When a variable measures rather than counts, the whole machinery changes. Individual values stop carrying probability, intervals take over, and the cumulative distribution function becomes the object you actually compute with. This lesson makes that transition carefully, starting from data you can see.
Commit to a prediction about a single instant.
A bus arrives at a random time $T$ in the next 10 minutes. Without looking ahead answer both: what is $P(T = 4)$ exactly, and if that probability is 0 for every instant, how can the bus arrive at all?
Estimate interval probabilities from a histogram.
From data · intervals are what you can measure
We just saw how a discrete distribution is estimated by dividing each frequency by the number of trials. That raises a question: a continuous measurement almost never repeats, so counting how often each value occurred would give a table of ones, and what replaces it? This card answers it → group the data into class intervals and estimate the probability of an interval by its relative frequency.
For a continuous variable the useful question is never "how likely is this exact value" but "how likely is a value in this range". A histogram answers exactly that question from data.
Worked through. The reaction times of 200 students, in seconds, grouped into five classes:
| class | $0.1 \le t < 0.2$ | $0.2 \le t < 0.3$ | $0.3 \le t < 0.4$ | $0.4 \le t < 0.5$ | $0.5 \le t < 0.6$ |
|---|---|---|---|---|---|
| frequency | 20 | 50 | 70 | 40 | 20 |
| rel. freq. | 0.10 | 0.25 | 0.35 | 0.20 | 0.10 |
So $P(0.2 \leq T < 0.4) \approx \dfrac{50 + 70}{200} = \dfrac{120}{200} = 0.6$, adding the two class frequencies before dividing.
The five relative frequencies sum to $0.10 + 0.25 + 0.35 + 0.20 + 0.10 = 1.00$, for the same reason as in the discrete case: every observation falls in exactly one class.
For continuous data, estimate $P(a \leq X < b)$ by the relative frequency of that range in a histogram.; A range cutting through a class needs an explicit even-spread assumption, and you must state it.
Pause, copy the histogram estimate of an interval probability, and the note that a range cutting through a class requires a stated even-spread assumption, into your book.
Quick check: Using the reaction-time table above, estimate $P(T \geq 0.4)$.
Establish why a single value carries no probability.
We just saw that continuous data are handled by intervals. That raises a question: what happens to the probability of a single value as the interval around it shrinks? This card answers it → it shrinks to zero, which makes every point probability 0 and makes strict and non-strict inequalities interchangeable.
Narrow the class width and watch the reaction-time estimate for a fixed target value:
| interval about $0.35$ | width $0.1$ | width $0.01$ | width $0.001$ | $\to 0$ |
|---|---|---|---|---|
| estimated probability | $0.35$ | $\approx 0.035$ | $\approx 0.0035$ | $\to 0$ |
The probability is roughly proportional to the width of the interval, so as the width goes to zero so does the probability. In the limit a single point carries none at all:
The consequence you will use constantly. Because endpoints contribute nothing, including or excluding them changes nothing:
This is emphatically not true for a discrete variable, where $P(2 \leq X \leq 5)$ and $P(2 < X < 5)$ generally differ by $p(2) + p(5)$.
True or false: If $X$ is a continuous random variable, then $P(X = 3) = 0$ means the value 3 can never occur.
Define the CDF and list the properties it must have.
The CDF · the function you actually compute with
We just saw that only intervals carry probability for a continuous variable. That raises a question: there are infinitely many intervals, so we cannot tabulate them, and what single function encodes them all? This card answers it → the cumulative distribution function, from which every interval probability follows by one subtraction.
$F(x)$ accumulates all the probability lying at or below $x$. Every interval probability comes straight out of it:
And because point probabilities vanish, the first of those is also $P(a \leq X \leq b)$, $P(a < X < b)$ and $P(a \leq X < b)$. One subtraction covers all four.
The properties every CDF has:
$F(x) = P(X \leq x)$; $P(a < X \leq b) = F(b) - F(a)$; $P(X > a) = 1 - F(a)$.; $F$ is non-decreasing, runs from 0 to 1, and for a continuous variable has no jumps.
Pause, copy the CDF definition, the subtraction rule $F(b) - F(a)$, the complement rule $1 - F(a)$, and the four properties of $F$, into your book.
Fill the blanks: A continuous variable has $F(1) = 0.2$, $F(2) = 0.5$ and $F(4) = 0.9$. Give each answer as a decimal.
$P(1 < X \leq 4) =$
$P(X > 2) =$
$P(2 \leq X \leq 4) =$
Work three examples, then read the traps.
Worked examples · 3 in a row, reveal as you go
A continuous random variable has $F(x) = 0$ for $x < 0$, $F(x) = \dfrac{x^2}{9}$ for $0 \leq x \leq 3$, and $F(x) = 1$ for $x > 3$. Verify this is a valid CDF.
For the same $F$, find $P(1 < X \leq 2)$, $P(X > 2)$ and $P(X = 2)$.
Describe the graph of the CDF above on the interval $0 \leq x \leq 3$, naming the features an examiner expects to see marked.
Common errors · traps that cost marks
Odd one out: Three of these are properties every cumulative distribution function has. Which one is not?
Work the quick-fire set, then revisit your prediction.
Quick-fire practice · estimate, subtract, check
Using the reaction-time table, estimate $P(T < 0.3)$ and $P(0.3 \leq T < 0.5)$.
$F(2) = 0.3$ and $F(5) = 0.85$ for a continuous variable. Find $P(2 < X \leq 5)$, $P(X > 5)$ and $P(X = 5)$.
Explain why $P(3 < X < 5) = P(3 \leq X \leq 5)$ holds for a continuous variable but generally fails for a discrete one.
Someone proposes $F(x) = \dfrac{x}{4}$ for $0 \leq x \leq 6$ as a CDF. Give one reason it cannot be one.
A CDF graph has a vertical step of height $0.2$ at $x = 4$. What does that tell you about the variable?
Match each expression to what it computes:
Earlier you were asked for $P(T = 4)$ and how the bus can arrive if every instant has probability zero. $P(T = 4) = 0$, and so does every other single instant. The resolution is that probability zero is not impossibility. Probability attaches to intervals, not to points: $P(3.9 < T < 4.1)$ is a perfectly ordinary positive number, and it shrinks toward 0 only as the window closes. The bus arrives in an interval, always; it just never arrives at a point you could have named in advance with positive probability.
The CDF answers every interval question by subtraction, but it does not show you the shape of the distribution — where values bunch up and where they thin out. In the next lesson you differentiate $F$ to recover that shape. The result is the probability density function, and the histogram from card 02 turns out to have been a picture of it all along.
Apply the lesson methods, then compare each response with its comprehensive answer.
Use the visible short-answer practice below during this lesson. After completing the focus area, use Checkpoint 3 for the checkpoint question bank.
Q1. The masses, in grams, of 500 apples are grouped as follows.
| mass | $100 \le m < 120$ | $120 \le m < 140$ | $140 \le m < 160$ | $160 \le m < 180$ |
|---|---|---|---|---|
| frequency | 60 | 175 | 190 | 75 |
(a) Estimate $P(120 \leq M < 160)$. (b) Estimate $P(M \geq 150)$, stating clearly any assumption you make. (c) Explain why $P(M = 150)$ is 0 even though several apples in the data set may weigh close to 150 g. (3 marks)
Q2. A continuous random variable $X$ has cumulative distribution function $F(x) = 0$ for $x < 1$, $F(x) = \dfrac{x - 1}{4}$ for $1 \leq x \leq 5$, and $F(x) = 1$ for $x > 5$.
(a) Verify that $F$ is a valid CDF, checking every property. (b) Find $P(2 < X \leq 4)$ and $P(X > 3)$. (c) Sketch $F$ for $0 \leq x \leq 6$ and describe the shape of the graph on $1 \leq x \leq 5$ in one sentence. (3 marks)
Q3. A student is shown a graph of a function $G$ that rises from 0 to 1, is non-decreasing throughout, and contains a single vertical jump of height $0.3$ at $x = 2$. They conclude: "This cannot be a cumulative distribution function, because a CDF has to be continuous."
(a) Explain why the student's conclusion is wrong. (b) State precisely what the jump tells you about the underlying random variable. (c) Describe a realistic context in which a random variable would produce exactly this kind of CDF, and justify why the jump appears where it does. (3 marks)
Drill 1: $P(T < 0.3) \approx 0.10 + 0.25 = 0.35$. $P(0.3 \leq T < 0.5) \approx 0.35 + 0.20 = 0.55$.
Drill 2: $P(2 < X \leq 5) = 0.85 - 0.30 = 0.55$. $P(X > 5) = 1 - 0.85 = 0.15$. $P(X = 5) = 0$, because $X$ is continuous.
Drill 3: For continuous $X$, $P(X = 3) = P(X = 5) = 0$, so adding or removing the endpoints changes nothing and the two probabilities are equal. For discrete $X$ the endpoints usually carry positive probability, and the two differ by exactly $p(3) + p(5)$.
Drill 4: At $x = 6$ it gives $F(6) = \tfrac{6}{4} = 1.5 > 1$, and a CDF is a probability so it can never exceed 1. (Equivalently, $F$ reaches 1 at $x = 4$ and would have to stay there.)
Drill 5: A jump of height $0.2$ at $x = 4$ means $P(X = 4) = 0.2$, a single value carrying positive probability. So the variable is not continuous at that point: it is either discrete or a mixed variable with a discrete component at 4.
Q1 (3 marks): (a) $P(120 \leq M < 160) \approx \dfrac{175 + 190}{500} = \dfrac{365}{500} = 0.73$ [1]. (b) The value 150 splits the class $140 \leq m < 160$ in half. Assuming the 190 apples in that class are spread evenly across it, about 95 lie at or above 150. So $P(M \geq 150) \approx \dfrac{95 + 75}{500} = \dfrac{170}{500} = 0.34$. The assumption must be stated to earn full marks [1]. (c) Mass is a continuous quantity, so $P(M = 150)$ means the probability of a mass of exactly 150.000... g, with infinite precision. The probability of an interval shrinks in proportion to its width, so a single point has probability 0. Apples recorded as "150 g" are really apples in a small interval around 150, which does have positive probability [1].
Q2 (3 marks): (a) $F(1) = \tfrac{0}{4} = 0$ and $F(5) = \tfrac{4}{4} = 1$, so the pieces join without a jump; $F' = \tfrac{1}{4} > 0$ on $[1, 5]$, so $F$ is non-decreasing; $0 \leq F \leq 1$ everywhere; and $F \to 0$ as $x \to -\infty$, $F \to 1$ as $x \to +\infty$. Valid CDF ✓ [1]. (b) $P(2 < X \leq 4) = F(4) - F(2) = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}$. $P(X > 3) = 1 - F(3) = 1 - \tfrac{2}{4} = \tfrac{1}{2}$ [1]. (c) The graph is flat at 0 up to $x = 1$, rises as a straight line of gradient $\tfrac{1}{4}$ from $(1, 0)$ to $(5, 1)$, then is flat at 1 beyond $x = 5$. A constant gradient means the probability is spread evenly across $[1, 5]$ [1].
Q3 (3 marks): (a) The student has confused a property of continuous random variables with a property of CDFs in general. Every CDF must be non-decreasing and run from 0 to 1, but it need not be continuous: a discrete variable's CDF is a staircase, and jumps are exactly how it encodes probability. $G$ satisfies every genuine requirement, so it is a perfectly good CDF [1]. (b) A jump of height $0.3$ at $x = 2$ means $P(X = 2) = 0.3$. Since $G$ also rises continuously elsewhere, $X$ is a mixed variable: it puts a lump of probability $0.3$ on the single value 2 and spreads the remaining $0.7$ continuously [1]. (c) One good context: the time a customer waits on hold, where $0.3$ of callers are answered immediately at exactly the 2-second automated pickup and the rest wait a genuinely variable time. The jump sits at 2 because that is the one waiting time produced by a fixed mechanism rather than by chance, so a whole block of probability piles up on that single value. Rainfall in a day is another, with an atom at exactly 0 for dry days [1].
Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.
Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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