Continuous Random Variables and Distributions
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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).
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.
2 P(a < X < b) = ∫ab f(x) dx
3 ∫−∞∞ f(x) dx = 1
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).
Could g(x) = kx² for 0 ≤ x ≤ 3 (and 0 otherwise) ever be a probability density function?
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.
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.
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).
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.
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.
E(aX + b) = aE(X) + b Var(aX + b) = a²Var(X)
X has p.d.f. f(x) = (3x/56)(5 − x) for 0 ≤ x ≤ 4. Find E(X) and Var(X).
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 x². Write out x·f(x) or x²·f(x) fully before you start integrating, rather than trying to do it in your head.
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.)
X has p.d.f. f(x) = −(2/9)x² + (10/9)x − 8/9 for 1 ≤ x ≤ 4. Find the mode.
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.
Lower quartile Q₁: F(Q₁) = 0.25
Upper quartile Q₃: F(Q₃) = 0.75
nth percentile Pₙ: F(Pₙ) = n/100
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.
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.
X has p.d.f. f(x) = x/8 for 0 ≤ x ≤ 4. Find the median and mode, and comment on skew.
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).
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)
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].
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).
What to Memorise
Your final-hour, formula-sheet-level reference.
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.
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.
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.
It's easy to accidentally integrate f(x) itself instead of x·f(x). Always write out the full integrand before integrating.
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.
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.
You don't need mean, median AND mode every time — stating one clear inequality (e.g. "median < mode, so negatively skewed") is sufficient justification.
- 4 · Mode, Median, Percentiles & Skewness
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