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Further Mathematics

Continuous Random Variables and Distributions

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Edexcel FS2 · Chapter 3

Continuous Distributions

The big idea: when a random variable can take literally any value in a range (like height, or waiting time), you can't ask "what's the probability it's exactly this?" — that's basically zero. Instead you use a curve, and probability becomes the area under that curve.

Summary

The whole chapter in one scroll — read this first, then dive into the detail below.

  • A continuous random variable has infinitely many possible values, so we use a probability density function, f(x), instead of a probability mass function.
  • Two rules make something a valid p.d.f: it's never negative, and the total area under it equals 1.
  • Probabilities are found by integrating f(x) between two limits — the area under the curve.
  • The cumulative distribution function, F(x), is the "running total" of probability up to a point: F(x) = P(X ≤ x).
  • Mean and variance for continuous variables use the same logic as discrete ones, just with integrals instead of sums.
  • The mode is where the graph peaks; the median is where F(x) = 0.5; comparing them tells you the skew.
  • The continuous uniform distribution U[a, b] is the "flat rectangle" special case — every value in the range is equally likely.

1 · What Even Is a Continuous Random Variable?

Setting the foundation before any formulas show up.

Think about height. If I asked "what's the probability someone is exactly 175.0000000... cm tall?" — the honest answer is basically zero, because height can take an infinite number of decimal values. There's always a "more precise" measurement hiding underneath. This is the core difference from something like rolling a die, where the outcomes are countable and separate (1, 2, 3, 4, 5, 6).

  Discrete random variable
Finite/countable values (like dice rolls). Uses a probability mass function. Probabilities found by summing. Graph looks like separate bars.
  Continuous random variable
Infinite values within a range (like height, time, mass). Uses a probability density function. Probabilities found by integrating. Graph is a smooth curve.

Because any single exact value has probability zero, we stop asking "what's the chance of exactly this value?" and instead ask "what's the chance the value falls in this range?" That's what the probability density function, f(x), is built for — it doesn't give you a probability directly, it gives you a "density" that you turn into a probability by finding an area underneath it.

The three defining rules of a p.d.f.
1   f(x) ≥ 0  for all x ∈ ℝ
2   P(a < X < b) = ∫ab f(x) dx
3   ∫−∞ f(x) dx = 1
In plain words: (1) density can never be negative — there's no such thing as "negative chance." (2) To find the probability of landing between two values, integrate the function between those limits — you're literally calculating an area. (3) Because the variable has to land somewhere, the total area under the whole curve must add up to exactly 1 (i.e. 100%).
Why this matters for exam questions

The most common opening question in this chapter is "show that f(x) is/isn't a valid p.d.f." You are checking both conditions: never negative, AND integrates to 1. Missing either one loses marks — students often check only the area and forget to comment on positivity (or vice versa).

Worked Example — Checking a valid p.d.f.

Could g(x) = kx² for 0 ≤ x ≤ 3 (and 0 otherwise) ever be a probability density function?

Step 1 — check positivity
If k > 0, then kx² ≥ 0 for all x in the range. ✓ Positivity is fine (as long as k is positive).
Step 2 — force the total area to equal 1
03 kx² dx = [kx³/3]03 = 9k
Step 3 — solve for k
9k = 1  ⇒  k = 1/9
Conclusion
So g(x) = x²/9 on [0,3] is a valid p.d.f., but only for this specific value of k.
Practice Question 1
A random variable has p.d.f. f(x) = 4x + 1 for 0 ≤ x ≤ 3, and 0 otherwise. Is this a valid probability density function? Justify your answer.
Practice Question 2
A random variable X has p.d.f. f(x) = c(4 − x) for 0 ≤ x ≤ 4, and 0 otherwise. Find the value of c.

2 · The Cumulative Distribution Function

The "running total" version of the p.d.f.

Imagine you had to find P(1 < X < 3), then P(2 < X < 5), then P(0.5 < X < 4) — integrating from scratch every single time would get tedious fast. The cumulative distribution function, F(x), solves this by pre-calculating "the probability of being less than or equal to x" for every value of x, so you can just plug numbers in afterwards, like a lookup table.

Definition
F(x) = P(X ≤ x) = ∫−∞x f(x′) dx′
Notice the lowercase/capital convention: f (p.d.f.) is lowercase, F (c.d.f.) is uppercase. Also notice we use a "dashed" variable x′ inside the integral — that's just to avoid confusing the variable we're integrating over with the value x we're integrating up to.
Building a c.d.f. in pieces

When a p.d.f. is defined in separate sections (e.g. one rule for 0 ≤ x ≤ 1, another for 1 < x ≤ 2), you integrate each section separately. Because the c.d.f. must keep increasing smoothly (it's cumulative — it can never go backwards), you carry forward the value of F from the end of the previous section as your starting point for the next one.

Worked Example — Building a c.d.f. from two pieces

X has p.d.f. f(x) = x/2 for 0 ≤ x ≤ 1, and f(x) = x³/5 for 1 < x ≤ 2. Fully specify F(x).

Before the range starts
For x < 0: F(x) = 0 (no probability accumulated yet)
First section — integrate from 0
For 0 ≤ x ≤ 1: F(x) = ∫0x x′/2 dx′ = x²/4
Second section — start from F(1), not from zero
For 1 < x ≤ 2: F(x) = F(1) + ∫1x x′³/5 dx′ = 1/4 + x⁴/20 − 1/20
After the range ends
For x > 2: F(x) = 1 (all the probability has been used up)
Sanity check you should always run

A finished c.d.f. must satisfy: (1) it equals 0 before the variable's range begins, (2) it equals 1 after the range ends, and (3) the pieces must join up smoothly — plugging the upper limit of one section into it should give the same value as plugging the lower limit into the next section. If your c.d.f. jumps or doesn't reach 1, you've made an arithmetic slip somewhere.

Practice Question 1
A continuous random variable X has c.d.f. F(x) = (x + 2)/6 for −2 ≤ x ≤ 4 (with F(x)=0 below and F(x)=1 above). Find P(X > 1).
Practice Question 2
Using the p.d.f. f(x) = 1/3 for 0 ≤ x ≤ 1, and f(x) = 8x³/45 for 1 < x ≤ 2, verify that F(2) = 1.

3 · Mean and Variance

Same ideas as discrete random variables — swap sums for integrals.

If you've done discrete distributions, this will feel very familiar. Where a discrete mean is Σxp(x), the continuous mean simply replaces the sum with an integral and the probability mass function with the density function. The logic — "multiply each value by how likely it is, then add it all up" — hasn't changed at all.

Mean (expectation)
E(X) = μ = ∫−∞ x f(x) dx
The "balance point" of the distribution — where it would balance if you cut the curve out of cardboard.
Variance
Var(X) = σ² = ∫−∞ (x − μ)² f(x) dx = ∫−∞ x² f(x) dx − μ²
In practice you almost always use the second version: Var(X) = E(X²) − [E(X)]². It's much less algebra than expanding (x − μ)² by hand.
Expectation of a function, and linear transformations
E(g(X)) = ∫−∞ g(x) f(x) dx

E(aX + b) = aE(X) + b     Var(aX + b) = a²Var(X)
These last two are your shortcut tools — if a question asks about a linear transformation of X (like "find E(3X − 2)"), you never need to re-integrate from scratch. Just transform the mean/variance you already have.
Worked Example — Finding E(X) and Var(X)

X has p.d.f. f(x) = (3x/56)(5 − x) for 0 ≤ x ≤ 4. Find E(X) and Var(X).

Step 1 — set up E(X), remembering to multiply by an extra x
E(X) = ∫04 (3x²/56)(5 − x) dx = (3/56)[5x³/3 − x⁴/4]04 = 16/7
Step 2 — find E(X²) first (multiply by x² this time)
E(X²) = ∫04 (3x³/56)(5 − x) dx = (3/56)[5x⁴/4 − x⁵/5]04 = 216/35
Step 3 — combine using the shortcut formula
Var(X) = E(X²) − [E(X)]² = 216/35 − (16/7)² = 216/35 − 256/49 ≈ 8.276
Common mistake

For E(X), students often forget to multiply f(x) by an extra x before integrating — you're integrating x·f(x), not just f(x). Similarly for E(X²), you multiply by . Write out x·f(x) or x²·f(x) fully before you start integrating, rather than trying to do it in your head.

Practice Question 1
X has p.d.f. f(x) = x/8 for 0 ≤ x ≤ 4. Find E(X).
Practice Question 2
Given that E(X) = 5 and Var(X) = 2, find E(4X − 3) and Var(4X − 3).

4 · Mode, Median, Percentiles & Skewness

Three ways to describe "the middle" — and what happens when a distribution leans one way.

A The mode

The mode is simply the x-value where f(x) is at its largest — the "peak" of the curve, or the most densely-packed value. To find it, differentiate f(x), set the derivative to 0, and solve — exactly like finding a maximum point in regular calculus. (If f(x) is a straight increasing or decreasing line rather than a curve, the mode is just found by inspection at whichever endpoint gives the largest value — no differentiation needed.)

Worked Example — Finding the mode

X has p.d.f. f(x) = −(2/9)x² + (10/9)x − 8/9 for 1 ≤ x ≤ 4. Find the mode.

Step 1 — differentiate
f′(x) = −(4/9)x + 10/9
Step 2 — set to zero and solve
−(4/9)x + 10/9 = 0 ⇒ x = 10/4 = 5/2

B The median and percentiles

The median is the value that splits the area under the curve exactly in half — 50% of the probability lies below it, 50% above. Because the c.d.f. already tells you "probability up to this point," finding the median is just a matter of solving F(m) = 0.5 for m. Quartiles and percentiles work exactly the same way, just with a different target value.

Using F(x) to find key values
Median m:  F(m) = 0.5
Lower quartile Q₁:  F(Q₁) = 0.25
Upper quartile Q₃:  F(Q₃) = 0.75
nth percentile Pₙ:  F(Pₙ) = n/100
Always check which "piece" you need first

If F(x) is defined in sections, don't blindly plug 0.5 into the first piece! Check the value of F at the boundary first (e.g. is F(1) > 0.5 or < 0.5?) to work out which section of the c.d.f. actually contains the median, then solve within that section only.

C Skewness

Skewness describes whether a distribution leans towards the small values or the large values.

  • Positive skew ("bottom heavy") — the peak (mode) is at a lower x-value, and there's a long tail stretching out to the right, towards higher values. Ordering: mode < median < mean.
  • No skew — the distribution is symmetrical (mode = median = mean).
  • Negative skew ("top heavy") — the peak is at a higher x-value, with a long tail stretching out to the left. Ordering: mean < median < mode.
The exam shortcut

You almost never need all three measures — comparing just one pair (e.g. median vs. mode) is enough to justify the direction of skew. If median < mode, that's negative skew; if median > mode, that's positive skew. Just remember the tail points in the direction opposite to where the mode sits.

Worked Example — Median, mode, and skew together

X has p.d.f. f(x) = x/8 for 0 ≤ x ≤ 4. Find the median and mode, and comment on skew.

Find F(x) by inspection (integrating x/8)
F(x) = x²/16  for 0 ≤ x ≤ 4
Solve F(m) = 0.5 for the median
m²/16 = 0.5 ⇒ m² = 8 ⇒ m = 2√2 ≈ 2.83
Mode — since f(x) = x/8 is an increasing straight line on [0,4], the max is at the right-hand endpoint
Mode = 4
Compare
Since 2√2 < 4, median < mode ⇒ the distribution is negatively skewed.
Practice Question 1
X has c.d.f. F(x) = x³/64 for 0 ≤ x ≤ 4. Find the median of X.
Practice Question 2
For a distribution, the mean is 12, the median is 10, and the mode is 8. What type of skew does this distribution have?

5 · The Continuous Uniform Distribution

The "flat rectangle" special case — every outcome equally likely.

This is the continuous equivalent of rolling a fair die — except instead of a few discrete outcomes, any value between two limits, a and b, is equally likely. Picture a perfectly flat, horizontal line between a and b — that flat line is the p.d.f. Since the area under it must equal 1, and the "rectangle" has width (b − a), the height must be 1/(b − a).

X ~ U[a, b]
f(x) = 1/(b − a)  for a ≤ x ≤ b,  0 otherwise

E(X) = (a + b)/2
Var(X) = (b − a)² / 12

F(x) = 0  (x < a)    F(x) = (x − a)/(b − a)  (a ≤ x ≤ b)    F(x) = 1  (x > b)
E(X) makes intuitive sense — the mean of a perfectly flat, symmetric shape is just the midpoint of the range. You could derive every one of these results yourself using the integration methods from earlier in the chapter — they're not "new rules to memorise blindly," they're the general formulas applied to a rectangle.
Real-world contexts examiners love

Waiting times: if buses arrive every 10 minutes but you don't know when the next one comes, your wait time ~ U[0, 10].
Rounding errors: if a measurement is rounded to the nearest cm, the rounding error R ~ U[−0.5, 0.5].

Worked Example — Using U[a,b] formulas directly

A bus arrives randomly and uniformly between 0 and 10 minutes after you arrive at the stop, i.e. X ~ U[0,10]. Find E(X), Var(X), and P(X > 7).

Mean — just the midpoint
E(X) = (0 + 10)/2 = 5 minutes
Variance — plug into the formula
Var(X) = (10 − 0)²/12 = 100/12 = 25/3 ≈ 8.33
Probability — use F(x) = (x − a)/(b − a)
P(X > 7) = 1 − F(7) = 1 − (7 − 0)/10 = 1 − 0.7 = 0.3
Practice Question 1
A rounding error R ~ U[−0.5, 0.5]. Find Var(R) and P(R > 0.2).
Practice Question 2
X ~ U[2, 12]. Find E(X) and the median of X. What does this tell you about the skew of a uniform distribution?

What to Memorise

Your final-hour, formula-sheet-level reference.

p.d.f. conditions
f(x) ≥ 0 and ∫f(x)dx = 1 over the full range
Probability from a p.d.f.
P(a < X < b) = ∫ab f(x) dx
Definition of c.d.f.
F(x) = P(X ≤ x) = ∫−∞x f(x′) dx′
Relationship between f and F
F(x) is the integral of f(x); f(x) is the derivative of F(x)
Mean
E(X) = ∫x f(x) dx
Variance (fast version)
Var(X) = E(X²) − [E(X)]²
Linear transformations
E(aX+b) = aE(X)+b,  Var(aX+b) = a²Var(X)
Median / quartiles / percentiles
Solve F(x) = 0.5 / 0.25 / 0.75 / n/100
Mode
Where f(x) is greatest — differentiate f(x), set to 0, solve (or inspect endpoints if linear)
Skew rule
Positive: mode < median < mean.  Negative: mean < median < mode
Uniform distribution X~U[a,b]
f(x) = 1/(b−a),  E(X) = (a+b)/2,  Var(X) = (b−a)²/12
Uniform c.d.f.
F(x) = (x−a)/(b−a) for a ≤ x ≤ b

Concepts Checklist

Tick these off honestly — if you hesitate on any, go back and re-read that section.

Exam Tips

Where marks are actually lost — and how to protect yours.

"Show it's a valid p.d.f." needs BOTH conditions

Examiners want to see you explicitly comment on positivity and show the area equals 1. A numerical answer alone without the positivity comment often drops a mark.

"Fully specify" or "in full" means ALL pieces

If a question says "find F(x) for all x ∈ ℝ," you must write the 0-before and 1-after pieces too, not just the middle section(s). Missing the boundary pieces is one of the most common lost marks in this topic.

Don't forget the extra x (or x²) in E(X) / E(X²)

It's easy to accidentally integrate f(x) itself instead of x·f(x). Always write out the full integrand before integrating.

Piecewise medians — check which section first

Before solving F(x) = 0.5, evaluate F at the section boundary to confirm which piece actually contains the median. Solving in the wrong piece gives a value outside the valid range — a red flag that you should double check your work.

Continuity ⇒ strict and non-strict inequalities behave the same

Because X is continuous, P(X < a) = P(X ≤ a) — there's no "gap" to worry about like there is with discrete variables. This is genuinely useful: it means you can substitute directly into F(x) without adjusting for strict/non-strict boundaries.

One comparison is enough for skew

You don't need mean, median AND mode every time — stating one clear inequality (e.g. "median < mode, so negatively skewed") is sufficient justification.

Continuous Distributions · Edexcel FS2, Chapter 3 · Revision guide built for offline study
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