1. Starter
The probability distribution of a discrete random variable X is given by:
P(X = r) = 30kr for r = 3, 5, 7
P(X = r) = 0 otherwise
What is the value of k?
Hence or otherwise, calculate P(X ≥ 5)
Calculate the area under the curve f(x) = ⅜(1 + x2) between 0 and 1.

2. Continuous Random Variables
Learning Objectives: Understand the difference between a discrete and continuous random variable Able to determine whether a function is a probability density function Able to calculate probabilities using a p.d.f.

3. Discrete vs Continuous
Discrete variables – can take specific values
e.g. shoe size, number on a die, etc.
Continuous variables – can take any value
e.g. weight of a baby, height of students, time taken to run 100m, etc.

4. Discrete Random Variables
D.R.V. – uses a probability distribution to describe the possible values

5. Continuous Random Variables
C.R.V. – described by a probability density function (p.d.f)
f(x) ≥ 0 for all x.
Area under the curve must equal 1, i.e.
= 1

6. Continuous Random Variables
P(X = r) = 0
Therefore:
P(a < X < b) = P(a ≤ X < b) = P(a < X ≤ b) = P(a ≤ X ≤ b)

7. Continuous Random Variables
We can find probabilities by integrating f(x) between certain limits, i.e.
P(a ≤ X ≤ b) =

8. Continuous Random Variables
Example: A continuous random variable X has the probability density function given by
f(x) =
show that f(x) has the properties of a p.d.f.
Find P(1.5 ≤ X ≤ 2) {
⅔x for 1 ≤ X ≤ 2
otherwise

9. f(x) ≥ 0 for all x since ⅔x > 0 for x > 0
= 1 2 1 2 1

10. P(1.5 ≤ X ≤ 2) = ⅔ [½x2] = ⅓ [x2] = ⅓ x (4 – 2.25) = 0.583 (3dp) 2 1.5

11. { The continuous random variable X has the p.d.f. given by: f(x) =
where k is a constant
Find the value of k
Find P(0.3 ≤ X ≤ 0.6)
Find P(|X| < 0.2) {
k(1 + x2) for -1 ≤ X ≤ 1
otherwise

12. {
k(1 + x2) for -1 ≤ X ≤ 1
otherwise
f(x) =
a) Find the value of k.

13. {
⅜(1 + x2) for -1 ≤ X ≤ 1
otherwise
f(x) =
b) Find P(0.3 ≤ X ≤ 0.6)

14. {
⅜(1 + x2) for -1 ≤ X ≤ 1
otherwise
f(x) =
c) Find P(|X| < 0.2)