Pearson Edexcel International Advanced Level Mathematics: Discrete random variables (YMA01) – Revision Notes

These revision notes are aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics (YMA01) specification, Issue 3, April 2019, unit S1 section 5 (topics 5.1 to 5.4). The specification states that S1 is an optional unit for IAS Mathematics and Further Mathematics and for IAL Mathematics and Further Mathematics. Read the study guide for the full explanations and examples, then test yourself with the practice questions.

Related pages: the Edexcel A Level Mathematics course page, the printable checklist and the free 10-minute diagnostics.

Key ideas by topic

Topic	Remember
5.1	X is the random variable, x is a value it takes. Discrete means separate, listable values
5.2	p(x) = P(X = x). F(x₀) = P(X ≤ x₀). Σ p(x) = 1
5.3	E(X) = Σ x p(x). Var(X) = E(X²) − (E(X))². Linear rules for aX + b
5.4	Uniform on 1 to n: p(x) = 1/n, mean (n + 1)/2, variance (n² − 1)/12

5.1 and 5.2: distributions and the CDF

A valid probability function has every p(x) between 0 and 1 and a total of exactly 1.

Formulas for whole-number values

Want	Write
P(X ≤ a)	F(a)
P(X < a)	F(a − 1)
P(X > a)	1 − F(a)
P(X ≥ a)	1 − F(a − 1)
P(a ≤ X ≤ b)	F(b) − F(a − 1)
P(a < X ≤ b)	F(b) − F(a)
p(x)	F(x) − F(x − 1)

Method: finding an unknown constant

1. Write Σ p(x) = 1 using every value in the table or formula.
2. Solve for the constant.
3. Check each probability lies between 0 and 1.

Method: building F from p

1. Keep a running total down the table.
2. The final total must be 1.
3. Write F as a table or state the value for each x.

5.3: expectation and variance

Formulas

Result	Formula
Mean	E(X) = Σ x p(x)
Expected value of g(X)	E(g(X)) = Σ g(x) p(x)
Second moment	E(X²) = Σ x² p(x)
Variance	Var(X) = E(X²) − (E(X))²
Linear mean	E(aX + b) = aE(X) + b
Linear variance	Var(aX + b) = a² Var(X)

Method: variance from a table

1. Add a row for x p(x) and total it to get E(X).
2. Add a row for x² p(x) and total it to get E(X²).
3. Subtract the square of the mean.
4. If asked, take the positive square root for the standard deviation.

Small worked reminder

x = 0, 1, 2, 3 with p(x) = 0.1, 0.2, 0.4, 0.3.

E(X)   = 0 + 0.2 + 0.8 + 0.9        = 1.9
E(X²)  = 0 + 0.2 + 1.6 + 2.7        = 4.5
Var(X) = 4.5 − 1.9² = 4.5 − 3.61    = 0.89

5.4: the discrete uniform distribution

Quantity	Value for x = 1, 2, …, n
P(X = x)	1/n
E(X)	(n + 1)/2
Var(X)	(n² − 1)/12

For consecutive whole numbers that start somewhere else, subtract to get back to 1, n and use the linear rules. For values 3 to 8, let Y = X − 2. Then Y is uniform on 1 to 6, E(X) = 3.5 + 2 = 5.5 and Var(X) = 35/12.

Method: a uniform variable

1. Count the values. This count is n.
2. If the values do not start at 1, subtract a constant to make them run from 1 to n.
3. Write the mean and variance for the shifted variable.
4. Undo the shift: add the constant to the mean and leave the variance unchanged.
5. Probabilities are each 1/n, so a probability for a range is the number of values in that range divided by n.

Worked reminders

A constant and the CDF

p(x) = k(x + 1) for x = 0, 1, 2, 3. Find k, then P(X ≥ 2).

k(1 + 2 + 3 + 4) = 10k = 1      so k = 1/10
F(1) = (1 + 2)/10 = 3/10
P(X ≥ 2) = 1 − F(1) = 7/10

Two unknowns from the mean

X takes 2, 4, 6 with probabilities a, b, 0.2 and E(X) = 4.2.

a + b + 0.2 = 1           so a + b = 0.8
2a + 4b + 1.2 = 4.2       so a + 2b = 1.5
Subtract: b = 0.7, then a = 0.1
Check: 0.2 + 2.8 + 1.2 = 4.2

A linear change of a uniform variable

X is uniform on 1 to 6 and Y = 3X + 1.

E(X) = 3.5, Var(X) = 35/12
E(Y)   = 3(3.5) + 1 = 11.5
Var(Y) = 9 × 35/12  = 105/4 = 26.25

Working out n from the variance

X is uniform on 1 to n and Var(X) = 4.

(n² − 1)/12 = 4  gives  n² = 49, so n = 7
E(X) = (7 + 1)/2 = 4

Command words and what to show

Phrase in the question	What to write
Show that …	Every line of working, ending on the stated result
Find the value of k	The equation Σ p(x) = 1, then the solution
Find the cumulative distribution function	A table or list of F(x) for each x, ending at 1
Find E(X) and Var(X)	Both totals, then the subtraction for the variance
Write down	The result only, usually from a stated formula

Examiners cannot reward an answer they cannot follow, so set out the sum for E(X) and E(X²) in full. Keep fractions exact until the last line unless the question asks for a decimal.

Must-know distinctions

1. E(X²) is not (E(X))². Their difference is the variance.
2. P(X ≥ a) uses F(a − 1). P(X > a) uses F(a).
3. Adding a constant moves the mean but leaves the variance alone. Multiplying by a constant changes both, and the variance changes by a².
4. p(x) is the probability at one value. F(x) is the running total up to and including it.
5. Var(X) is on a squared scale. The standard deviation is back on the scale of X.

Quick self-test

1. State whether each variable is discrete: (a) the number of siblings a person has, (b) a person’s height.
2. p(x) = kx for x = 1, 2, 3, 4. Find k.
3. X takes values 1, 2, 3, 4 with F(1) = 0.2, F(2) = 0.5, F(3) = 0.8 and F(4) = 1. Find P(X ≥ 3) and p(3).
4. X takes the values 1, 2, 3 with probabilities 0.4, 0.4, 0.2. Find E(X) and Var(X).
5. E(X) = 3 and Var(X) = 2. Find E(5 − 2X) and Var(5 − 2X).
6. E(X) = 3 and E(X²) = 13. Find Var(X).
7. X is uniform on 1, 2, …, 20. Find E(X) and Var(X).
8. X is uniform on 1, 2, …, n with E(X) = 8. Find n and Var(X).
9. X is chosen at random from the whole numbers 3 to 8. Find E(X) and Var(X).
10. True or false: Var(2X) = 2 Var(X). Give the correct statement.
11. A fair spinner has 8 equal sectors numbered 1 to 8. Find P(X ≥ 6) and F(5).

Answers

1. (a) Discrete. (b) Not discrete, because height can take any value in an interval.
2. 10k = 1, so k = 1/10.
3. P(X ≥ 3) = 1 − F(2) = 0.5. p(3) = F(3) − F(2) = 0.3.
4. E(X) = 0.4 + 0.8 + 0.6 = 1.8. E(X²) = 0.4 + 1.6 + 1.8 = 3.8. Var(X) = 3.8 − 3.24 = 0.56.
5. E = 5 − 2(3) = −1. Var = (−2)² × 2 = 8.
6. Var(X) = 13 − 9 = 4.
7. E(X) = 21/2 = 10.5. Var(X) = 399/12 = 33.25.
8. (n + 1)/2 = 8 gives n = 15. Var(X) = (225 − 1)/12 = 56/3.
9. E(X) = 5.5. Var(X) = 35/12.
10. False. Var(2X) = 4 Var(X).
11. P(X ≥ 6) = 3/8. F(5) = 5/8.

Where marks are usually lost

• Using 1 − F(a) when the question says “at least a”.
• Not showing the equation Σ p(x) = 1 before solving for a constant.
• Dropping the squares in E(X²) and just reusing E(X).
• Subtracting E(X) instead of (E(X))² when finding the variance.
• Multiplying the variance by a instead of a² in Var(aX + b).
• Adding b to the variance in Var(aX + b).
• Applying (n² − 1)/12 to values such as 3 to 8 without shifting them first.
• Rounding E(X) early and carrying the error into the variance.
• Leaving a final answer as an unsimplified fraction when a decimal or simplified fraction is clearly expected.
• Not stating the final line in terms of the question, such as the probability or the mean asked for.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, first teaching September 2018, published by Pearson. These notes follow S1 section 5, topics 5.1 to 5.4.