E(X) & Var(X) (Continuous)

E(X)is the expected value, or mean, of a random variable X
- E(X) is the same as the population mean so can also be denoted by µ
Var (X) is the variance of the continuous random variable X
- Standard deviation is the square root of the variance

$E (X) = μ = \int_{- \infty}^{\infty} x f (x) d x$

$Var (X) = σ^{2} = \int_{- \infty}^{\infty} x^{2} f (x) d x - μ^{2}$

This is equivalent to $Σ x^{2} P (X = x) - μ^{2}$ for discrete random variables
Be careful about confusing $E (X^{2})$ and ${[E (X)]}^{2}$
- $E (X^{2}) = \int_{- \infty}^{\infty} x^{2} f (x) d x$ “mean of the squares”
- ${[E (X)]}^{2} = {[\int_{- \infty}^{\infty} x f (x) d x]}^{2}$ “square of the mean”
- If you are happy with the difference between these and how to calculate them the variance formula becomes very straightforward

$Var (X) = E (X^{2}) - {[E (X)]}^{2}$

$E (g (X)) = \int_{- \infty}^{\infty} g (x) f (x) d x$
In particular:
- $E (X^{2}) = \int_{- \infty}^{\infty} x^{2} f (x) d x$ as seen above
If $g (X) = a X + b$ (a linear function) then
- $E (g (X)) = E (a X + b) = a E (X) + b$
- $Var (g (X)) = Var (a X + b) = a^{2} Var (X)$

A continuous random variable, X, is modelled by the probability distribution function f(x), such that

$f (x) = \{\begin{cases} 1.5 x^{2} (1 - 0.5) & 0 \leq x \leq 2 \\ 0 & otherwise \end{cases}$

(a)

Find

E (X)

(b)

Find

Var (X)

(a)

Find

E (X)

1-3-2-ial-fig1-we-solution-part-1

(b)

Find

Var (X)

A sketch of the graph of y = f(x) can highlight any symmetrical properties which can help reduce the work involved in finding the mean and variance
Take care with awkward values and negatives – use the memory features on your calculator and avoid rounding until your final answer (if rounding at all!)
The formulae for $E (X) = μ$ and $V a r (X) = σ^{2}$ are given in the formulae booklet

E(X) & Var(X) (Continuous) (Edexcel International A Level Maths: Statistics 2)