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EDEXCEL INTERNATIONAL A LEVEL MATHEMATICS • S2

THE PROBABILITY FIELD

Enter the second region of the statistical universe — where discrete events become distributions, continuous variables become probability landscapes, and evidence begins to challenge assumptions.

PROBABILITY FIELD ONLINE
S2
PROBABILITY FIELD
B(n,p)
DISCRETE EVENTS
f(x)
CONTINUOUS FIELD
H₀ / H₁
INFERENCE SIGNAL
STATISTICAL UNIVERSE // REGION 02

Navigate the Probability Field

S2 develops the statistical systems introduced earlier, moving from discrete probability models into continuous distributions, normal approximations and formal hypothesis testing.

FIELD 01 • DISCRETE DISTRIBUTIONS

The Binomial and Poisson Distributions

S2 begins by developing the probability distributions introduced in S1. The Binomial and Poisson distributions can be used to model different types of discrete random variables.

Binomial Distribution

A random variable X has a Binomial distribution when there is a fixed number of independent trials, each trial has two possible outcomes, and the probability of success remains constant.

X ~ B(n, p)

Conditions for a Binomial Model

Binomial Probability

P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ

Mean and Variance

E(X) = np
Var(X) = np(1 − p)

Modelling with the Binomial Distribution

A model should only be used when its assumptions are reasonable. In examination questions, you may need to explain whether a Binomial model is appropriate.

Worked Example

A machine produces components. The probability that a component is defective is 0.08. A random sample of 20 components is selected. Let X be the number of defective components.

Each component has two possible outcomes: defective or non-defective.

The probability of a defective component remains constant at 0.08.

Assuming the components are independent:

X ~ B(20, 0.08)

Mean = 20 × 0.08 = 1.6

Variance = 20 × 0.08 × 0.92 = 1.472

Poisson Distribution

The Poisson distribution can be used to model the number of times an event occurs in a fixed interval of time, length, area or another suitable interval.

X ~ Po(λ)

Poisson Probability

P(X = r) = e⁻λ λʳ ÷ r!

Mean and Variance

E(X) = λ
Var(X) = λ
Key idea:

For a Poisson distribution, the mean and variance are equal.

Additive Property

If two independent Poisson random variables are added, the result is also Poisson.

X ~ Po(λ₁), Y ~ Po(λ₂)
X + Y ~ Po(λ₁ + λ₂)

Poisson Approximation to the Binomial

Under suitable conditions, a Binomial distribution can be approximated by a Poisson distribution.

B(n,p) ≈ Po(np)

The approximation is particularly useful when n is large and p is small.

Exam reminder:

Always consider whether the assumptions of the model are reasonable before using an approximation.

FIELD 02 • CONTINUOUS VARIABLES

Continuous Random Variables

Unlike a discrete random variable, a continuous random variable can take any value within an interval.

Probability Density Function

A probability density function, or PDF, describes how probability is distributed over the possible values of X.

Conditions for a PDF:
  • f(x) ≥ 0 for all possible values of x.
  • The total area under the curve is 1.

Finding Probabilities

For a continuous random variable, probability is represented by area under the probability density function.

P(a < X < b) = ∫ₐᵇ f(x) dx

Probability at a Single Point

P(X = a) = 0

Cumulative Distribution Function

The cumulative distribution function gives the probability that X is less than or equal to a particular value.

F(x) = P(X ≤ x)

Relationship Between PDF and CDF

F(x) = ∫₋∞ˣ f(t) dt
f(x) = dF(x) / dx

Worked Example

A continuous random variable X has density f(x) = 2x for 0 ≤ x ≤ 1. Find P(X ≤ 0.5).

P(X ≤ 0.5) = ∫₀⁰·⁵ 2x dx

= [x²]₀⁰·⁵

= 0.25

P(X ≤ 0.5) = 0.25

Simple Polynomial Density Functions

S2 questions may involve probability density functions defined by simple polynomial expressions over a specified interval.

You may need to use the condition that the total probability is 1 to find an unknown constant.

∫ f(x) dx = 1

FIELD 03 • DISTRIBUTION LANDSCAPE

Continuous Distributions

Mean of a Continuous Random Variable

The mean of a continuous random variable is found by weighting each possible value by its probability density.

E(X) = ∫ x f(x) dx

Second Moment

E(X²) = ∫ x² f(x) dx

Variance

Var(X) = E(X²) − [E(X)]²

Standard Deviation

σ = √Var(X)

Mode

The mode is the value of X at which the probability density function reaches its maximum.

Median

The median is the value m for which half of the probability lies below m and half lies above m.

F(m) = 0.5

Quartiles

Quartiles divide the distribution into four equal probability sections.

F(Q₁) = 0.25
F(Q₃) = 0.75

Worked Example

Suppose the cumulative distribution function is known. To find the median, set the cumulative probability equal to 0.5 and solve for x.

Set:

F(x) = 0.5

Solve the resulting equation for x.

The solution is the median.

Continuous Uniform Distribution

A continuous uniform distribution has constant probability density throughout a specified interval.

X ~ U(a,b)

Probability Density

f(x) = 1 ÷ (b − a)

Mean

E(X) = (a + b) ÷ 2

Variance

Var(X) = (b − a)² ÷ 12
Key idea:

The uniform distribution gives equal probability density across its entire interval.

FIELD 04 • NORMAL TRANSFORMATION

Normal Approximations

The Normal distribution can be used to approximate certain Binomial and Poisson distributions when appropriate.

Normal Distribution

X ~ N(μ, σ²)

Binomial to Normal

If X ~ B(n,p), the approximating Normal distribution has:

μ = np
σ² = np(1 − p)

Poisson to Normal

If X ~ Po(λ), a suitable Normal approximation has:

μ = λ
σ² = λ

Continuity Correction

A Binomial or Poisson random variable is discrete, whereas the Normal distribution is continuous. A continuity correction adjusts the boundary by 0.5.

Examples:

P(X ≥ 10) becomes P(Y > 9.5).

P(X ≤ 20) becomes P(Y < 20.5).

P(10 ≤ X ≤ 20) becomes P(9.5 < Y < 20.5).

Standardisation

Z = (X − μ) ÷ σ

Worked Example

X ~ B(100, 0.4). Approximate P(X ≥ 45) using a Normal distribution.

Mean:

μ = np = 100 × 0.4 = 40

Variance:

σ² = np(1 − p) = 100 × 0.4 × 0.6 = 24

Therefore use Y ~ N(40, 24).

Apply continuity correction:

P(X ≥ 45) ≈ P(Y > 44.5)

Then standardise using:

Z = (44.5 − 40) / √24

FIELD 05 • INFERENCE SIGNAL

Hypothesis Tests

Hypothesis testing provides a formal method for deciding whether sample evidence is sufficiently strong to reject a statistical claim.

Null Hypothesis

The null hypothesis, H₀, represents the claim or assumption being tested.

Alternative Hypothesis

The alternative hypothesis, H₁, represents the competing claim.

Significance Level

The significance level is the probability of rejecting H₀ when H₀ is actually true.

Common significance levels: 5% and 1%

Critical Region

The critical region contains values of the test statistic that would lead to rejection of H₀.

Binomial Hypothesis Tests

Binomial models can be used to test claims about the probability of success p.

Worked Example Structure

Step 1: State H₀ and H₁.

Step 2: Identify the distribution under H₀.

Step 3: Find the relevant probability or critical value.

Step 4: Compare the result with the significance level.

Step 5: State the conclusion in context.

Poisson Hypothesis Tests

A Poisson distribution can be used to test claims about the mean rate λ.

Testing Procedure

Important exam language:

Do not simply say “H₀ is true” when a test result is not significant. Say that there is insufficient evidence to reject H₀.

Normal Approximation in Hypothesis Testing

When appropriate, a Normal approximation can be used for a Binomial hypothesis test. The continuity correction must be applied when making the approximation.

FIELD TEST • MIXED PRACTICE

S2 Mixed Practice

These questions combine the major S2 techniques rather than separating practice by topic.

1. State the four conditions required for a Binomial model.
Show answer

Fixed number of trials, two possible outcomes per trial, constant probability of success and independent trials.

2. X ~ Po(6). Find E(X) and Var(X).
Show answer

E(X) = 6 and Var(X) = 6.

3. Explain when a Binomial distribution may be approximated by a Poisson distribution.
Show answer

When n is sufficiently large and p is sufficiently small, using λ = np.

4. A continuous random variable has PDF f(x). What must the total area under the PDF equal?
Show answer

The total area must equal 1.

5. What is P(X = 4) for a continuous random variable?
Show answer

P(X = 4) = 0.

6. State the relationship between a PDF f(x) and its CDF F(x).
Show answer

F(x) is obtained by integrating f(x), and f(x) is the derivative of F(x).

7. X ~ U(2, 10). Find E(X).
Show answer

E(X) = (2 + 10) / 2 = 6.

8. X ~ B(100, 0.2). State the mean and variance of the Normal approximation.
Show answer

Mean = np = 20.

Variance = np(1 − p) = 100 × 0.2 × 0.8 = 16.

9. When approximating P(X ≥ 30) using a Normal distribution, what continuity correction is required?
Show answer

Use P(Y > 29.5).

10. What should be stated when a hypothesis test does not provide sufficient evidence against H₀?
Show answer

There is insufficient evidence to reject H₀.

FIELD MEMORY CORE • QUICK REVIEW

S2 Key Formulae

X ~ B(n,p)
E(X) = np
Var(X) = np(1 − p)
X ~ Po(λ)
E(X) = Var(X) = λ
B(n,p) ≈ Po(np)
P(a < X < b) = ∫ₐᵇ f(x) dx
F(x) = P(X ≤ x)
f(x) = dF(x) / dx
E(X) = ∫x f(x) dx
Var(X) = E(X²) − [E(X)]²
X ~ U(a,b)
E(X) = (a + b) ÷ 2
Var(X) = (b − a)² ÷ 12
Z = (X − μ) ÷ σ
Continuity correction: ±0.5
S2 exam checklist:
  • Check the assumptions of a probability model.
  • Use the correct distribution.
  • Remember continuity correction for Normal approximations.
  • Use integration correctly for continuous variables.
  • State hypothesis-test conclusions in context.

THE PROBABILITY FIELD

Every observation enters as uncertainty. The field transforms uncertainty into measurable probability, distributions and statistical evidence.

EVENT → PROBABILITY → DISTRIBUTION → EVIDENCE