What is a continuous random variable in the context of Cambridge International A Levels Mathematics?

In Cambridge International A Levels Mathematics, a continuous random variable is a type of random variable that can take any value within a given range. Unlike discrete random variables, which have specific, countable outcomes, continuous random variables are associated with measurements and can assume an infinite number of possible values. Examples include variables like height, weight, and time.

How is the probability density function (PDF) used with continuous random variables?

The probability density function (PDF) is a fundamental concept in understanding continuous random variables. It describes the likelihood of a random variable taking on a specific value. For continuous random variables, the PDF is used to calculate probabilities over an interval, as the probability of the variable taking on any exact value is zero. The area under the PDF curve over a specific interval represents the probability that the random variable falls within that interval.

What is the difference between a probability density function (PDF) and a cumulative distribution function (CDF) for continuous random variables?

The probability density function (PDF) and the cumulative distribution function (CDF) are both used to describe continuous random variables, but they serve different purposes. The PDF provides the relative likelihood of the random variable taking on a specific value, while the CDF gives the probability that the random variable is less than or equal to a certain value. The CDF is obtained by integrating the PDF over the range of interest, and it is a non-decreasing function that ranges from 0 to 1.

Why is the concept of expected value important for continuous random variables?

The expected value of a continuous random variable is a crucial concept in Probability and Statistics 2, as it provides a measure of the central tendency or the 'average' outcome of the random variable. It is calculated by integrating the product of the variable and its probability density function over the entire range of possible values. The expected value helps in making predictions and informed decisions based on the random variable's behavior.

How do you calculate the variance of a continuous random variable?

The variance of a continuous random variable is a measure of how much the values of the variable spread out from the expected value. It is calculated by integrating the squared difference between the variable and its expected value, multiplied by the probability density function, over the entire range of possible values. The variance provides insight into the variability and dispersion of the random variable's outcomes, which is essential for understanding its behavior in Probability and Statistics 2.

Why is "Treating PDF as probability" a common mistake in Continuous random variables?

Why it happens: Confusion with discrete case. How to avoid it: For continuous: P(X = x) = 0. PDF gives DENSITY, not probability. Integrate to get probability. Source: 9709 Examiner Reports 2022-2024

Why is "Forgetting to verify $\int f = 1$" a common mistake in Continuous random variables?

Why it happens: Skipping check. How to avoid it: If question provides f with unknown constant, find it using f = 1. Source: 9709 Examiner Reports 2022-2024

Cambridge International A Levels Mathematics (9709)

Continuous random variables

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Short Notes - Continuous random variables

Detailed Study Notes

Detailed notes on Probability and Statistics 2 - Paper 6 for Cambridge International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Continuous Random Variables Study Notes — Cambridge International A Level Mathematics 9709 S2 (2024-2026 syllabus)

PDFs and CDFs. Probabilities and expectations via integration.

At a glance

PDF $f(x) \geq 0$ with $\int f = 1$ .
Probability = area under PDF.
$P(X = x) = 0$ for continuous.
CDF $F(x) = \int_{-\infty}^x f$ .
$E(X) = \int x f \, dx$ .

What you’ll learn

Mapped to the Cambridge International A Level 9709 syllabus (2024-2026).

S2.3.1 — Use PDFs.
S2.3.2 — Find CDF.
S2.3.3 — Compute $E(X)$ and $\text{Var}(X)$ .

Probability density functions

Area = probability.

PDF. $f(x) \geq 0$ for all $x$ , with $\int_{-\infty}^\infty f(x) dx = 1$ .

Probability. $P(a < X < b) = \int_a^b f(x) dx$ .

Note. For continuous RVs, $P(X = c) = 0$ at any single point.

Finding constants. If PDF given as $kg(x)$ , use $\int kg(x) dx = 1$ to find $k$ .

Example. $f(x) = kx$ for $0 \leq x \leq 4$ .

$\int_0^4 kx \, dx = k \cdot 8 = 1 \Rightarrow k = \frac{1}{8}$ .
$P(X > 2) = \int_2^4 \frac{x}{8} dx = \frac{1}{8} \cdot 6 = \frac{3}{4}$ .

Cambridge tip. Always verify $\int f = 1$ first.

For a continuous variable the density itself is not a probability — only areas are. The total area must be 1, which is what fixes any unknown constant in f.

The condition ∫f = 1 is what determines k. Here the region is a triangle, so ½ × base × height gives the same equation as the integral — a quick check worth doing.

PDF non-negative, integrates to 1.
Probability = area.
$P(X = x) = 0$ for continuous.

See the full worked example for continuous random variables →

CDF

$F(x) = P(X \leq x)$ .

Definition. $F(x) = P(X \leq x) = \int_{-\infty}^x f(t) dt$ .

Properties.

Non-decreasing.
$F(-\infty) = 0$ , $F(\infty) = 1$ .
$f(x) = F'(x)$ (PDF is derivative of CDF).

Probabilities from CDF. $P(a < X < b) = F(b) - F(a)$ .

Piecewise CDF. Usually has three pieces:

$F(x) = 0$ below support.
$F(x) = \int$ within support.
$F(x) = 1$ above support.

Example. $f(x) = \frac{x}{8}$ for $0 \leq x \leq 4$ . CDF:

$x < 0$ : $F(x) = 0$ .
$0 \leq x \leq 4$ : $F(x) = \int_0^x \frac{t}{8} dt = \frac{x^2}{16}$ .
$x > 4$ : $F(x) = 1$ .

Cambridge tip. Always write CDF as piecewise function.

F(x) accumulates the area under f up to x, so it starts at 0, ends at 1 and never decreases. Probabilities above a value are read as 1 − F.

F and f are two views of the same distribution: F accumulates area, f measures how fast that area is growing. Differentiating F is often the quickest route to f.

$F(x) = P(X \leq x)$ .
$F$ from 0 to 1.
$f = F'$ .

See the full worked example for continuous random variables →

Expectation and variance

Integrate $x f$ and $x^2 f$ .

Expectation. $E(X) = \int x f(x) \, dx$ over support.

$E(X^2)$ . $\int x^2 f(x) \, dx$ .

Variance. $\text{Var}(X) = E(X^2) - [E(X)]^2$ .

Linear functions. Same formulae as discrete case:

$E(aX + b) = aE(X) + b$ .
$\text{Var}(aX + b) = a^2 \text{Var}(X)$ .

Example. $f(x) = \frac{x}{8}$ on $[0, 4]$ .

$E(X) = \int_0^4 x \cdot \frac{x}{8} dx = \frac{1}{8} \cdot \frac{64}{3} = \frac{8}{3}$ .
$E(X^2) = \int_0^4 x^2 \cdot \frac{x}{8} dx = \frac{1}{8} \cdot 64 = 8$ .
$\text{Var}(X) = 8 - \frac{64}{9} = \frac{8}{9}$ .

Cambridge tip. Show all integration working.

The continuous versions of E(X) and E(X²) replace the sums of the discrete case with integrals. Var(X) = E(X²) − [E(X)]² is unchanged.

$E(X) = \int x f \, dx$ .
Variance via $E(X^2) - [E(X)]^2$ .
Same linear-transform rules.

See the full worked example for continuous random variables →

Quick recap

PDF non-negative, integrates to 1.
$P(X = x) = 0$ for continuous.
CDF from 0 to 1.
$E$ and $\text{Var}$ via integration.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

PDF requirements.
Probability from PDF integral.
CDF piecewise.
$E(X), E(X^2), \text{Var}(X)$ formulas.

How it’s examined

Continuous RVs appear every S2 — typically 10-15 marks. Most-tested: find constant + probability (7 marks), $E$ , $\text{Var}$ (8 marks), CDF (6 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/62 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Continuous random variables

Step-by-step solutions to past-paper-style questions on continuous random variables, written exactly the way a tutor would explain them at the board.

1Probability from PDF (7 marks)
Extended• Adapted from 9709/62 May/Jun 2024• PDF
Question
PDF: $f(x) = kx$ for $0 \leq x \leq 4$ , zero elsewhere. Find $k$ and $P(X > 2)$ . (7 marks)
Step-by-step solution
1. Step 1
  Total probability = 1.
  $\int_0^4 kx \, dx = 1 \Rightarrow k \cdot \dfrac{16}{2} = 1 \Rightarrow k = \dfrac{1}{8}$
2. Step 2
  Compute $P(X > 2)$ .
  $P(X > 2) = \int_2^4 \dfrac{x}{8} \, dx = \dfrac{1}{8} \cdot \dfrac{16 - 4}{2} = \dfrac{12}{16} = \dfrac{3}{4}$
Answer
$k = \frac{1}{8}$ ; $P(X > 2) = \frac{3}{4}$ .
2 $E(X)$ and $\text{Var}(X)$ from PDF (8 marks)
Extended• expectation
Question
PDF as above. Find $E(X)$ and $\text{Var}(X)$ . (8 marks)
Step-by-step solution
1. Step 1
  $E(X) = \int x f(x) \, dx$ .
  $E(X) = \int_0^4 x \cdot \dfrac{x}{8} \, dx = \dfrac{1}{8} \int_0^4 x^2 \, dx = \dfrac{1}{8} \cdot \dfrac{64}{3} = \dfrac{8}{3}$
2. Step 2
  $E(X^2) = \int x^2 f(x) \, dx$ .
  $E(X^2) = \int_0^4 x^2 \cdot \dfrac{x}{8} \, dx = \dfrac{1}{8} \int_0^4 x^3 \, dx = \dfrac{1}{8} \cdot 64 = 8$
3. Step 3
  $\text{Var}(X) = E(X^2) - [E(X)]^2$ .
  $\text{Var}(X) = 8 - \dfrac{64}{9} = \dfrac{72 - 64}{9} = \dfrac{8}{9}$
Answer
$E(X) = \frac{8}{3}$ ; $\text{Var}(X) = \frac{8}{9}$ .
3CDF from PDF (6 marks)
Extended• CDF
Question
PDF $f(x) = \frac{x}{8}$ for $0 \leq x \leq 4$ . Find the CDF $F(x)$ . (6 marks)
Step-by-step solution
1. Step 1
  For $0 \leq x \leq 4$ :
  $F(x) = \int_0^x \dfrac{t}{8} \, dt = \dfrac{x^2}{16}$
2. Step 2
  Piecewise.
  $F(x) = \begin{cases} 0, & x < 0 \\ \frac{x^2}{16}, & 0 \leq x \leq 4 \\ 1, & x > 4 \end{cases}$
Answer
$F(x) = 0$ for $x < 0$ ; $\frac{x^2}{16}$ for $0 \leq x \leq 4$ ; $1$ for $x > 4$ .

Key Formulae — Continuous random variables

The formulae you need to memorise for continuous random variables on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

PDF total probability
$\displaystyle\int_{-\infty}^{\infty} f(x) \, dx = 1$
When to use
Must hold for $f$ to be a valid PDF.
Probability from PDF
$P(a < X < b) = \displaystyle\int_a^b f(x) \, dx$
When to use
Probability that $X$ falls in interval.
CDF
$F(x) = P(X \leq x) = \displaystyle\int_{-\infty}^x f(t) \, dt$
When to use
Cumulative distribution function. Reverse: $f(x) = F'(x)$ .
Continuous expectation
$E(X) = \displaystyle\int x f(x) \, dx; \quad E(X^2) = \int x^2 f(x) \, dx$
When to use
Mean of continuous RV. Variance: $\text{Var}(X) = E(X^2) - [E(X)]^2$ .

Key Definitions and Keywords — Continuous random variables

Definitions to memorise and the exact keywords mark schemes credit for continuous random variables answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9709 sitting.

Continuous random variable
Examiner keyword
RV taking any value in an interval. Probability described by PDF, not point probabilities.
Probability density function (PDF)
Examiner keyword
Function $f(x) \geq 0$ with $\int f = 1$ . $P(a < X < b) = \int_a^b f$ .
Cumulative distribution function (CDF)
Examiner keyword
$F(x) = P(X \leq x)$ . Non-decreasing, from 0 to 1.

Common Mistakes and Misconceptions — Continuous random variables

The traps other students keep falling into on continuous random variables questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Treating PDF as probability
9709 Examiner Reports 2022-2024
Why it happens
Confusion with discrete case.
How to avoid it
For continuous: $P(X = x) = 0$ . PDF gives DENSITY, not probability. Integrate to get probability.
✕Forgetting to verify $\int f = 1$
9709 Examiner Reports 2022-2024
Why it happens
Skipping check.
How to avoid it
If question provides $f$ with unknown constant, find it using $\int f = 1$ .

Continuous random variables — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge International A Levels Mathematics (9709)

Continuous random variables

0% Complete

3 tasks remaining

Short Notes - Continuous random variables

Detailed Study Notes

Detailed notes on Probability and Statistics 2 - Paper 6 for Cambridge International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Continuous Random Variables Study Notes — Cambridge International A Level Mathematics 9709 S2 (2024-2026 syllabus)

PDFs and CDFs. Probabilities and expectations via integration.

At a glance

PDF $f(x) \geq 0$ with $\int f = 1$ .
Probability = area under PDF.
$P(X = x) = 0$ for continuous.
CDF $F(x) = \int_{-\infty}^x f$ .
$E(X) = \int x f \, dx$ .

What you’ll learn

Mapped to the Cambridge International A Level 9709 syllabus (2024-2026).

S2.3.1 — Use PDFs.
S2.3.2 — Find CDF.
S2.3.3 — Compute $E(X)$ and $\text{Var}(X)$ .

Probability density functions

Area = probability.

PDF. $f(x) \geq 0$ for all $x$ , with $\int_{-\infty}^\infty f(x) dx = 1$ .

Probability. $P(a < X < b) = \int_a^b f(x) dx$ .

Note. For continuous RVs, $P(X = c) = 0$ at any single point.

Finding constants. If PDF given as $kg(x)$ , use $\int kg(x) dx = 1$ to find $k$ .

Example. $f(x) = kx$ for $0 \leq x \leq 4$ .

$\int_0^4 kx \, dx = k \cdot 8 = 1 \Rightarrow k = \frac{1}{8}$ .
$P(X > 2) = \int_2^4 \frac{x}{8} dx = \frac{1}{8} \cdot 6 = \frac{3}{4}$ .

Cambridge tip. Always verify $\int f = 1$ first.

For a continuous variable the density itself is not a probability — only areas are. The total area must be 1, which is what fixes any unknown constant in f.

The condition ∫f = 1 is what determines k. Here the region is a triangle, so ½ × base × height gives the same equation as the integral — a quick check worth doing.

PDF non-negative, integrates to 1.
Probability = area.
$P(X = x) = 0$ for continuous.

See the full worked example for continuous random variables →

CDF

$F(x) = P(X \leq x)$ .

Definition. $F(x) = P(X \leq x) = \int_{-\infty}^x f(t) dt$ .

Properties.

Non-decreasing.
$F(-\infty) = 0$ , $F(\infty) = 1$ .
$f(x) = F'(x)$ (PDF is derivative of CDF).

Probabilities from CDF. $P(a < X < b) = F(b) - F(a)$ .

Piecewise CDF. Usually has three pieces:

$F(x) = 0$ below support.
$F(x) = \int$ within support.
$F(x) = 1$ above support.

Example. $f(x) = \frac{x}{8}$ for $0 \leq x \leq 4$ . CDF:

$x < 0$ : $F(x) = 0$ .
$0 \leq x \leq 4$ : $F(x) = \int_0^x \frac{t}{8} dt = \frac{x^2}{16}$ .
$x > 4$ : $F(x) = 1$ .

Cambridge tip. Always write CDF as piecewise function.

F(x) accumulates the area under f up to x, so it starts at 0, ends at 1 and never decreases. Probabilities above a value are read as 1 − F.

F and f are two views of the same distribution: F accumulates area, f measures how fast that area is growing. Differentiating F is often the quickest route to f.

$F(x) = P(X \leq x)$ .
$F$ from 0 to 1.
$f = F'$ .

See the full worked example for continuous random variables →

Expectation and variance

Integrate $x f$ and $x^2 f$ .

Expectation. $E(X) = \int x f(x) \, dx$ over support.

$E(X^2)$ . $\int x^2 f(x) \, dx$ .

Variance. $\text{Var}(X) = E(X^2) - [E(X)]^2$ .

Linear functions. Same formulae as discrete case:

$E(aX + b) = aE(X) + b$ .
$\text{Var}(aX + b) = a^2 \text{Var}(X)$ .

Example. $f(x) = \frac{x}{8}$ on $[0, 4]$ .

$E(X) = \int_0^4 x \cdot \frac{x}{8} dx = \frac{1}{8} \cdot \frac{64}{3} = \frac{8}{3}$ .
$E(X^2) = \int_0^4 x^2 \cdot \frac{x}{8} dx = \frac{1}{8} \cdot 64 = 8$ .
$\text{Var}(X) = 8 - \frac{64}{9} = \frac{8}{9}$ .

Cambridge tip. Show all integration working.

The continuous versions of E(X) and E(X²) replace the sums of the discrete case with integrals. Var(X) = E(X²) − [E(X)]² is unchanged.

$E(X) = \int x f \, dx$ .
Variance via $E(X^2) - [E(X)]^2$ .
Same linear-transform rules.

See the full worked example for continuous random variables →

Quick recap

PDF non-negative, integrates to 1.
$P(X = x) = 0$ for continuous.
CDF from 0 to 1.
$E$ and $\text{Var}$ via integration.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

PDF requirements.
Probability from PDF integral.
CDF piecewise.
$E(X), E(X^2), \text{Var}(X)$ formulas.

How it’s examined

Continuous RVs appear every S2 — typically 10-15 marks. Most-tested: find constant + probability (7 marks), $E$ , $\text{Var}$ (8 marks), CDF (6 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/62 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Continuous random variables

Step-by-step solutions to past-paper-style questions on continuous random variables, written exactly the way a tutor would explain them at the board.

1Probability from PDF (7 marks)
Extended• Adapted from 9709/62 May/Jun 2024• PDF
Question
PDF: $f(x) = kx$ for $0 \leq x \leq 4$ , zero elsewhere. Find $k$ and $P(X > 2)$ . (7 marks)
Step-by-step solution
1. Step 1
  Total probability = 1.
  $\int_0^4 kx \, dx = 1 \Rightarrow k \cdot \dfrac{16}{2} = 1 \Rightarrow k = \dfrac{1}{8}$
2. Step 2
  Compute $P(X > 2)$ .
  $P(X > 2) = \int_2^4 \dfrac{x}{8} \, dx = \dfrac{1}{8} \cdot \dfrac{16 - 4}{2} = \dfrac{12}{16} = \dfrac{3}{4}$
Answer
$k = \frac{1}{8}$ ; $P(X > 2) = \frac{3}{4}$ .
2 $E(X)$ and $\text{Var}(X)$ from PDF (8 marks)
Extended• expectation
Question
PDF as above. Find $E(X)$ and $\text{Var}(X)$ . (8 marks)
Step-by-step solution
1. Step 1
  $E(X) = \int x f(x) \, dx$ .
  $E(X) = \int_0^4 x \cdot \dfrac{x}{8} \, dx = \dfrac{1}{8} \int_0^4 x^2 \, dx = \dfrac{1}{8} \cdot \dfrac{64}{3} = \dfrac{8}{3}$
2. Step 2
  $E(X^2) = \int x^2 f(x) \, dx$ .
  $E(X^2) = \int_0^4 x^2 \cdot \dfrac{x}{8} \, dx = \dfrac{1}{8} \int_0^4 x^3 \, dx = \dfrac{1}{8} \cdot 64 = 8$
3. Step 3
  $\text{Var}(X) = E(X^2) - [E(X)]^2$ .
  $\text{Var}(X) = 8 - \dfrac{64}{9} = \dfrac{72 - 64}{9} = \dfrac{8}{9}$
Answer
$E(X) = \frac{8}{3}$ ; $\text{Var}(X) = \frac{8}{9}$ .
3CDF from PDF (6 marks)
Extended• CDF
Question
PDF $f(x) = \frac{x}{8}$ for $0 \leq x \leq 4$ . Find the CDF $F(x)$ . (6 marks)
Step-by-step solution
1. Step 1
  For $0 \leq x \leq 4$ :
  $F(x) = \int_0^x \dfrac{t}{8} \, dt = \dfrac{x^2}{16}$
2. Step 2
  Piecewise.
  $F(x) = \begin{cases} 0, & x < 0 \\ \frac{x^2}{16}, & 0 \leq x \leq 4 \\ 1, & x > 4 \end{cases}$
Answer
$F(x) = 0$ for $x < 0$ ; $\frac{x^2}{16}$ for $0 \leq x \leq 4$ ; $1$ for $x > 4$ .

Key Formulae — Continuous random variables

The formulae you need to memorise for continuous random variables on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

PDF total probability
$\displaystyle\int_{-\infty}^{\infty} f(x) \, dx = 1$
When to use
Must hold for $f$ to be a valid PDF.
Probability from PDF
$P(a < X < b) = \displaystyle\int_a^b f(x) \, dx$
When to use
Probability that $X$ falls in interval.
CDF
$F(x) = P(X \leq x) = \displaystyle\int_{-\infty}^x f(t) \, dt$
When to use
Cumulative distribution function. Reverse: $f(x) = F'(x)$ .
Continuous expectation
$E(X) = \displaystyle\int x f(x) \, dx; \quad E(X^2) = \int x^2 f(x) \, dx$
When to use
Mean of continuous RV. Variance: $\text{Var}(X) = E(X^2) - [E(X)]^2$ .

Key Definitions and Keywords — Continuous random variables

Continuous random variable
Examiner keyword
RV taking any value in an interval. Probability described by PDF, not point probabilities.
Probability density function (PDF)
Examiner keyword
Function $f(x) \geq 0$ with $\int f = 1$ . $P(a < X < b) = \int_a^b f$ .
Cumulative distribution function (CDF)
Examiner keyword
$F(x) = P(X \leq x)$ . Non-decreasing, from 0 to 1.

Common Mistakes and Misconceptions — Continuous random variables

✕Treating PDF as probability
9709 Examiner Reports 2022-2024
Why it happens
Confusion with discrete case.
How to avoid it
For continuous: $P(X = x) = 0$ . PDF gives DENSITY, not probability. Integrate to get probability.
✕Forgetting to verify $\int f = 1$
9709 Examiner Reports 2022-2024
Why it happens
Skipping check.
How to avoid it
If question provides $f$ with unknown constant, find it using $\int f = 1$ .

Continuous random variables — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Continuous random variables

Short Notes - Continuous random variables

Detailed Study Notes

At a glance

What you’ll learn

Probability density functions

CDF

Expectation and variance

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Probability from PDF (7 marks)

2E(X)E(X)E(X) and Var(X)\text{Var}(X)Var(X) from PDF (8 marks)

3CDF from PDF (6 marks)

PDF total probability

Probability from PDF

CDF

Continuous expectation

Continuous random variable

Probability density function (PDF)

Cumulative distribution function (CDF)

✕Treating PDF as probability

✕Forgetting to verify ∫f=1\int f = 1∫f=1

Continuous random variables — frequently asked questions

Continuous random variables

Short Notes - Continuous random variables

Detailed Study Notes

At a glance

What you’ll learn

Probability density functions

CDF

Expectation and variance

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Probability from PDF (7 marks)

2E(X)E(X)E(X) and Var(X)\text{Var}(X)Var(X) from PDF (8 marks)

3CDF from PDF (6 marks)

PDF total probability

Probability from PDF

CDF

Continuous expectation

Continuous random variable

Probability density function (PDF)

Cumulative distribution function (CDF)

✕Treating PDF as probability

✕Forgetting to verify ∫f=1\int f = 1∫f=1

Continuous random variables — frequently asked questions

2 $E(X)$ and $\text{Var}(X)$ from PDF (8 marks)

✕Forgetting to verify $\int f = 1$

2 $E(X)$ and $\text{Var}(X)$ from PDF (8 marks)

✕Forgetting to verify $\int f = 1$