2.3.2 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

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

$Var (X) = \int_{- \infty}^{\infty} x^{2} f (x) d x - {[E (X)]}^{2}$

This is equivalent to $Σ x^{2} P (X = x) - {[E (X)]}^{2}$ 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”
- ${[E (X)]}^{2} = {[\int_{- \infty}^{\infty} x^{2} f (x) d x]}^{2}$ “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

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

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

(i)

Find

E (X)

(ii)

Find

Var (X)

(i)

Find

E (X)

2-3-2-cie-fig1-we-solution_a

(ii)

Find

Var (X)

2-3-2-cie-fig1-we-solution_b

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!)

CIE A Level Maths: Probability & Statistics 2

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