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 {
120x2(1 – x) for 0 ≤ X ≤ 1
otherwise

5. As X was thousands of litres of petrol,{
120x2(1 – x) for 0 ≤ X ≤ 1
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

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
otherwise

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

8. { 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
otherwise

9. { 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

10. { 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

11. { 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 [ x ]
= ¾ {[(1 + √0.2)2 – (⅓ x (1 + √0.2)3)]
– [(1 - √0.2)2 – (⅓ x (1 - √0.2)3)]}
= ¾ ( – )
= (3dp) x3 3
1 + √0.2
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-5/2 + 1 {
2560x-7/2 for x ≥ 16
otherwise m 16

14. 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

15. 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
otherwise