Statistics 1 Revision Notes - Mr Barton Maths

⇒ The expected mean and variance for the score, X, are μ = 3 and σ 2 = 2
(b) Z = 3X – 6
⇒
E[Z] = E[3X – 6] = 3E[X] – 6 = 10 – 6 = 4
and
Var[Z] = Var[3X – 6] = 3 2 Var[X] = 9 ×
= 26
⇒ The expected mean and variance for the prize, $Z, are μ = 4 and σ 2 = 26
The discrete uniform distribution
Conditions for a discrete uniform distribution
• The discrete random variable X is defined over a set of n distinct values
• Each value is equally likely, with probability 1 / n .
Example: The random variable X is defined as the score on a single die. X is a discrete
uniform distribution on the set {1, 2, 3, 4, 5, 6}
The probability distribution is
Score 1 2 3 4 5 6
Probability
Expected mean and variance
For a discrete uniform random variable, X defined on the set {1, 2, 3, 4, ..., n},
X 1 2 3 4 ... ... n
Probability
By symmetry we can see that the Expected mean = μ = E[X] = (n + 1),
or μ = E[X] = ∑ = 1 × + 2 × + 3 × + … + n ×
= (1 + 2 + 3 + … + n) ×
= n(n + 1) × = (n + 1)
36 14/04/2013 Statistics 1 SDB