Statistics 1 Revision Notes - Mr Barton Maths

The expected variance,
Var[X] = σ 2 = E[X 2 ] – (E[X]) 2
= ∑
= (1 2 + 2 2 + 3 2 + … + n 2 ) × − 1
= n(n + 1)(2n + 1) × − (n + 1) 2 since Σ i 2 = n(n + 1)(2n + 1)
=
=
(n + 1){(8n + 4) − (6n + 6)}
(n + 1) (2n − 2)
⇒ Var[X] = σ 2 =
(n2 − 1)
These formulae can be quoted in an exam (if you learn them!).
Non-standard uniform distribution
The formulae can sometimes be used for non-standard uniform distributions.
Example: X is the score on a fair 10 sided spinner. Define Y = 5X + 3.
Find the mean and variance of Y.
Y is the distribution {8, 13, 18, … 53}, all with the same probability .
Solution: X is a discrete uniform distribution on the set {1, 2, 3, …, 10}
⇒
E[X] = (n + 1) = 5
and Var[X] =
(n2 − 1) =
= 8
⇒ E[Y] = E[5X + 3] = 5E[X] + 3 = 30
and
Var[X] = Var[5X + 3] = 5 2 Var[X] = 25 ×
= 206
⇒ mean and variance of Y are 27 and 206 .
14/04/2013 Statistics 1 SDB 37