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

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Presentation on theme: "Starter. Continuous Random Variables Learning Objectives: Able to calculate the probability of a CRV Able to calculate the expectation and variance of."— Presentation transcript:

1 Starter

2 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

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

4 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 120x 2 (1 – x)for 0 ≤ X ≤ 1 0 otherwise {

5 f(x) = E(X) = = = 120 = 120 [ - ] = 120 x 1/20 = 6 120x 2 (1 – x)for 0 ≤ X ≤ 1 0 otherwise { 1010 x 4 4 x 5 5 As X was thousands of litres of petrol, E(X) = 6000 litres.

6 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 {

7 Variance Discrete Random Variables: Var(X) = σ 2 = ∑x 2 P(X = x) – μ 2 Continuous Random Variables: Var(X) = σ 2 =

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

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

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

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

12 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) =

13 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/ x -7/2 for x ≥ 16 0 otherwise { m 16

14 ½ = 1024m -5/2 m -5/2 = m 5/2 = 2048 m = 21.1 (3sf) Therefore the median salary is $21,100 to the nearest $

15 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 ≥ otherwise {


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