OxfordAQA International AS and A-level Mathematics: Mean and variance of continuous variables and the exponential distribution (9660) – Revision Notes

These notes condense the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section S2.2 (second part: mean and variance of continuous random variables and of sums of independent variables) and section S2.3 (exponential distribution). The specification places this in International A-level Unit S2 (Statistics); it is not examined at International AS. For full explanations, read the study guide, then test yourself with the practice set. Pdf and cdf work is in Probability density and cumulative distribution functions. Course pages: hub and checklist. Unsure where to start? Take a free diagnostic.

S2.2 Formulae to know

Result	Formula
Mean	E(X) = μ = ∫ x f(x) dx
Function of X	E(g(X)) = ∫ g(x) f(x) dx
Variance	Var(X) = E(X²) − μ² (equal to E((X − μ)²))
Standard deviation	σ = √Var(X)
Linear change	E(aX + b) = aE(X) + b
Linear change	Var(aX + b) = a²Var(X)
Two independent variables	E(aX ± bY) = aE(X) ± bE(Y)
Two independent variables	Var(aX ± bY) = a²Var(X) + b²Var(Y)
Many independent variables	E(ΣXᵢ) = ΣE(Xᵢ) and Var(ΣXᵢ) = ΣVar(Xᵢ)

The integrals run over the range where f(x) is non-zero. The specification says these formulae are expected knowledge, so learn them rather than deriving them.

Method: mean and variance from a pdf

1. Write down f(x) and its range. Check that the area is 1 if you can do so quickly.
2. Calculate E(X) = ∫ x f(x) dx.
3. Calculate E(X²) = ∫ x² f(x) dx.
4. Var(X) = E(X²) − μ².
5. For other functions, put g(x) into the integral: E(g(X)) = ∫ g(x) f(x) dx.
6. Sense-check: the mean must lie inside the range of X, and the variance must be positive.

Worked reminder. f(x) = 3(1 − x)² for 0 ≤ x ≤ 1.

E(X)  = 3 ∫ x(1 − x)² dx = 3 ∫ (x − 2x² + x³) dx
      = 3 [x²/2 − 2x³/3 + x⁴/4] from 0 to 1
      = 3(1/2 − 2/3 + 1/4) = 3(1/12) = 1/4

E(X²) = 3 ∫ (x² − 2x³ + x⁴) dx = 3(1/3 − 1/2 + 1/5) = 3(1/30) = 1/10

Var(X) = 1/10 − (1/4)² = 1/10 − 1/16 = 3/80

Functions of X

Function	Mean	Variance
X + b	E(X) + b	Var(X)
aX	aE(X)	a²Var(X)
aX + b	aE(X) + b	a²Var(X)
5X²	5E(X²)	needs E(X⁴): integrate
10X⁻¹	10E(X⁻¹) by integration	100(E(X⁻²) − (E(X⁻¹))²)
100X⁻²	100E(X⁻²) by integration	needs E(X⁻⁴): integrate

The specification gives E(2X + 3), E(5X²), E(10X⁻¹), E(100X⁻²), Var(3X), Var(4X − 5) and Var(6X⁻¹) as examples of what can be set.

Must-know distinctions

• E(X²) is not (E(X))². The difference is Var(X).
• E(1/X) is not 1/E(X). Integrate (1/x) f(x).
• Var(aX + b) ignores b and squares a.
• Var(−X) = Var(X).
• X₁ + X₂ (two independent copies) has variance 2Var(X). 2X has variance 4Var(X).
• Var(X − Y) = Var(X) + Var(Y) for independent X and Y.

Worked reminder. E(X) = 12, Var(X) = 4; E(Y) = 5, Var(Y) = 9; X and Y independent.

E(3X − 2Y) = 3(12) − 2(5) = 26
Var(3X − 2Y) = 9(4) + 4(9) = 72

S2.3 Exponential distribution

Conditions. The events occur singly, independently, at random and at a constant average rate λ per unit of time (or length). The lengths of the intervals between Poisson events then have an exponential distribution. If the number of events in an interval of length t is Poisson with mean λt, the gap X to the next event is Exp(λ).

Result	Formula (x ≥ 0)
pdf	f(x) = λe^(−λx)
cdf	F(x) = 1 − e^(−λx)
Upper tail	P(X > x) = e^(−λx)
Interval	P(a < X < b) = e^(−λa) − e^(−λb)
Mean	E(X) = 1/λ
Variance	Var(X) = 1/λ²
Standard deviation	1/λ
Second moment	E(X²) = Var(X) + (E(X))² = 2/λ²

Proofs of the mean and variance are not expected.

Method for an exponential question

1. Identify the rate λ per unit. If you are given the mean m, then λ = 1/m.
2. Make sure the units of λ and x match. Convert if not.
3. Choose the form: P(X > x) = e^(−λx) for an upper tail, 1 − e^(−λx) for a lower tail, or a difference for an interval.
4. To find x for a given probability, set the expression equal to the probability and take natural logs.
5. For a median, solve F(x) = 0.5, which gives x = ln 2 / λ.

Worked reminder. X ~ Exp(0.25). Then P(X > 3) = e^(−0.75) = 0.472. The mean and standard deviation are both 4. Also E(X²) = 2/0.0625 = 32, so E(5X²) = 160.

Worked reminder (combined). Two independent waiting times T₁ and T₂ are each Exp(0.5), so each has mean 2 and variance 4.

E(T₁ + T₂) = 4         Var(T₁ + T₂) = 8        sd = 2.83
E(T₁ − T₂) = 0         Var(T₁ − T₂) = 8
E(3T₁) = 6             Var(3T₁) = 36

The sum of the two is not exponential. Use only the mean and variance results for it, and do not use the exponential cdf.

Poisson link. The number of events in time t is Poisson with mean λt. P(no events in time t) = e^(−λt) = P(X > t). Both routes give the same value.

Which tool, when

The question gives or asks	Use
A pdf and asks for a mean	Integrate x f(x) over the range
A pdf and asks for E(1/X) or E(X²)	Integrate g(x) f(x) with the right g
E(X) and Var(X) and asks about aX + b	The linear results, no integration
Two or more independent variables with known means and variances	Sum or difference formulae
Random events at a constant rate and a gap or distance	Exponential, with λ equal to the rate
A mean waiting time	λ = 1/mean
A count of events in an interval	Poisson with mean λt

Quick self-test

1. E(X) = 5 and Var(X) = 4. Find E(3X − 7) and Var(3X − 7).
2. E(X) = 10 and Var(X) = 5. Find E(X²).
3. X and Y are independent. E(X) = 6, Var(X) = 2, E(Y) = 4, Var(Y) = 3. Find E(X − 2Y) and Var(X − 2Y).
4. X has variance 0.5. Find the variance of X₁ + X₂ + X₃ + X₄ (independent copies of X) and of 4X.
5. f(x) = 3(1 − x)² for 0 ≤ x ≤ 1. Find E(X) without looking at the reminder above.
6. X ~ Exp(0.25). Find P(X > 2) to 3 decimal places.
7. X is exponential with mean 20. Find P(X < 10) to 3 decimal places.
8. Faults occur at random at an average rate of 3 per hour. Find the probability of no faults in a 20-minute period to 3 decimal places.
9. X ~ Exp(2). Find E(5X²).
10. F(x) = 1 − e^(−0.5x) for x ≥ 0. Find the median to 3 decimal places.
11. State the conditions on events that make the gaps between them exponential.
12. f(x) = 2x for 0 ≤ x ≤ 1. Find E(1/X).

Answers

1. E = 3(5) − 7 = 8; Var = 9 × 4 = 36.
2. E(X²) = Var(X) + (E(X))² = 5 + 100 = 105.
3. E = 6 − 2(4) = −2; Var = 2 + 4(3) = 14.
4. Sum: 4 × 0.5 = 2. For 4X: 16 × 0.5 = 8.
5. 1/4.
6. e^(−0.5) = 0.607.
7. λ = 1/20 = 0.05; 1 − e^(−0.5) = 0.393.
8. Rate per minute = 3/60 = 0.05, so λt = 0.05 × 20 = 1 and the probability is e^(−1) = 0.368.
9. E(X²) = 2/λ² = 2/4 = 0.5, so E(5X²) = 2.5.
10. 1 − e^(−0.5x) = 0.5 gives x = 2 ln 2 = 1.386.
11. Events occur singly, independently, and at a constant average rate.
12. ∫ (1/x)(2x) dx from 0 to 1 = ∫ 2 dx = 2.

Where marks are usually lost

• Stopping at E(X²) and forgetting to subtract μ² (or subtracting μ instead of μ²).
• Using the variance formula with the wrong sign for a difference of independent variables.
• Treating E(g(X)) as g(E(X)) for non-linear g.
• Using the wrong limits, such as integrating over a range where f is zero.
• Mixing units, such as a rate per hour with a time in minutes.
• Quoting the mean of an exponential distribution as λ rather than 1/λ.
• Using the lower-tail formula when the question asks for P(X > x), or the reverse.
• Rounding too early in an exponential calculation. Keep unrounded values until the final line.
• Not showing the integral or the substitution when a question says “show that” or asks for exact values.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA: sections S2.2 (second part) and S2.3.