 6 and 30 and deduce the general results.

i.e.

n = 6	E(X) = \frac{7}{2}	V(X) = \frac{35}{12}
n = 10	E(X) = 11	V(X) = \frac{33}{4} \left( = \frac{99}{12}\right)
n	E(X) = ?	V(X) = ?

The number 99 looks 'clumsy' but it can be written as 100 − 1 = 10² − 1, which suggests V(X) = \frac{n^2-1}{12}, although E(X) = \frac{n+1}{2} is simpler to see.

You will need the series summation results:

	1 + 2 + 3 + ... + n = (n + 1)	(1)
and
	1² + 2² + 3² + ... + n² = (n + 1)(2n + 1)	(2)

to understand the proofs of the general results.

E(X)	= 1 × \frac{1}{n} + 2 × \frac{1}{n} + 3 × \frac{1}{n} + ... + n × \frac{1}{n}
	= (1 + 2 + 3 + ... + n) × \frac{1}{n}, using (1),
	= \frac{n}{2}(n + 1) × \frac{1}{n}
E(X)	= \frac{n + 1}{2}

Using equation (2), we can also show that

V(X) = \frac{n^2 - 1}{12}

Worked Examples

Exercises