Starter

Continuous Random Variables Learning Objectives: Able to calculate the probability of a CRV Able to calculate the expectation and variance of a CRV Able to calculate the median of a CRV

Expectation Discrete Random Variables: E(X) = μ = ∑xP(X = x) Continuous Random Variables: E(X) = μ =

Expectation At a garage, the weekly demand for petrol, X, in thousands of litres, can be modelled by the probability density function: f(x) = Find the mean weekly demand for petrol { 120x2(1 – x) for 0 ≤ X ≤ 1 0 otherwise

As X was thousands of litres of petrol, { 120x2(1 – x) for 0 ≤ X ≤ 1 0 otherwise f(x) = E(X) = = = 120 = 120 [ - ] = 120 x 1/20 = 6 As X was thousands of litres of petrol, E(X) = 6000 litres. x4 4 x5 5 1

The amount of savings, Y, in thousands of dollars, of a random selection of men can be modelled by the p.d.f.: f(x) = Calculate the mean savings for men. { ¾(3 - x)(x – 5) for 3 ≤ X ≤ 5 0 otherwise

Variance Discrete Random Variables: Var(X) = σ2 = ∑x2P(X = x) – μ2 Continuous Random Variables: Var(X) = σ2 =

{ Variance For the CRV X with p.d.f. defined by f(x) = Find: a) the mean b) the variance c) P(μ – σ ≤ X ≤ μ + σ) { ¾x(2 – x) for 0 ≤ X ≤ 2 0 otherwise

{ ¾x(2 – x) for 0 ≤ X ≤ 2 f(x) = 0 otherwise E(X) = = ¾ [ - ] = ¾ [ - ] = ¾ x 4/3 = 1 2x3 3 x4 4 2

{ ¾x(2 – x) for 0 ≤ X ≤ 2 0 otherwise f(x) = b) Var (X) = = = ¾ [ - ] - 1 = (¾ x 8/5) - 1 = 0.2 2x4 4 x5 5 2

{ ¾x(2 – x) for 0 ≤ X ≤ 2 0 otherwise x3 3 f(x) = c) P(μ – σ ≤ X ≤ μ + σ) = P(1 - √0.2 ≤ X ≤ 1 + √0.2) = = ¾ [ x2 - ] = ¾ {[(1 + √0.2)2 – (⅓ x (1 + √0.2)3)] – [(1 - √0.2)2 – (⅓ x (1 - √0.2)3)]} = ¾ (1.0841 – 0.2493) = 0.626 (3dp) x3 3 1 + √0.2 1 - √0.2

Median The median divides the p.d.f. into two equal halves Therefore, the median, m, of a CRV is that value for which P(X ≤ m) =

Find the median salary (to the nearest $100) of the p.d.f. given by f(x) = P(X ≤ m) = ½ = 2560 [ (-⅖)x-5/2 ] ½ = [(2560 x (-⅖)m-5/2) – (2560 x (-⅖) x 16-5/2) ½ = -1024m-5/2 + 1 { 2560x-7/2 for x ≥ 16 0 otherwise m 16

½ = 1024m-5/2 m-5/2 = m5/2 = 2048 m = 21.1 (3sf) Therefore the median salary is $21,100 to the nearest $100. 1 2048

A computer ink cartridge has a life of X hours A computer ink cartridge has a life of X hours. The variable X is modelled by a p.d.f. f(x) = Find the median lifetime of these cartridges { 400x-2 for x ≥ 400 0 otherwise